course Phy 202 nUx緻➃ߥassignment #020
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15:11:30 Query introductory set #1, 1-8 Explain how we calculate the magnitude and direction of the electrostatic force on a given charge at a given point of the x-y plane point due to a given point charge at the origin.
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RESPONSE --> ok
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15:13:02 Query introductory set #1, 1-8 Explain how we calculate the magnitude and direction of the electrostatic force on a given charge at a given point of the x-y plane point due to a given point charge at the origin.
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RESPONSE --> The magnitude of the force on a test charge Q is F = k q1 * Q / r^2, where q1 is the charge at the origin. The force is one of repulsion if q1 and Q are of like sign, attraction if the signs are unlike. The force is therefore directly away from the origin or directly toward the origin To find the direction of this displacement vector, find arctan(y / x), adding 180 deg if x is negative. If the q1 and Q are like charges then this is the direction of the field. If q1 and Q are unlike then the direction of the field is opposite this direction. The angle of the field would therefore be 180 degrees greater or less than the angle.
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15:13:37 ** The magnitude of the force on a test charge Q is F = k q1 * Q / r^2, where q1 is the charge at the origin. The force is one of repulsion if q1 and Q are of like sign, attraction if the signs are unlike. The force is therefore directly away from the origin (if q1 and Q are of like sign) or directly toward the origin (if q1 and Q are of unlike sign). To find the direction of this displacement vector we find arctan(y / x), adding 180 deg if x is negative. If the q1 and Q are like charges then this is the direction of the field. If q1 and Q are unlike then the direction of the field is opposite this direction. The angle of the field would therefore be 180 degrees greater or less than this angle.**
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RESPONSE --> I got this question right for the most part I just didn't add some of the things to the end of the summary.
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15:13:44 ** The magnitude of the force on a test charge Q is F = k q1 * Q / r^2, where q1 is the charge at the origin. The force is one of repulsion if q1 and Q are of like sign, attraction if the signs are unlike. The force is therefore directly away from the origin (if q1 and Q are of like sign) or directly toward the origin (if q1 and Q are of unlike sign). To find the direction of this displacement vector we find arctan(y / x), adding 180 deg if x is negative. If the q1 and Q are like charges then this is the direction of the field. If q1 and Q are unlike then the direction of the field is opposite this direction. The angle of the field would therefore be 180 degrees greater or less than this angle.**
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RESPONSE --> ok
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15:13:48 Explain how we calculate the magnitude and direction of the electric field at a given point of the x-y plane point due to a given point charge at the origin.
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RESPONSE --> ok
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15:14:27 Explain how we calculate the magnitude and direction of the electric field at a given point of the x-y plane point due to a given point charge at the origin.
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RESPONSE --> The magnitude of the force on a test charge Q is F = k q1 * Q / r^2, where q1 is the charge at the origin. The electric field is therefore F / Q = k q1 / r^2. The direction is the direction of the force experienced by a positive test charge. The electric field is therefore directly away from the origin (if q1 is positive) or directly toward the origin (if q1 is negative). The direction of the electric field is in the direction of the displacement vector from the origin to the point if q1 is positive, and opposite to this direction if q1 is negative. To find the direction of this displacement vector find arctan(y / x), adding 180 deg if x is negative. If q1 is positive then this is the direction of the field. If q1 is negative then the direction of the field is opposite this direction, 180 degrees more or less than the calculated angle.
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15:14:57 ** The magnitude of the force on a test charge Q is F = k q1 * Q / r^2, where q1 is the charge at the origin. The electric field is therefore F / Q = k q1 / r^2. The direction is the direction of the force experienced by a positive test charge. The electric field is therefore directly away from the origin (if q1 is positive) or directly toward the origin (if q1 is negative). The direction of the electric field is in the direction of the displacement vector from the origin to the point if q1 is positive, and opposite to this direction if q1 is negative. To find the direction of this displacement vector we find arctan(y / x), adding 180 deg if x is negative. If q1 is positive then this is the direction of the field. If q1 is negative then the direction of the field is opposite this direction, 180 degrees more or less than the calculated angle. **
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RESPONSE --> looks to be a good answer and it looks as if I have answered the question correctly
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ʝ~{Q׆z{ assignment #021 KTL^l|¼} Physics II 07-25-2006 UƳX assignment #021 KTL^l|¼} Physics II 07-25-2006
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15:15:55 Query introductory set #1, 9-16 Explain how to find the potential difference in volts between two given points on the x axis, due to a given charge at the origin.
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RESPONSE --> ok
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15:17:06 Query introductory set #1, 9-16 Explain how to find the potential difference in volts between two given points on the x axis, due to a given charge at the origin.
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RESPONSE --> Potential difference is the work per Coulomb of charge moved between the two points. To find this work multiply the average force on a Coulomb of charge by the displacement from the first point to the second.
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15:17:13 ** Potential difference is the work per Coulomb of charge moved between the two points. To find this work you can multiply the average force on a Coulomb of charge by the displacement from the first point to the second. You can find an approximate average force by finding the force on a 1 Coulomb test charge at the two points and averaging the two forces. Multiplying this ave force by the displacement gives an approximate potential difference. Since the force is not a linear function of distance from the given charge, if the ratio of the two distances from the test charge is not small the approximation won't be particularly good. The approximation can be improved to any desired level of accuracy by partitioning the displacement between charges into smaller intervals of displacement and calculating the work done over each. The total work required is found by adding up the contributions from all the subintervals. University Physics students should understand how this process yields the exact work, which is the integral of the force function F(x) = k Q / x^2 between the two x values, yielding total work W = k * Q * 1 Coulomb ( 1 / x1 - 1 / x2) and potential difference V = k * Q ( 1 / x1 - 1 / x2). **
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RESPONSE --> OK
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15:17:21 ** Potential difference is the work per Coulomb of charge moved between the two points. To find this work you can multiply the average force on a Coulomb of charge by the displacement from the first point to the second. You can find an approximate average force by finding the force on a 1 Coulomb test charge at the two points and averaging the two forces. Multiplying this ave force by the displacement gives an approximate potential difference. Since the force is not a linear function of distance from the given charge, if the ratio of the two distances from the test charge is not small the approximation won't be particularly good. The approximation can be improved to any desired level of accuracy by partitioning the displacement between charges into smaller intervals of displacement and calculating the work done over each. The total work required is found by adding up the contributions from all the subintervals. University Physics students should understand how this process yields the exact work, which is the integral of the force function F(x) = k Q / x^2 between the two x values, yielding total work W = k * Q * 1 Coulomb ( 1 / x1 - 1 / x2) and potential difference V = k * Q ( 1 / x1 - 1 / x2). **
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RESPONSE --> ok
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15:17:26 Explain how to find the potential difference between two points given the magnitude and direction of the uniform electric field between those points.
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RESPONSE --> ok
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15:18:06 ** The work per Coulomb done between the two points is equal to the product of the electric field E and the displacement `dr. Thus for constant field E we have V = E * `dr. **
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RESPONSE --> * The work per Coulomb done between the two points is equal to the product of the electric field E and the displacement `dr. Thus for constant field E we have V = E * `dr
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15:18:14 ** The work per Coulomb done between the two points is equal to the product of the electric field E and the displacement `dr. Thus for constant field E we have V = E * `dr. **
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RESPONSE --> ok
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15:18:17 Explain how to find the average electric field between two points given a specific charge and the work done on the charge by the electric field as the charge moves between the points.
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RESPONSE -->
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15:18:53 Explain how to find the average electric field between two points given a specific charge and the work done on the charge by the electric field as the charge moves between the points.
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RESPONSE --> Get ave force from work and distance: F_ave = `dW / `ds. Get ave electric field from work and charge: E_ave = F / q.
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15:19:25 ** You get ave force from work and distance: F_ave = `dW / `ds. You get ave electric field from work and charge: E_ave = F / q. An alternative: Find potential difference `dV = `dW / q. Ave electric field is Eave = `dV / `ds **
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RESPONSE --> I got the first part of the question but I am not sure that that is all that is needed. ok
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15:19:33 ** You get ave force from work and distance: F_ave = `dW / `ds. You get ave electric field from work and charge: E_ave = F / q. An alternative: Find potential difference `dV = `dW / q. Ave electric field is Eave = `dV / `ds **
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RESPONSE --> ok
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15:19:36 In your own words explain the meaning of the electric field.
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RESPONSE -->
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15:19:54 In your own words explain the meaning of the electric field.
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RESPONSE --> concentration of the electrical force
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15:20:11 STUDENT RESPONSE AND INSTRUCTOR COMMENT: electric field is the concentration of the electrical force ** That's a pretty good way to put it. Very specifically electric field measures the magnitude and direction of the electrical force, per unit of charge, that would be experienced by a charge placed at a given point. **
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RESPONSE --> I put it in a good way
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15:20:18 STUDENT RESPONSE AND INSTRUCTOR COMMENT: electric field is the concentration of the electrical force ** That's a pretty good way to put it. Very specifically electric field measures the magnitude and direction of the electrical force, per unit of charge, that would be experienced by a charge placed at a given point. **
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RESPONSE --> ok
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15:20:21 In your own words explain the meaning of voltage.
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RESPONSE -->
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15:21:09 ** Voltage is the work done per unit of charge in moving charge from one point to another. **
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RESPONSE --> Voltage is the work done per unit of charge in moving charge from one point to another
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pxnvVIHyܒ assignment #022 KTL^l|¼} Physics II 07-25-2006 EړˊͰځ]ͺ assignment #022 KTL^l|¼} Physics II 07-25-2006
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15:21:49 Query problem set 1 #'s 17-24 If we know the initial KE of a particle, its charge and the uniform electric field in which it moves, then if the net force on the particle is due only to the electric field, how do we find the KE after the particle has moved through a given displacement?
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RESPONSE -->
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15:22:25 Query problem set 1 #'s 17-24 If we know the initial KE of a particle, its charge and the uniform electric field in which it moves, then if the net force on the particle is due only to the electric field, how do we find the KE after the particle has moved through a given displacement?
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RESPONSE --> find the Force by the relationship, q*E. Next, use the Force found to find the work done: `dW = F * `ds By the relationship `dW +`dKE = 0, then find `dKE, which we combine with KE0 to get KEf.
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15:22:34 ** GOOD STUDENT SOLUTION: Given KE0, q, E, `ds: First we can find the Force by the relationship, q*E. Next, we can use the Force found to find the work done: `dW = F * `ds By the relationship `dW +`dKE = 0, we can then find `dKE, which we combine with KE0 to get KEf. **
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RESPONSE --> ok looks good
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15:22:40 ** GOOD STUDENT SOLUTION: Given KE0, q, E, `ds: First we can find the Force by the relationship, q*E. Next, we can use the Force found to find the work done: `dW = F * `ds By the relationship `dW +`dKE = 0, we can then find `dKE, which we combine with KE0 to get KEf. **
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RESPONSE --> ok
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15:23:10 If we know the charge transferred between two points, the time and the average power necessary to accomplish the transfer, how do we find the potential difference between the points?
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RESPONSE --> The potential difference is found from the work done and the charge. Potential difference, or voltage, is work / charge, in Joules / Coulomb. find the work from the power and the time, since power = work / time
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15:23:18 ** The potential difference is found from the work done and the charge. Potential difference, or voltage, is work / charge, in Joules / Coulomb. We find the work from the power and the time, since power = work / time. **
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RESPONSE --> ok
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15:23:52 Explain how we can use the flux picture to determine the electric field due to a point charge Q at a distance r from the charge.
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RESPONSE --> Flux = 4pikQ Flux = area of sphere * electric field = 4 pi r^2 * E k is 9.0 x 10^9 N m^2/C^2 We have 4 pi r^2 * E = 4 pi k Q so E = 4 pi k Q / ( 4 pi r^2) = k Q / r^2
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15:24:24 STUDENT RESPONSE AND INSTRUCTOR COMMENT: Flux = 4pikQ Flux = area of sphere * electric field = 4 pi r^2 * E k is 9.0 x 10^9 N m^2/C^2 We have 4 pi r^2 * E = 4 pi k Q so E = 4 pi k Q / ( 4 pi r^2) = k Q / r^2 INSTRUCTOR COMMENT: ** Note that the sphere is centered at the charge Q and passes thru the point at distance r so the radius of the sphere is r. Note also that this works because the electric field is radial from Q and hence always perpendicular to the sphere. **
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RESPONSE --> ok looks to be a good student answer
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15:25:04 Explain how we can use the flux picture to determine the electric field due to a charge Q uniformly distributed over a straight line of length L, at a distance r << L from that line but not close to either end.
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RESPONSE --> imagine a circular cylinder around a long segment of the wire; determine the charge on the segment. Total flux is 4 pi k * charge. By the symmetry of the situation the electric field has a very nearly constant magnitude over the curved surface of the cylinder (for an infinite wire the field would be absolutely constant). Almost all of the flux exits the curved surface of the cylinder and is at every point perpendicular to this surface (for an infinite wire all the flux would exit thru the curved surface and would be exactly perpendicular). So you can find flux / area, which is the field. You get E = flux / area = 4 pi k Q / ( 2 pi r * L) = 2 k Q / L.
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15:25:12 ** imagine a circular cylinder around a long segment of the wire; determine the charge on the segment. Total flux is 4 pi k * charge. By the symmetry of the situation the electric field has a very nearly constant magnitude over the curved surface of the cylinder (for an infinite wire the field would be absolutely constant). Almost all of the flux exits the curved surface of the cylinder and is at every point perpendicular to this surface (for an infinite wire all the flux would exit thru the curved surface and would be exactly perpendicular). So you can find flux / area, which is the field. You get E = flux / area = 4 pi k Q / ( 2 pi r * L) = 2 k Q / L. **
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RESPONSE --> ok
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{`_}fw䱅ȞߐK assignment #023 KTL^l|¼} Physics II 07-25-2006 U猵byݤŇ assignment #023 KTL^l|¼} Physics II 07-25-2006
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15:26:09 If we know the number of conduction electrons in a wire, the length of the wire and the average drift velocity of the electrons how to we determine the current in the wire?
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RESPONSE -->
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15:26:51 If we know the number of conduction electrons in a wire, the length of the wire and the average drift velocity of the electrons how to we determine the current in the wire?
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RESPONSE --> find the ratio of the length `dL/L, and multiply this ratio by the number of electrons in the entire length `dL, to find the number of electrons for that small increment of length and time (1 sec). multiply that number of electrons by the charge of an electron to find the current flowing past a point at any given second.
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15:26:59 GOOD STUDENT SOLUTION: Given: # of electrons, L, vAve of the drift: From the velocity, we find that the electrons will drift a certain `dL per second. We find the ratio of the length `dL/L, and multiply this ratio by the number of electrons in the entire length `dL, to find the number of electrons for that small increment of length and time (1 sec). We can then multiply that number of electrons by the charge of an electron to find the current flowing past a point at any given second. **
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RESPONSE --> ok looking good
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15:27:05 GOOD STUDENT SOLUTION: Given: # of electrons, L, vAve of the drift: From the velocity, we find that the electrons will drift a certain `dL per second. We find the ratio of the length `dL/L, and multiply this ratio by the number of electrons in the entire length `dL, to find the number of electrons for that small increment of length and time (1 sec). We can then multiply that number of electrons by the charge of an electron to find the current flowing past a point at any given second. **
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RESPONSE --> ok
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15:27:08 For a given potential difference across two otherwise identical wires, why is the current through the longer wire less than that through the shorter wire?
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RESPONSE -->
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15:27:42 For a given potential difference across two otherwise identical wires, why is the current through the longer wire less than that through the shorter wire?
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RESPONSE --> The potential gradient, which is the potential difference per unit of length, will be higher for the shorter length. The potential gradient is the electric field, which is what exerts the accelerating force on the electrons. So in the shorter wire the electrons are accelerated by a greater average net force and hence build more velocity between collisions. With greater average drift velocity, more electrons therefore pass a given point in a given time interval
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15:27:49 ** The potential gradient, which is the potential difference per unit of length, will be higher for the shorter length. The potential gradient is the electric field, which is what exerts the accelerating force on the electrons. So in the shorter wire the electrons are accelerated by a greater average net force and hence build more velocity between collisions. With greater average drift velocity, more electrons therefore pass a given point in a given time interval. **
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RESPONSE --> ok
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15:27:55 ** The potential gradient, which is the potential difference per unit of length, will be higher for the shorter length. The potential gradient is the electric field, which is what exerts the accelerating force on the electrons. So in the shorter wire the electrons are accelerated by a greater average net force and hence build more velocity between collisions. With greater average drift velocity, more electrons therefore pass a given point in a given time interval. **
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RESPONSE --> ok
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15:27:58 For a given potential difference across two otherwise identical wires, why is the current through the thicker wire greater than that through the thinner wire?
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RESPONSE -->
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15:28:43 ** The key is that more electrons are available per unit length in the thicker wire. The potential gradient (i.e., the electric field) is the same because the length is the same, so more electrons respond to the same field. GOOD STUDENT SOLUTION: If we know the diameters of both the wires (d1 and d2), we know that the cross-sectional area of the second diameter is (d2/d1)^2 times the cross-sectional area of the first wire. This means that the second wire will have (d2/d1)^2 times as many charge carriers per unit length. This also means that the current in the second wire will be (d2/d1)^2 times that of the first. Therfore the thicker wire will have a greater current.
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RESPONSE --> ok
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15:28:49 ** The key is that more electrons are available per unit length in the thicker wire. The potential gradient (i.e., the electric field) is the same because the length is the same, so more electrons respond to the same field. GOOD STUDENT SOLUTION: If we know the diameters of both the wires (d1 and d2), we know that the cross-sectional area of the second diameter is (d2/d1)^2 times the cross-sectional area of the first wire. This means that the second wire will have (d2/d1)^2 times as many charge carriers per unit length. This also means that the current in the second wire will be (d2/d1)^2 times that of the first. Therfore the thicker wire will have a greater current.
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RESPONSE --> ok
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15:28:52 If we know the length of a uniform wire and the potential difference between its ends, how do we calculate the average net force exerted on a conduction electron within the wire?
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RESPONSE -->
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15:29:35 If we know the length of a uniform wire and the potential difference between its ends, how do we calculate the average net force exerted on a conduction electron within the wire?
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RESPONSE --> First determine the work done and then divide it by the distance to get the average net force
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15:29:41 STUDENT RESPONSE WITH INSTRUCTOR COMMENT: First determine the work done and then divide it by the distance to get the average net force INSTRUCTOR COMMENT: The work done on an electron is the product of its charge and the potential difference. Having this information we can then do as you indicate. GOOD STUDENT SOLUTION: From the charge and voltage (potential difference) we can find how much work is done over the entire length of the wire. We multiply the voltage by the charge to get work done over the full length. We know that F = W/s, so we can divide the work we found by the distance and this will give us the amount of force.
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RESPONSE --> ok
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15:29:47 STUDENT RESPONSE WITH INSTRUCTOR COMMENT: First determine the work done and then divide it by the distance to get the average net force INSTRUCTOR COMMENT: The work done on an electron is the product of its charge and the potential difference. Having this information we can then do as you indicate. GOOD STUDENT SOLUTION: From the charge and voltage (potential difference) we can find how much work is done over the entire length of the wire. We multiply the voltage by the charge to get work done over the full length. We know that F = W/s, so we can divide the work we found by the distance and this will give us the amount of force.
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RESPONSE --> ok
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15:29:53 If we know the voltage and the resistance in a circuit, how do we find the current, and how do we use this result to then reason out the power required to maintain the current?
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RESPONSE -->
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15:30:21 If we know the voltage and the resistance in a circuit, how do we find the current, and how do we use this result to then reason out the power required to maintain the current?
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RESPONSE --> find the current by solving the equation I= V/R for I which divided the volts by the resistance of a circuit and to maintain it you multiply the volts by the current determined before.
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15:30:58 ** Reasoning in terms of units: Power is work per unit of time. P = `dW / `dt, in J / s. Current is voltage / resistance. I = V / R, in C / s. Voltage is work per unit of charge: V = `dW / Q, in J / C. If we multiply voltage in J / C by current in C / s we get power in J / s. CORRECT STUDENT SOLUTION: you find the current by solving the equation I= V/R for I which divided the volts by the resistance of a circuit and to maintain it you multiply the volts by the current determined before.
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RESPONSE --> ok i didn't write out all of the first part of the answer but i hope that i have done enough in order to get credit
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15:31:05 ** Reasoning in terms of units: Power is work per unit of time. P = `dW / `dt, in J / s. Current is voltage / resistance. I = V / R, in C / s. Voltage is work per unit of charge: V = `dW / Q, in J / C. If we multiply voltage in J / C by current in C / s we get power in J / s. CORRECT STUDENT SOLUTION: you find the current by solving the equation I= V/R for I which divided the volts by the resistance of a circuit and to maintain it you multiply the volts by the current determined before.
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RESPONSE --> ok
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15:31:09 Describe the effect of a magnetic field on a current. Note the relative directions of the magnetic field, the current and the force exerted on the current. Note whether a sustained current experiences a sustained force.
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RESPONSE -->
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15:31:26 Describe the effect of a magnetic field on a current. Note the relative directions of the magnetic field, the current and the force exerted on the current. Note whether a sustained current experiences a sustained force.
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RESPONSE --> uniform magnetic field B oriented perpendicular to a current I in a straight current carrier of length L exerts a force equal to I * L * B on the current. This force is perpendicular to the magnetic field and to the current by the right-hand rule where I is crossed with B. If B and the current make angle theta then the force is I * L * B * sin(theta). Again the right-hand rule applies.
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15:31:31 ** A uniform magnetic field B oriented perpendicular to a current I in a straight current carrier of length L exerts a force equal to I * L * B on the current. This force is perpendicular to the magnetic field and to the current by the right-hand rule where I is crossed with B. If B and the current make angle theta then the force is I * L * B * sin(theta). Again the right-hand rule applies. **
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RESPONSE --> ok
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۰wj쑮| assignment #024 KTL^l|¼} Physics II 07-25-2006 yN̈ljaz{K׳ assignment #024 KTL^l|¼} Physics II 07-25-2006
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15:55:45 Query problem set 3 #'2 1-6. How do we determine the current in the circuit and the voltage across each resistor when we know the voltage across a series combination of two known resistances?
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RESPONSE -->
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15:56:10 Query problem set 3 #'2 1-6. How do we determine the current in the circuit and the voltage across each resistor when we know the voltage across a series combination of two known resistances?
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RESPONSE --> To get the current calculate I = V / R, where R is the sum of the two resistances. To get the voltage across each resistor calculate V = I * R for each resistor
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15:56:16 ** To get the current calculate I = V / R, where R is the sum of the two resistances. To get the voltage across each resistor calculate V = I * R for each resistor. **
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RESPONSE --> ok
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15:56:22 ** To get the current calculate I = V / R, where R is the sum of the two resistances. To get the voltage across each resistor calculate V = I * R for each resistor. **
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RESPONSE --> ok
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15:56:34 How do we determine the current and voltage across each resistor when we know the voltage across a parallel combination of two known resistances?
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RESPONSE --> The voltage across both resistors is the same and is equal to the voltage across the combination. The current in each resistor is calculated by I = V / R. The total current is the sum of the two currents.
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15:56:41 ** The voltage across both resistors is the same and is equal to the voltage across the combination. The current in each resistor is calculated by I = V / R. The total current is the sum of the two currents. **
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RESPONSE --> ok
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15:57:12 A series circuit contains a capacitor of known capacitance and a resistor of known resistance. The capacitor was originally uncharged before the source voltage was applied, and is in the process of being charged by the source. If we know the charge on the capacitor, how do we find the current through circuit?
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RESPONSE --> The voltage across the capacitor is equal to the charge divided by capacitance. The voltage across the capacitor opposes the voltage of the source. Since the voltage drop around the complete circuit must be zero, the voltage across the resistor is the difference between source voltage and the voltage across the capacitor. Dividing the voltage across the resistor by the resistance we obtain the current thru the circuit
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15:57:18 The voltage across the capacitor is equal to the charge divided by capacitance. The voltage across the capacitor opposes the voltage of the source. Since the voltage drop around the complete circuit must be zero, the voltage across the resistor is the difference between source voltage and the voltage across the capacitor. Dividing the voltage across the resistor by the resistance we obtain the current thru the circuit.
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RESPONSE --> ok
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15:58:16 If we know the capacitance and initial charge on a capacitor in series with a resistor of known resistance then how to we find the approximate time required for the capacitor to discharge 1% of its charge through the circuit?
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RESPONSE --> From capacitance and initial charge find the voltage. From the voltage and the resistance find the current. Take 1% of the initial charge and divide it by the current to get the approximate time required to discharge 1% of the charge. This result is a slight underestimate of the time required since as the capacitor discharges the current decreases.
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15:58:21 ** From capacitance and initial charge we find the voltage. From the voltage and the resistance we find the current. We take 1% of the initial charge and divide it by the current to get the approximate time required to discharge 1% of the charge. } This result is a slight underestimate of the time required since as the capacitor discharges the current decreases. **
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RESPONSE --> ok
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LJ͗՞ޅR assignment #025 KTL^l|¼} Physics II 07-25-2006 MץUYzg assignment #025 KTL^l|¼} Physics II 07-25-2006
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15:59:28 Query problem 16.15 charges 6 microC on diagonal corners, -6 microC on other diagonal corners of 1 m square; force on each. What is the magnitude and direction of the force on the positive charge at the lower left-hand corner?
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RESPONSE -->
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16:00:11 Query problem 16.15 charges 6 microC on diagonal corners, -6 microC on other diagonal corners of 1 m square; force on each. What is the magnitude and direction of the force on the positive charge at the lower left-hand corner?
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RESPONSE --> The charges which lie 1 meter apart are unlike and therefore exert attractive forces; these forces are each .324 Newtons. This is calculated using Coulomb's Law: F = 9 * 10^9 N m^2/C^2 * ( 6 * 10^-6 C) * ( 6 * 10^-6 C) / ( 1 m)^2 = 324 * 10^-3 N = .324 N. Charges across a diagonal are like and separated by `sqrt(2) meters = 1.414 meters, approx, and exert repulsive forces of .162 Newtons. This repulsive force is calculated using Coulomb's Law: F = 9 * 10^9 N m^2/C^2 * ( 6 * 10^-6 C) * ( 6 * 10^-6 C) / ( 1.414 m)^2 = 162 * 10^-3 N = .162 N. The charge at the lower left-hand corner therefore experiences a force of .324 Newtons to the right, a force of .324 Newtons straight upward and a force of .162 Newtons at 45 deg down and to the left (at angle 225 deg with respect to the standard positive x axis, which we take as directed toward the right). This latter force has components Fy = .162 N sin(225 deg) = -.115 N, approx, and Fx = .162 N cos(225 deg) = -.115 N. The total force in the x direction is therefore -.115 N + .324 N = .21 N, approx; the total force in the y direction is -.115 N + .324 N = .21 N
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16:01:05 ** The charges which lie 1 meter apart are unlike and therefore exert attractive forces; these forces are each .324 Newtons. This is calculated using Coulomb's Law: F = 9 * 10^9 N m^2/C^2 * ( 6 * 10^-6 C) * ( 6 * 10^-6 C) / ( 1 m)^2 = 324 * 10^-3 N = .324 N. Charges across a diagonal are like and separated by `sqrt(2) meters = 1.414 meters, approx, and exert repulsive forces of .162 Newtons. This repulsive force is calculated using Coulomb's Law: F = 9 * 10^9 N m^2/C^2 * ( 6 * 10^-6 C) * ( 6 * 10^-6 C) / ( 1.414 m)^2 = 162 * 10^-3 N = .162 N. The charge at the lower left-hand corner therefore experiences a force of .324 Newtons to the right, a force of .324 Newtons straight upward and a force of .162 Newtons at 45 deg down and to the left (at angle 225 deg with respect to the standard positive x axis, which we take as directed toward the right). This latter force has components Fy = .162 N sin(225 deg) = -.115 N, approx, and Fx = .162 N cos(225 deg) = -.115 N. The total force in the x direction is therefore -.115 N + .324 N = .21 N, approx; the total force in the y direction is -.115 N + .324 N = .21 N, approx. Thus the net force has magnitude `sqrt( (.21 N)^2 + (.21 N)^2) = .29 N at an angle of tan^-1( .21 N / .21 N) = tan^-1(1) = 45 deg. The magnitude and direction of the force on the negative charge at the lower right-hand corner is obtained by a similar analysis, which would show that this charge experiences forces of .324 N to the left, .324 N straight up, and .162 N down and to the right. The net force is found by standard vector methods to be about .29 N up and to the left. **
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RESPONSE --> shew that is an incrediblely long problem and I am not sure that I have said all of the right things but I said some of the things that are on the answers
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16:01:33 query univ 21.68 (22.52 10th edition) 5 nC at the origin, -2 nC at (4 cm, 0). If 6 nC are placed at (4cm, 3cm), what are the components of the resulting force?
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RESPONSE --> don't do
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16:01:39 ** The 5 nC charge lies at distance sqrt( 4^2 + 3^2) cm from the 6 nC charge, which is 3 cm from the charge at (4 cm ,0). The force exerted on the 6 nC charge by the two respective forces have magnitudes .00011 N and .00012 N respectively. The x and y components of the force exerted by the charge at the origin are in the proportions 4/5 and 3/5 to the total charge, giving respective components .000086 N and .000065 N. The force exerted by the charge at (4 cm, 0) is in the negative y direction. So the y component of the resultant are .000065 N - .00012 N = -.000055 N and its x component is .000086 N. **
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RESPONSE --> ok
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16:01:47 Query univ 21.76 (10th edition 22.60) quadrupole (q at (0,a), (0, -a), -2q at origin). For y > a what is the magnitude and direction of the electric field at (0, y)?
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RESPONSE --> dont do
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16:01:52 ** The magnitude of the field due to the charge at a point is k q / r^2. For a point at coordinate y on the y axis, for y > a, we have distances r = y-a, y+a and y. The charges at these distances are respectively q, q and -2q. So the field is k*q/(y - a)^2 + k*q/(y + a)^2 - 2k*q/y^2 = 2*k*q*(y^2 + a^2)/((y + a)^2*(y - a)^2) - 2*k*q/y^2 = 2*k*q* [(y^2 + a^2)* y^2 - (y+a)^2 ( y-a)^2) ] / ( y^2 (y + a)^2*(y - a)^2) = 2*k*q* [y^4 + a^2 y^2 - (y^2 - a^2)^2 ] / ( y^2 (y + a)^2*(y - a)^2) = 2*k*q* [y^4 + a^2 y^2 - y^4 + 2 y^2 a^2 - a^4 ] / ( y^2 (y + a)^2*(y - a)^2) = 2*k*q* [ 3 a^2 y^2 - a^4 ] / ( y^2 (y + a)^2*(y - a)^2) . For large y the denominator is close to y^6 and the a^4 in the numerator is insignifant compared to a^2 y^2 so the expression becomes 6 k q a^2 / y^4, which is inversely proportional to y^4. **
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RESPONSE --> ok
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16:01:58 query univ 22.102 annulus in yz plane inner radius R1 outer R2, charge density `sigma.What is a total electric charge on the annulus?
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RESPONSE --> don't do
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16:02:03 ** The total charge on the annulus is the product Q = sigma * A = sigma * (pi R2^2 - pi R1^2). To find the field at distance x along the x axis, due to the charge in the annulus, we first find the field due to a thin ring of charge: The charge in a thin ring of radius r and ring thickness `dr is the product `dQ = 2 pi r `dr * sigma of ring area and area density. From any small segment of this ring the electric field at a point of the x axis would be directed at angle arctan( r / x ) with respect to the x axis. By either formal trigonometry or similar triangles we see that the component parallel to the x axis will be in the proportion x / sqrt(x^2 + r^2) to the magnitude of the field from this small segment. By symmetry only the x component of the field will remain when we sum over the entire ring. So the field due to the ring will be in the same proportion to the expression k `dQ / (x^2 + r^2). Thus the field due to this thin ring will be magnitude of field due to thin ring: k `dQ / (x^2 + r^2) * x / sqrt (x^2 + r^2) = 2 pi k r `dr * x / (x^2 + r^2)^(3/2). Summing over all such thin rings, which run from r = R1 to r = R2, we obtain the integral magnitude of field = integral ( 2 pi k r x / (x^2 + r^2)^(3/2) with respect to r, from R1 to R2). Evaluating the integral we find that magnitude of field = 2* pi k *x* | 1 /sqrt(x^2 + r1^2) - 1 / sqrt(x^2 + r2^2) | The direction of the field is along the x axis. If the point is close to the origin then x is close to 0 and x / sqrt(x^2 + r^2) is approximately equal to x / r, for any r much larger than x. This is because the derivative of x / sqrt(x^2 + r^2) with respect to x is r^2 / (x^2+r^2)^(3/2), which for x = 0 is just 1/r, having no x dependence. So at small displacement `dx from the origin the field strength will just be some constant multiple of `dx. **
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RESPONSE --> ok
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xJƘwO assignment #026 KTL^l|¼} Physics II 07-25-2006 ϋ}尮 assignment #026 KTL^l|¼} Physics II 07-25-2006
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16:03:04 Query problem 16.32. field 745 N/C midway between two equal and opposite point charges separated by 16 cm. What is the magnitude of each charge?
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RESPONSE -->
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16:04:03 Query problem 16.32. field 745 N/C midway between two equal and opposite point charges separated by 16 cm. What is the magnitude of each charge?
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RESPONSE --> If the magnitude of the charge is q then the field contribution of each charge is k q / r^2, with r = 8 cm = .08 meters. Since both charges contribute equally to the field, with the fields produced by both charges being in the same direction (on any test charge at the midpoint one force is of repulsion and the other of attraction, and the charges are on opposite sides of the midpoint), the field of either charge has magnitude 1/2 (745 N/C) = 373 N/C. Thus E = 373 N/C and E = k q / r^2. knowing k, E and r so solving for q to obtain q = E * r^2 / k = 373 N/C * (.08 m)^2 / (9 * 10^9 N m^2 / C^2) = 373 N/C * .0064 m^2 / (9 * 10^9 N m^2 / C^2) = 2.6 * 10^-10 C, approx.
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16:04:18 ** If the magnitude of the charge is q then the field contribution of each charge is k q / r^2, with r = 8 cm = .08 meters. Since both charges contribute equally to the field, with the fields produced by both charges being in the same direction (on any test charge at the midpoint one force is of repulsion and the other of attraction, and the charges are on opposite sides of the midpoint), the field of either charge has magnitude 1/2 (745 N/C) = 373 N/C. Thus E = 373 N/C and E = k q / r^2. We know k, E and r so we solve for q to obtain q = E * r^2 / k = 373 N/C * (.08 m)^2 / (9 * 10^9 N m^2 / C^2) = 373 N/C * .0064 m^2 / (9 * 10^9 N m^2 / C^2) = 2.6 * 10^-10 C, approx. **
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RESPONSE --> ok looks to be of some good.
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16:04:23 ** If the magnitude of the charge is q then the field contribution of each charge is k q / r^2, with r = 8 cm = .08 meters. Since both charges contribute equally to the field, with the fields produced by both charges being in the same direction (on any test charge at the midpoint one force is of repulsion and the other of attraction, and the charges are on opposite sides of the midpoint), the field of either charge has magnitude 1/2 (745 N/C) = 373 N/C. Thus E = 373 N/C and E = k q / r^2. We know k, E and r so we solve for q to obtain q = E * r^2 / k = 373 N/C * (.08 m)^2 / (9 * 10^9 N m^2 / C^2) = 373 N/C * .0064 m^2 / (9 * 10^9 N m^2 / C^2) = 2.6 * 10^-10 C, approx. **
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RESPONSE --> ok
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16:04:27 If the charges are represented by Q and -Q, what is the electric field at the midpoint?
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RESPONSE -->
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16:04:48 If the charges are represented by Q and -Q, what is the electric field at the midpoint?
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RESPONSE --> The field would be 2 k Q / r^2, where r=.08 meters and the factor 2 is because there are two charges of magnitude Q both at the same distance from the point
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16:04:52 ** this calls for a symbolic expression in terms of the symbol Q. The field would be 2 k Q / r^2, where r=.08 meters and the factor 2 is because there are two charges of magnitude Q both at the same distance from the point. **
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RESPONSE --> ok
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16:05:01 query univ 22.30 (10th edition 23.30). Cube with faces S1 in xz plane, S2 top, S3 right side, S4 bottom, S5 front, S6 back on yz plane. E = -5 N/(C m) x i + 3 N/(C m) z k. What is the flux through each face of the cube, and what is the total charge enclosed by the cube?
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RESPONSE --> dont do
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16:05:05 **** Advance correction of a common misconception: Flux is not a vector quantity so your flux values will not be multiples of the i, j and k vectors. The vectors normal to S1, S2, ..., S6 are respectively -j, k, j, -k, i and -i. For any face the flux is the dot product of the field with the normal vector, multiplied by the area. The area of each face is (.3 m)^2 = .09 m^2 So we have: For S1 the flux is (-5 x N / (C m) * i + 3 z N / (C m) k ) dot (-j) * .09 m^2 = 0. For S2 the flux is (-5 x N / (C m) * i + 3 z N / (C m) k ) dot ( k) * .09 m^2 = 3 z N / (C m) * .09 m^2. For S3 the flux is (-5 x N / (C m) * i + 3 z N / (C m) k ) dot ( j) * .09 m^2 = 0. For S4 the flux is (-5 x N / (C m) * i + 3 z N / (C m) k ) dot (-k) * .09 m^2 = -3 z N / (C m) * .09 m^2. For S5 the flux is (-5 x N / (C m) * i + 3 z N / (C m) k ) dot ( i) * .09 m^2 = -5 x N / (C m) * .09 m^2. For S6 the flux is (-5 x N / (C m) * i + 3 z N / (C m) k ) dot (-i) * .09 m^2 = 5 x N / (C m) * .09 m^2. On S2 and S4 we have z = .3 m and z = 0 m, respectively, giving us flux .027 N m^2 / C on S2 and flux 0 on S4. On S5 and S6 we have x = .3 m and x = 0 m, respectively, giving us flux -.045 N m^2 / C on S5 and flux 0 on S6. The total flux is therefore .027 N m^2 / C - .045 N m^2 / C = -.018 N m^2 / C. Since the total flux is 4 pi k Q, where Q is the charge enclosed by the surface, we have 4 pi k Q = -.018 N m^2 / C and Q = -.018 N m^2 / C / (4 pi k) = -.018 N m^2 / C / (4 pi * 9 * 10^9 N m^2 / C^2) = -1.6 * 10^-13 C, approx. **
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RESPONSE --> ok
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16:05:29 ** The electric field inside either shell must be zero, so the charge enclosed by any sphere concentric with the shells and lying within either shell must be zero, and the field is zero for a < r < b and for c < r < d. Thus the total charge on the inner surface of the innermost shell is zero, since this shell encloses no charge. The entire charge 2q of the innermost shell in concentrated on its outer surface. For any r such that b < r < c the charge enclosed by the corresponding sphere is the 2 q of the innermost shell, so that the electric field is 4 pi k * 2q / r^2 = 8 pi k q / r^2. Considering a sphere which encloses the inner but not the outer surface of the second shell we see that this sphere must contain the charge 2q of the innermost shell. Since this sphere is within the conducting material the electric field on this sphere is zero and the net flux thru this sphere is zero. Thus the total charge enclosed by this sphere is zero. Since the charge enclosed by the sphere includes the 2q of the innermost shell, the sphere must also enclose a charge -2 q, which by symmetry must be evenly distributed on the inner surface of the second shell. Any sphere which encloses both shells must enclose the total charge of both shells, which is 6 q. Since we have 2q on the innermost shell and -2q on the inner surface of the second shell the charge on the outer surface of this shell must be 6 q. For any r such that d < r the charge enclosed by the corresponding sphere is the 6 q of the two shells, so that the electric field is 4 pi k * 6q / r^2 = 24 pi k q / r^2. **
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RESPONSE --> ok
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16:05:35 query univ 23.46 (23.34 10th edition). Long conducting tube inner radius a, outer b. Lin chg density `alpha. Line of charge, same density along axis.
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RESPONSE --> dont do
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16:05:40 **The Gaussian surfaces appropriate to this configuration are cylinders of length L which are coaxial with the line charge. The symmetries of the situation dictate that the electric field is everywhere radial and hence that the field passes through the curved surface of each cylinder at right angle to that surface. The surface area of the curved portion of any such surface is 2 pi r L, where r is the radius of the cylinder. For r < a the charge enclosed by the Gaussian surface is L * alpha so that the flux is charge enclosed = 4 pi k L * alpha and the electric field is electric field = flux / area = 4 pi k L * alpha / (2 pi r L ) = 2 k alpha / r. For a < r < b, a Gaussian surface of radius r lies within the conductor so the field is zero (recall that if the field wasn't zero, the free charges inside the conductor would move and we wouldn't be in a steady state). So the net charge enclosed by this surface is zero. Since the line charge enclosed by the surface is L * alpha, the inner surface of the conductor must therefore contain the equal and opposite charge -L * alpha, so that the inner surface carries charge density -alpha. For b < r the Gaussian surface encloses both the line charge and the charge of the cylindrical shell, each of which has charge density alpha, so the charge enclosed is 2 L * alpha and the electric field is radial with magnitude 4 pi k * 2 L * alpha / (2 pi r L ) = 4 k alpha / r. Since the enclosed charge that of the line charge (L * alpha) as well as the inner surface of the shell (L * (-alpha) ), which the entire system carries charge L * alpha, we have line charge + charge on inner sphere + charge on outer sphere = alpha * L, we have alpha * L - alpha * L + charge on outer sphere = alpha * L, so charge on outer sphere = 2 alpha * L, so the outer surface of the shell has charge density 2 alpha. **
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RESPONSE --> ok
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jnݨЋzU}} assignment #027 KTL^l|¼} Physics II 07-25-2006 p}Xg܄ assignment #027 KTL^l|¼} Physics II 07-25-2006
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16:06:49 Note that the solutions given here use k notation rather than epsilon0 notation. The k notation is more closely related to the geometry of Gauss' Law and is consistent with the notation used in the Introductory Problem Sets. However you can easily switch between the two notations, since k = 1 / (4 pi epsilon0), or alternatively epsilon0 = 1 / (4 pi k).
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RESPONSE -->
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16:07:00 Note that the solutions given here use k notation rather than epsilon0 notation. The k notation is more closely related to the geometry of Gauss' Law and is consistent with the notation used in the Introductory Problem Sets. However you can easily switch between the two notations, since k = 1 / (4 pi epsilon0), or alternatively epsilon0 = 1 / (4 pi k).
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RESPONSE -->
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16:07:07 Query text problem 17.18 potential 2.5 * 10^-15 m from proton; PE of two protons at this separation in a nucleus. What is the electrostatic potential at a distance of 2.5 * 10^-15 meters?
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RESPONSE -->
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16:07:53 Query text problem 17.18 potential 2.5 * 10^-15 m from proton; PE of two protons at this separation in a nucleus. What is the electrostatic potential at a distance of 2.5 * 10^-15 meters?
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RESPONSE --> q = 1.60*10^-19C=charge on proton V = kq/r = 9.0*10^9N*m^2/C^2(1.60*10^-19C) / (2.5*10^-15m) = 5.8*10^5V. PE=(1.60*10^-19C)(5.8*10^5V) = 9.2*10^-14 J.
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16:08:03 STUDENT SOLUTION: For a part, to determine the electric potential a distance fo 2.5810^-15m away from a proton, I simply used the equation V = k q / r for electric potential for point charge: q = 1.60*10^-19C=charge on proton V = kq/r = 9.0*10^9N*m^2/C^2(1.60*10^-19C) / (2.5*10^-15m) = 5.8*10^5V. Part B was the more difficult portion of the problem. You have to consider a system that consists of two protons 2.5*10^-5m apart. The work done against the electric field to assemble these charges is W = qV. The potential energy is equal to the work done against the field. PE=(1.60*10^-19C)(5.8*10^5V) = 9.2*10^-14 J.
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RESPONSE --> ok looks good
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16:08:20 query univ phy 23.58 (24.58 10th edition). Geiger counter: long central wire 145 microns radius, hollow cylinder radius 1.8 cm. What potential difference between the wire in the cylinder will produce an electric field of 20,000 volts/m at 1.2 cm from the wire?
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RESPONSE --> dont do
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16:08:34 ** The voltage V_ab is obtained by integrating the electric field from the radius of the central wire to the outer radius. From this we determine that E = Vab / ln(b/a) * 1/r, where a is the inner radius and b the outer radius. If E = 20,000 V/m at r = 1.2 cm then Vab = E * r * ln(b/a) = 20,000 V/m * ln(1.8 cm / .0145 cm) * .012 m = 1157 V. **
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RESPONSE --> ok
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16:08:41 Query univ 23.78 (24.72 10th edition). Rain drop radius .65 mm charge -1.2 pC. What is the potential at the surface of the rain drop?
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RESPONSE --> don't do
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16:08:45 STUDENT RESPONSE FOLLOWED BY SOLUTION: The problem said that V was 0 at d = inifinity, which I understnad to mean that as we approach the raindrop from infinity, the potential differencegrows from 0, to some amount at the surface of the raindrop. Because water molecules are more positive on one side that the other, they tend to align in a certain direction. Since positive charges tend to drift toward negative charge, I would think that the raindrop, with its overall negative charge, has molecules arranged so that their more positive sides are pointing toward the center and negative sides will be alighed along the surface of the raindrop. Probably all wrong. I tried several differnet integrand configuraitons but never found one that gave me an answer in volts. SOLUTION: You will have charge Q = -1.2 * 10^-12 C on the surface of a sphere of radius .00065 m. The field is therefore E = k Q / r^2 = 9 * 10^9 N m^2 / C^2 * (-1.2 * 10^-12 C) / r^2 = -1.08 * 10^-2 N m^2 / C / r^2. Integrating the field from infinity to .00065 m we get (-1.08 * 10^-2 N m^2 / C) / (.00065 m) = -16.6 N m / C = -16.6 V. If two such drops merge they form a sphere with twice the volume and hence 2^(1/3) times the radius, and twice the charge. The surface potential is proportional to charge and inversely proportional to volume. So the surface potential will be 2 / 2^(1/3) = 2^(2/3) times as great as before. The surface potential is therefore 16.6 V * 2^(2/3) = -26.4 volts, approx.. **
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RESPONSE --> ok
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oٽ`vv_yͭ assignment #028 KTL^l|¼} Physics II 07-25-2006 ȈbHޗƁ~ assignment #028 KTL^l|¼} Physics II 07-25-2006
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16:09:36 Query magnetic fields produced by electric currents. What evidence do we have that electric currents produce magnetic fields?
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RESPONSE -->
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16:10:04 Query magnetic fields produced by electric currents. What evidence do we have that electric currents produce magnetic fields?
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RESPONSE --> We do have evidence that electric currents produce magnetic fields. This is observed in engineering when laying current carrying wires next to each other. The current carrying wires produce magnetic fields that may affect other wires or possibly metal objects that are near them. This is evident in the video experiment. When Dave placed the metal ball near the coil of wires and turned the generator to produce current in the wires the ball moved toward the coil. This means that there was an attraction toward the coil which in this case was a magnetic field.
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16:10:12 STUDENT RESPONSE: We do have evidence that electric currents produce magnetic fields. This is observed in engineering when laying current carrying wires next to each other. The current carrying wires produce magnetic fields that may affect other wires or possibly metal objects that are near them. This is evident in the video experiment. When Dave placed the metal ball near the coil of wires and turned the generator to produce current in the wires the ball moved toward the coil. This means that there was an attraction toward the coil which in this case was a magnetic field. INSTRUCTOR COMMENT: Good observations. A very specific observation that should be included is that a compass placed over a conducting strip or wire initially oriented in the North-South direction will be deflected toward the East-West direction. **
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RESPONSE --> ok
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16:10:18 STUDENT RESPONSE: We do have evidence that electric currents produce magnetic fields. This is observed in engineering when laying current carrying wires next to each other. The current carrying wires produce magnetic fields that may affect other wires or possibly metal objects that are near them. This is evident in the video experiment. When Dave placed the metal ball near the coil of wires and turned the generator to produce current in the wires the ball moved toward the coil. This means that there was an attraction toward the coil which in this case was a magnetic field. INSTRUCTOR COMMENT: Good observations. A very specific observation that should be included is that a compass placed over a conducting strip or wire initially oriented in the North-South direction will be deflected toward the East-West direction. **
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RESPONSE --> ok
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16:10:21 How does the direction of an electric current compare with the direction of the magnetic field that results?
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RESPONSE -->
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16:10:34 How does the direction of an electric current compare with the direction of the magnetic field that results?
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RESPONSE --> The direction of the magnetic field relative to the direction of the electric current can be described using the right hand rule. This means simply using your right hand as a model you hold it so that your thumb is extended and your four fingers are flexed as if you were holding a cylinder. In this model, your thumb represents the direction of the electric current in the wire and the flexed fingers represent the direction of the magnetic field due to the current. In the case of the experiment the wire was in a coil so the magnetic field goes through the hole in the middle in one direction.
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16:10:39 ** GOOD STUDENT RESPONSE: The direction of the magnetic field relative to the direction of the electric current can be described using the right hand rule. This means simply using your right hand as a model you hold it so that your thumb is extended and your four fingers are flexed as if you were holding a cylinder. In this model, your thumb represents the direction of the electric current in the wire and the flexed fingers represent the direction of the magnetic field due to the current. In the case of the experiment the wire was in a coil so the magnetic field goes through the hole in the middle in one direction. **
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RESPONSE --> ok
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16:10:46 ** GOOD STUDENT RESPONSE: The direction of the magnetic field relative to the direction of the electric current can be described using the right hand rule. This means simply using your right hand as a model you hold it so that your thumb is extended and your four fingers are flexed as if you were holding a cylinder. In this model, your thumb represents the direction of the electric current in the wire and the flexed fingers represent the direction of the magnetic field due to the current. In the case of the experiment the wire was in a coil so the magnetic field goes through the hole in the middle in one direction. **
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RESPONSE --> ok
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16:11:34 Query problem 17.35 What would be the area of a .20 F capacitor if plates are separated by 2.2 mm of air?
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RESPONSE --> If a parallel-plate capacitor holds charge Q and the plates are separated by air, which has dielectric constant very close to 1, then the electric field between the plates is E = 4 pi k sigma = 4 pi k Q / A, obtained by applying Gauss' Law to each plate. The voltage between the plates is therefore V = E * d, where d is the separation. The capacitance is C = Q / V = Q / (4 pi k Q / A * d) = A / (4 pi k d). Solving this formula C = A / (4 pi k d) for A for the area get A = 4 pi k d * C If capacitance is C = .20 F and the plates are separated by 2.2 mm = .0022 m therefore have A = 4 pi k d * C = 4 pi * 9 * 10^9 N m^2 / C^2 * .20 C / V * .0022 m = 5 * 10^7 N m^2 / C^2 * C / ( J / C) * m = 5 * 10^7 N m^2 / (N m) * m = 5 * 10^7 m^2.
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16:11:39 ** If a parallel-plate capacitor holds charge Q and the plates are separated by air, which has dielectric constant very close to 1, then the electric field between the plates is E = 4 pi k sigma = 4 pi k Q / A, obtained by applying Gauss' Law to each plate. The voltage between the plates is therefore V = E * d, where d is the separation. The capacitance is C = Q / V = Q / (4 pi k Q / A * d) = A / (4 pi k d). Solving this formula C = A / (4 pi k d) for A for the area we get A = 4 pi k d * C If capacitance is C = .20 F and the plates are separated by 2.2 mm = .0022 m we therefore have A = 4 pi k d * C = 4 pi * 9 * 10^9 N m^2 / C^2 * .20 C / V * .0022 m = 5 * 10^7 N m^2 / C^2 * C / ( J / C) * m = 5 * 10^7 N m^2 / (N m) * m = 5 * 10^7 m^2. **
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RESPONSE --> ok
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16:12:15 Query problem 17.50 charge Q on capacitor; separation halved as dielectric with const K inserted; by what factor does the energy storage change? Compare the new electric field with the old.
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RESPONSE --> Electric field is independent of separation, as long as we don't have some huge separation. Voltage is work / unit charge to move from one plate to the other, which is force / unit charge * distance between plates, or electric field * distance. That is, V = E * d. Capacitance is Q / V, ration of charge to voltage. Energy stored is .5 Q^2 / C, which is just the work required to move charge Q across the plates with the 'average' voltage .5 Q / C (also obtained by integrating `dW = `dQ * V = `dq * q / C from q=0 to q = Q). The dielectric increases capacitance by reducing the electric field, which thereby reduces the voltage between plates. The electric field will be 1 / k times as great, meaning 1/k times the voltage at any given separation. C will increase by factor k due to dielectric, and will also increase by factor 2 due to halving of the distance. This is because the electric field is independent of the distance between plates, so halving the distance will halve the voltage between the plates. Since C = Q / V, this halving of the denominator will double C. Thus the capacitance increases by factor 2 k, which will decrease .5 Q^2 / C, the energy stored, by factor 2 k
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16:12:19 Note that the problem in the latest version of the text doubles rather than halves the separation. The solution for the halved separation, given here, should help you assess whether your solution was correct, and if not should help you construct a detailed and valid self-critique. ** For a capacitor we know the following: Electric field is independent of separation, as long as we don't have some huge separation. Voltage is work / unit charge to move from one plate to the other, which is force / unit charge * distance between plates, or electric field * distance. That is, V = E * d. Capacitance is Q / V, ration of charge to voltage. Energy stored is .5 Q^2 / C, which is just the work required to move charge Q across the plates with the 'average' voltage .5 Q / C (also obtained by integrating `dW = `dQ * V = `dq * q / C from q=0 to q = Q). The dielectric increases capacitance by reducing the electric field, which thereby reduces the voltage between plates. The electric field will be 1 / k times as great, meaning 1/k times the voltage at any given separation. C will increase by factor k due to dielectric, and will also increase by factor 2 due to halving of the distance. This is because the electric field is independent of the distance between plates, so halving the distance will halve the voltage between the plates. Since C = Q / V, this halving of the denominator will double C. Thus the capacitance increases by factor 2 k, which will decrease .5 Q^2 / C, the energy stored, by factor 2 k. **
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RESPONSE --> ok
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16:13:07 Query introductory problems set 54 #'s 1-7. Explain how to obtain the magnetic field due to a given current through a small current segment, and how the position of a point with respect to the segment affects the result.
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RESPONSE --> IL is the source. The law is basically an inverse square law and the angle theta between IL and the vector r from the source to the point also has an effect so that the field is B = k ' I L / r^2 * sin(theta)
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16:13:11 ** IL is the source. The law is basically an inverse square law and the angle theta between IL and the vector r from the source to the point also has an effect so that the field is B = k ' I L / r^2 * sin(theta). **
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RESPONSE --> ok
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16:13:53 Explain how to obtain the magnetic field due to a circular loop at the center of the loop.
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RESPONSE --> Each small increment `dL of the loop is a source I `dL. The vector from `dL to the center of the loop has magnitude r, where r is the radius of the loop, and is perpendicular to the loop so the contribution of increment * `dL to the field is k ' I `dL / r^2 sin(90 deg) = I `dL / r^2, where r is the radius of the loop. The field is either upward or downward by the right-hand rule, depending on whether the current runs counterclockwise or clockwise, respectively. The field has this direction regardless of where the increment is located. The sum of the fields from all the increments therefore has magnitude B = sum(k ' I `dL / r^2), where the summation occurs around the entire loop. I and r are constants so the sum is B = k ' I / r^2 sum(`dL). The sum of all the length increments around the loop is the circumference 2 pi r of the loop so we have B = 2 pi r k ' I / r^2 = 2 pi k ' I / r
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16:13:58 ** For current running in a circular loop: Each small increment `dL of the loop is a source I `dL. The vector from `dL to the center of the loop has magnitude r, where r is the radius of the loop, and is perpendicular to the loop so the contribution of increment * `dL to the field is k ' I `dL / r^2 sin(90 deg) = I `dL / r^2, where r is the radius of the loop. The field is either upward or downward by the right-hand rule, depending on whether the current runs counterclockwise or clockwise, respectively. The field has this direction regardless of where the increment is located. The sum of the fields from all the increments therefore has magnitude B = sum(k ' I `dL / r^2), where the summation occurs around the entire loop. I and r are constants so the sum is B = k ' I / r^2 sum(`dL). The sum of all the length increments around the loop is the circumference 2 pi r of the loop so we have B = 2 pi r k ' I / r^2 = 2 pi k ' I / r. **
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RESPONSE --> ok
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16:14:05 query univ 24.50 (25.36 10th edition). Parallel plates 16 cm square 4.7 mm apart connected to a 12 volt battery. What is the capacitance of this capacitor?
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RESPONSE --> dont do
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16:14:09 ** Fundamental principles include the fact that the electric field is very neary constant between parallel plates, the voltage is equal to field * separation, electric field from a single plate is 2 pi k sigma, the work required to displace a charge is equal to charge * ave voltage, and capacitance is charge / voltage. Using these principles we reason out the problem as follows: If the 4.7 mm separation experiences a 12 V potential difference then the electric field is E = 12 V / (4.7 mm) = 12 V / (.0047 m) = 2550 V / m, approx. Since the electric field of a plane charge distribution with density sigma is 2 pi k sigma, and since the electric field is created by two plates with equal opposite charge density, the field of the capacitor is 4 pi k sigma. So we have 4 pi k sigma = 2250 V / m and sigma = 2250 V / m / (4 pi k) = 2250 V / m / (4 pi * 9 * 10^9 N m^2 / C^2) = 2.25 * 10^-8 C / m^2. The area of the plate is .0256 m^2 so the charge on a plate is .0256 m^2 * 2.25 * 10^-8 C / m^2 = 5.76 * 10^-10 C. The capacitance is C = Q / V = 5.67 * 10^-10 C / (12 V) = 4.7 * 10^-11 C / V = 4.7 * 10^-11 Farads. The energy stored in the capacitor is equal to the work required to move this charge from one plate to another, starting with an initially uncharged capacitor. The work to move a charge Q across an average potential difference Vave is Vave * Q. Since the voltage across the capacitor increases linearly with charge the average voltage is half the final voltage, so we have vAve = V / 2, with V = 12 V. So the energy is energy = vAve * Q = 12 V / 2 * (5.76 * 10^-10 C) = 3.4 * 10^-9 V / m * C. Since the unit V / m * C is the same as J / C * C = J, we see that the energy is 3.4 * 10^-9 J. Pulling the plates twice as far apart while maintaining the same voltage would cut the electric field in half (the voltage is the same but charge has to move twice as far). This would imply half the charge density, half the charge and therefore half the capacitance. Since we are moving only half the charge through the same average potential difference we use only 1/2 the energy. Note that the work to move charge `dq across the capacitor when the charge on the capacitor is `dq * V = `dq * (q / C), so to obtain the work required to charge the capacitor we integrate q / C with respect to q from q = 0 to q = Q, where Q is the final charge. The antiderivative is q^2 / ( 2 C ) so the definite integral is Q^2 / ( 2 C). This is the same result obtained using average voltage and charge, which yields V / 2 * Q = (Q / C) / 2 * Q = Q^2 / (2 C) Integration is necessary for cylindrical and spherical capacitors and other capacitors which are not in a parallel-plate configuration. **
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RESPONSE --> ok
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16:14:16 query univ 24.51 (25.37 10th edition). Parallel plates 16 cm square 4.7 mm apart connected to a 12 volt battery. If the battery remains connected and plates are pulled to separation 9.4 mm then what are the capacitance, charge of each plate, electric field, energy stored?
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RESPONSE --> dont do
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16:14:21 The potential difference between the plates is originally 12 volts. 12 volts over a 4.7 mm separation implies electric field = potential gradient = 12 V / (.0047 m) = 2500 J / m = 2500 N / C, approx.. The electric field is E = 4 pi k * sigma = 4 pi k * Q / A so we have Q = E * A / ( 4 pi k) = 2500 N / C * (.16 m)^2 / (4 * pi * 9 * 10^9 N m^2 / C^2) = 5.7 * 10^-7 N / C * m^2 / (N m^2 / C^2) = 5.7 * 10^-10 C, approx.. The energy stored is E = 1/2 Q V = 1/2 * 5.6 * 10^-10 C * 12 J / C = 3.36 * 10^-9 J. If the battery remains connected then if the plate separation is doubled the voltage will remain the same, while the potential gradient and hence the field will be halved. This will halve the charge on the plates, giving us half the capacitance. So we end up with a charge of about 2.8 * 10^-10 C, and a field of about 1250 N / C. The energy stored will also be halved, since V remains the same but Q is halved.
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RESPONSE --> ok
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16:14:26 query univ 24.68 (25.52 10th edition). Solid conducting sphere radius R charge Q. What is the electric-field energy density at distance r < R from the center of the sphere? What is the electric-field energy density at distance r > R from the center of the sphere?
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RESPONSE --> dont do
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16:14:36 ** The idea is that we have to integrate the energy density over all space. We'll do this by finding the total energy in a thin spherical shell of radius r and thickness `dr, using this result to obtain an expression we integrate from R to infinity, noting that the field of the conducting sphere is zero for r < R. Then we can integrate to find the work required to assemble the charge on the surface of the sphere and we'll find that the two results are equal. Energy density, defined by dividing the energy .5 C V^2 required to charge a parallel-plate capacitor by the volume occupied by its electric field, is Energy density = .5 C V^2 / (volume) = .5 C V^2 / (d * A), where d is the separation of the plates and A the area of the plates. Since C = epsilon0 A / d and V = E * d this gives us .5 epsilon0 A / d * (E * d)^2 / (d * A) = .5 epsilon0 E^2 so that Energy density = .5 epsilon0 E^2, or in terms of k Energy density = 1 / (8 pi k) E^2, Since your text uses epsilon0 I'll do the same on this problem, where the epsilon0 notation makes a good deal of sense: For the charged sphere we have for r > R E = Q / (4 pi epsilon0 r^2), and therefore energy density = .5 epsilon0 E^2 = .5 epsilon0 Q^2 / (16 pi^2 epsilon0^2 r^4) = Q^2 / (32 pi^2 epsilon0 r^4). The energy density between r and r + `dr is nearly constant if `dr is small, with energy density approximately Q^2 / (32 pi^2 epsilon0 r^4). The volume of space between r and r + `dr is approximately A * `dr = 4 pi r^2 `dr. The expression for the energy lying between distance r and r + `dr is therefore approximately energy density * volume = Q^2 / (32 pi^2 epsilon0 r^4) * 4 pi r^2 `dr = Q^2 / (8 pi epsilon0 r^2) `dr. This leads to a Riemann sum over radius r; as we let `dr approach zero we approach an integral with integrand Q^2 / (8 pi epsilon0 r^2), integrated with respect to r. To get the energy between two radii we therefore integrate this expression between those two radii. If we integrate this expression from r = R to infinity we get the total energy of the field of the charged sphere. This integral gives us Q^2 / (8 pi epsilon0 R), which is the same as k Q^2 / (2 R). The work required to bring a charge `dq from infinity to a sphere containing charge q is k q / R `dq, leading to the integral of k q / R with respect to q. If we integrate from q = 0 to q = Q we get the total work required to charge the sphere. Our antiderivative is k (q^2 / 2) / r. If we evaluate this antiderivative at lower limit 0 and upper limit Q we get k Q^2 / (2 R). Since k = 1 / (4 pi epsilon0) the work is k Q^2 / (2 R) = k Q^2 / (8 pi epsilon0 R). So the energy in the field is equal to the work required to assemble the charge distribution. **
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RESPONSE --> ok
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טC{I鏸 assignment #029 KTL^l|¼} Physics II 07-25-2006 Þ{|e assignment #029 KTL^l|¼} Physics II 07-25-2006
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16:15:37 Query introductory problem set 54 #'s 8-13 Explain how to determine the magnetic flux of a uniform magnetic field through a plane loop of wire, and explain how the direction of the field and the direction of a line perpendicular to the plane of the region affect the result.
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RESPONSE -->
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16:15:58 Query introductory problem set 54 #'s 8-13 Explain how to determine the magnetic flux of a uniform magnetic field through a plane loop of wire, and explain how the direction of the field and the direction of a line perpendicular to the plane of the region affect the result.
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RESPONSE --> To do this we need to simply find the area of the plane loop of wire. If we are given the radius we can find the area using Pi * r ^2 Then we multiply the area of the loop (In square meters ) by the strength of the field (in tesla). This will give us the strength of the flux if the plane of the loop is perpendicular to the field. If the perpendicular to the loop is at some nonzero angle with the field, then we multiply the previous result by the cosine of the angle.
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16:16:04 To do this we need to simply find the area of the plane loop of wire. If we are given the radius we can find the area using Pi * r ^2 Then we multiply the area of the loop (In square meters ) by the strength of the field (in tesla). This will give us the strength of the flux if the plane of the loop is perpendicular to the field. If the perpendicular to the loop is at some nonzero angle with the field, then we multiply the previous result by the cosine of the angle.
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RESPONSE --> ok
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16:16:10 To do this we need to simply find the area of the plane loop of wire. If we are given the radius we can find the area using Pi * r ^2 Then we multiply the area of the loop (In square meters ) by the strength of the field (in tesla). This will give us the strength of the flux if the plane of the loop is perpendicular to the field. If the perpendicular to the loop is at some nonzero angle with the field, then we multiply the previous result by the cosine of the angle.
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RESPONSE --> ok
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16:16:14 Explain how to determine the average rate of change of magnetic flux due to a uniform magnetic field through a plane loop of wire, as the loop is rotated in a given time interval from an orientation perpendicular to the magnetic field to an orientation parallel to the magnetic field.
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RESPONSE -->
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16:16:31 Explain how to determine the average rate of change of magnetic flux due to a uniform magnetic field through a plane loop of wire, as the loop is rotated in a given time interval from an orientation perpendicular to the magnetic field to an orientation parallel to the magnetic field.
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RESPONSE --> The first thing that we need to do is again use Pi * r ^ 2 to find the area of the loop. Then we multiply the area of the loop (m^2) by the strength of the field (testla) to find the flux when the loop is perpendicular to the field. Then we do the same thing for when the loop is parallel to the field, and since the cos of zero degrees is zero, the flux when the loop is parallel to the field is zero. This makes sense because at this orientation the loop will pick up none of the magnetic field. So now we have Flux 1 and Flux 2 being when the loop is perpendicular and parallel, respectively. So if we subtract Flux 2 from flux 1 and divide this value by the given time in seconds, we will have the average rate of change of magnetic flux. If we use MKS units this value will be in Tesla m^2 / sec = volts.
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16:16:40 ** EXPLANATION BY STUDENT: The first thing that we need to do is again use Pi * r ^ 2 to find the area of the loop. Then we multiply the area of the loop (m^2) by the strength of the field (testla) to find the flux when the loop is perpendicular to the field. Then we do the same thing for when the loop is parallel to the field, and since the cos of zero degrees is zero, the flux when the loop is parallel to the field is zero. This makes sense because at this orientation the loop will pick up none of the magnetic field. So now we have Flux 1 and Flux 2 being when the loop is perpendicular and parallel, respectively. So if we subtract Flux 2 from flux 1 and divide this value by the given time in seconds, we will have the average rate of change of magnetic flux. If we use MKS units this value will be in Tesla m^2 / sec = volts. **
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RESPONSE --> ok looks good
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16:16:45 ** EXPLANATION BY STUDENT: The first thing that we need to do is again use Pi * r ^ 2 to find the area of the loop. Then we multiply the area of the loop (m^2) by the strength of the field (testla) to find the flux when the loop is perpendicular to the field. Then we do the same thing for when the loop is parallel to the field, and since the cos of zero degrees is zero, the flux when the loop is parallel to the field is zero. This makes sense because at this orientation the loop will pick up none of the magnetic field. So now we have Flux 1 and Flux 2 being when the loop is perpendicular and parallel, respectively. So if we subtract Flux 2 from flux 1 and divide this value by the given time in seconds, we will have the average rate of change of magnetic flux. If we use MKS units this value will be in Tesla m^2 / sec = volts. **
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RESPONSE --> ok
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16:16:49 Explain how alternating current is produced by rotating a coil of wire with respect to a uniform magnetic field.
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RESPONSE -->
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16:17:38 ** First note that the current I is different for diferent faces. The resistance of the block is proportional to the distance between faces and inversely proportional to the area, so current is proportional to the area and inversely proportional to the distance between faces. Current density is proportional to current and inversely proportional to the area of the face, so current density is proportional to area and inversely proportional to the distance between faces and to area, leaving current inversely proportional to distance between faces. For the faces measuring d x 2d we have resistance R = rho * L / A = rho * (3d) / (2 d^2) = 3 / 2 rho / d so current is I = V / R = V / (3/2 rho / d) = 2d V / (3 rho). Current density is I / A = (2 d V / (3 rho) ) / (2 d^2) = V / (3 rho d) = 1/3 V / (rho d). For the faces measuring d x 3d we have resistance R = rho * L / A = rho * (2d) / (3 d^2) = 2 / 3 rho / d so current is I = V / R = V / (2/3 rho / d) = 3 d V / (2 rho). Current density is I / A = (3 d V / (2 rho) ) / (3 d^2) = V / (2 rho d) = 1/2 V / (rho d). For the faces measuring 3d x 2d we have resistance R = rho * L / A = rho * (d) / (6 d^2) = 1 / 6 rho / d so current is I = V / R = V / (1/6 rho / d) = 6 d V / (rho). Current density is I / A = (6 d V / (rho) ) / (6 d^2) = V / (rho d) = V / (rho d). Max current density therefore occurs when the voltage is applied to the largest face. **
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RESPONSE --> ok.
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dx|ߕīK assignment #030 KTL^l|¼} Physics II 07-25-2006 ̒}ߢǗ^ assignment #030 KTL^l|¼} Physics II 07-25-2006
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16:18:30 Query introductory problem set 54 #'s 14-18. Explain whether, and if so how, the force on a charged particle due to the field between two capacitor plates is affected by its velocity.
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RESPONSE -->
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16:19:03 Query introductory problem set 54 #'s 14-18. Explain whether, and if so how, the force on a charged particle due to the field between two capacitor plates is affected by its velocity.
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RESPONSE --> There is a force due to the electric field between the plates, but the effect of an electric field does not depend on velocity. The plates of a capacitor do not create a magnetic field.
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16:19:14 ** There is a force due to the electric field between the plates, but the effect of an electric field does not depend on velocity. The plates of a capacitor do not create a magnetic field. **
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RESPONSE --> ok looks good
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16:19:21 ** There is a force due to the electric field between the plates, but the effect of an electric field does not depend on velocity. The plates of a capacitor do not create a magnetic field. **
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RESPONSE --> ok
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16:19:25 Explain whether, and if so how, the force on a charged particle due to the magnetic field created by a wire coil is affected by its velocity.
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RESPONSE -->
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16:20:01 Explain whether, and if so how, the force on a charged particle due to the magnetic field created by a wire coil is affected by its velocity.
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RESPONSE --> A wire coil does create a magnetic field perpendicular to the plane of the coil. If the charged particle moves in a direction perpendicular to the coil then a force F = q v B is exerted by the field perpendicular to both the motion of the particle and the direction of the field. The precise direction is determined by the right-hand rule
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16:20:12 ** A wire coil does create a magnetic field perpendicular to the plane of the coil. If the charged particle moves in a direction perpendicular to the coil then a force F = q v B is exerted by the field perpendicular to both the motion of the particle and the direction of the field. The precise direction is determined by the right-hand rule. **
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RESPONSE --> ok
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16:20:21 ** A wire coil does create a magnetic field perpendicular to the plane of the coil. If the charged particle moves in a direction perpendicular to the coil then a force F = q v B is exerted by the field perpendicular to both the motion of the particle and the direction of the field. The precise direction is determined by the right-hand rule. **
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RESPONSE --> ok
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16:20:24 Explain how the net force changes with velocity as a charged particle passes through the field between two capacitor plates, moving perpendicular to the constant electric field, in the presence of a constant magnetic field oriented perpendicular to both the velocity of the particle and the field of the capacitor.
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RESPONSE -->
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16:20:45 Explain how the net force changes with velocity as a charged particle passes through the field between two capacitor plates, moving perpendicular to the constant electric field, in the presence of a constant magnetic field oriented perpendicular to both the velocity of the particle and the field of the capacitor.
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RESPONSE --> At low enough velocities the magnetic force F = q v B is smaller in magnitude than the electrostatic force F = q E. At high enough velocities the magnetic force is greater in magnitude than the electrostatic force. At a certain specific velocity, which turns out to be v = E / B, the magnitudes of the two forces are equal. If the perpendicular magnetic and electric fields exert forces in opposite directions on the charged particle then when the magnitudes of the forces are equal the net force on the particle is zero and it passes through the region undeflected
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16:20:54 ** At low enough velocities the magnetic force F = q v B is smaller in magnitude than the electrostatic force F = q E. At high enough velocities the magnetic force is greater in magnitude than the electrostatic force. At a certain specific velocity, which turns out to be v = E / B, the magnitudes of the two forces are equal. If the perpendicular magnetic and electric fields exert forces in opposite directions on the charged particle then when the magnitudes of the forces are equal the net force on the particle is zero and it passes through the region undeflected. **
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RESPONSE --> ok looks alright
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16:21:06 ** At low enough velocities the magnetic force F = q v B is smaller in magnitude than the electrostatic force F = q E. At high enough velocities the magnetic force is greater in magnitude than the electrostatic force. At a certain specific velocity, which turns out to be v = E / B, the magnitudes of the two forces are equal. If the perpendicular magnetic and electric fields exert forces in opposite directions on the charged particle then when the magnitudes of the forces are equal the net force on the particle is zero and it passes through the region undeflected. **
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RESPONSE --> ok
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16:21:09 Query General Physics Problem (formerly 20.32, but omitted from new version). This problem is not assigned but you should solve it now: If an electron is considered to orbit a proton in a circular orbit of radius .529 * 10^-10 meters (the electron doesn't really move around the proton in a circle; the behavior of this system at the quantum level does not actually involve a circular orbit, but the result obtained from this assumption agrees with the results of quantum mechanics), the electron's motion constitutes a current along its path. What is the field produced at the location of the proton by the current that results from this 'orbit'? To obtain an answer you might want to first answer the two questions: 1. What is the velocity of the electron? 2. What therefore is the current produced by the electron? How did you calculate the magnetic field produced by this current?
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RESPONSE -->
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16:21:35 Query General Physics Problem (formerly 20.32, but omitted from new version). This problem is not assigned but you should solve it now: If an electron is considered to orbit a proton in a circular orbit of radius .529 * 10^-10 meters (the electron doesn't really move around the proton in a circle; the behavior of this system at the quantum level does not actually involve a circular orbit, but the result obtained from this assumption agrees with the results of quantum mechanics), the electron's motion constitutes a current along its path. What is the field produced at the location of the proton by the current that results from this 'orbit'? To obtain an answer you might want to first answer the two questions: 1. What is the velocity of the electron? 2. What therefore is the current produced by the electron? How did you calculate the magnetic field produced by this current?
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RESPONSE --> Setting centripetal force = Coulomb attraction for the orbital radius, which is .529 Angstroms = .529 * 10^-10 meters, we have m v^2 / r = k q1 q2 / r^2 so that v = sqrt(k q1 q2 / (m r) ). Evaluating for k, with q1 = q2 = fundamental charge and m = mass of the electron we obtain v = 2.19 * 10^6 m/s. The circumference of the orbit is `dt = 2 pi r so the time required to complete an orbit is `dt = 2 pi r / v, which we evaluate for the v obtained above. We find that `dt = 1.52 * 10^-16 second. Thus the current is I = `dq / `dt = q / `dt, where q is the charge of the electron. Simplifying we get I = .00105 amp, approx.. The magnetic field due to a .00105 amp current in a loop of radius .529 Angstroms is B = k ' * 2 pi r I / r^2 = 2 pi k ' I / r = 12.5 Tesla.
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16:21:47 **If you know the orbital velocity of the electron and orbital radius then you can determine how long it takes to return to a given point in its orbit. So the charge of 1 electron 'circulates' around the orbit in that time interval. Current is charge flowing past a point / time interval. Setting centripetal force = Coulomb attraction for the orbital radius, which is .529 Angstroms = .529 * 10^-10 meters, we have m v^2 / r = k q1 q2 / r^2 so that v = sqrt(k q1 q2 / (m r) ). Evaluating for k, with q1 = q2 = fundamental charge and m = mass of the electron we obtain v = 2.19 * 10^6 m/s. The circumference of the orbit is `dt = 2 pi r so the time required to complete an orbit is `dt = 2 pi r / v, which we evaluate for the v obtained above. We find that `dt = 1.52 * 10^-16 second. Thus the current is I = `dq / `dt = q / `dt, where q is the charge of the electron. Simplifying we get I = .00105 amp, approx.. The magnetic field due to a .00105 amp current in a loop of radius .529 Angstroms is B = k ' * 2 pi r I / r^2 = 2 pi k ' I / r = 12.5 Tesla. **
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RESPONSE --> looks good
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16:21:54 query univ 27.60 (28.46 10th edition). cyclotron 3.5 T field.
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RESPONSE --> don't do
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16:22:15 **If you know the orbital velocity of the electron and orbital radius then you can determine how long it takes to return to a given point in its orbit. So the charge of 1 electron 'circulates' around the orbit in that time interval. Current is charge flowing past a point / time interval. Setting centripetal force = Coulomb attraction for the orbital radius, which is .529 Angstroms = .529 * 10^-10 meters, we have m v^2 / r = k q1 q2 / r^2 so that v = sqrt(k q1 q2 / (m r) ). Evaluating for k, with q1 = q2 = fundamental charge and m = mass of the electron we obtain v = 2.19 * 10^6 m/s. The circumference of the orbit is `dt = 2 pi r so the time required to complete an orbit is `dt = 2 pi r / v, which we evaluate for the v obtained above. We find that `dt = 1.52 * 10^-16 second. Thus the current is I = `dq / `dt = q / `dt, where q is the charge of the electron. Simplifying we get I = .00105 amp, approx.. The magnetic field due to a .00105 amp current in a loop of radius .529 Angstroms is B = k ' * 2 pi r I / r^2 = 2 pi k ' I / r = 12.5 Tesla. **
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RESPONSE --> ok
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16:22:19 query univ 27.60 (28.46 10th edition). cyclotron 3.5 T field.
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RESPONSE -->
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16:22:39 What is the radius of orbit for a proton with kinetic energy 2.7 MeV?
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RESPONSE --> We know that the centripetal force for an object moving in a circle is F = m v^2 / r. In a magnetic field perpendicular to the velocity this force is equal to the magnetic force F = q v B. So we have m v^2 / r = q v B so that r = m v / (q B). A proton with ke 2.7 MeV = 2.7 * 10^6 * (1.6 * 10^-19 J) = 3.2 * 10^-13 J has velocity such that v = sqrt(2 KE / m) = sqrt(2 * 3.2 * 10^-13 J / (1.67 * 10^-27 kg) ) = 2.3 * 10^7 m/s approx.. So we have r = m v / (q B) = 1.67 * 10^-27 kg * 2.3 * 10^7 m/s / (1.6 * 10^-19 C * 3.5 T) = .067 m approx.
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16:22:44 ** We know that the centripetal force for an object moving in a circle is F = m v^2 / r. In a magnetic field perpendicular to the velocity this force is equal to the magnetic force F = q v B. So we have m v^2 / r = q v B so that r = m v / (q B). A proton with ke 2.7 MeV = 2.7 * 10^6 * (1.6 * 10^-19 J) = 3.2 * 10^-13 J has velocity such that v = sqrt(2 KE / m) = sqrt(2 * 3.2 * 10^-13 J / (1.67 * 10^-27 kg) ) = 2.3 * 10^7 m/s approx.. So we have r = m v / (q B) = 1.67 * 10^-27 kg * 2.3 * 10^7 m/s / (1.6 * 10^-19 C * 3.5 T) = .067 m approx. **
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RESPONSE --> ok
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16:23:07 What is the radius of orbit for a proton with kinetic energy 5.4 MeV?
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RESPONSE --> Doubling the KE of the proton increases velocity by factor sqrt(2) and therefore increases the radius of the orbit by the same factor. We end up with a radius of about .096 m
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16:23:12 ** Doubling the KE of the proton increases velocity by factor sqrt(2) and therefore increases the radius of the orbit by the same factor. We end up with a radius of about .096 m. **
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RESPONSE --> ok
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16:23:37 query 20.73 (was 28.52) rail gun bar mass m with current I across rails, magnetic field B perpendicular to loop formed by bars and rails What is the expression for the magnitude of the force on the bar, and what is the direction of the force?
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RESPONSE --> The length of the bar is given as L. So the force is I L B, since the current and field are perpendicular. The acceleration of the bar is therefore a = I L B / m. If the distance required to achieve a given velocity is `ds and initial velocity is 0 then vf^2 = v0^2 + 2 a `ds gives us ds = (vf^2 - v0^2) / (2 a) = vf^2 / (2 a). If v stands for the desired final velocity this is written `ds = v^2 / (2 a). In terms of I, L, B and m we have `ds = v^2 / (2 I L B / m) = m v^2 / (2 I L B). Note that we would get the same expression using KE: since (neglecting dissipative losses) we have `dKE = `dW = F `ds we have `ds = `dKE / F = 1/2 m v^2 / (I L B). For the given quantities we get `ds = 25 kg * (1.12 * 10^4 m/s)^2 / (2 * 2000 amps * .50 Tesla * .5 meters) = 3.2 * 10^6 meters, or about 3200 km
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16:23:44 ** The length of the bar is given as L. So the force is I L B, since the current and field are perpendicular. The acceleration of the bar is therefore a = I L B / m. If the distance required to achieve a given velocity is `ds and initial velocity is 0 then vf^2 = v0^2 + 2 a `ds gives us ds = (vf^2 - v0^2) / (2 a) = vf^2 / (2 a). If v stands for the desired final velocity this is written `ds = v^2 / (2 a). In terms of I, L, B and m we have `ds = v^2 / (2 I L B / m) = m v^2 / (2 I L B). Note that we would get the same expression using KE: since (neglecting dissipative losses) we have `dKE = `dW = F `ds we have `ds = `dKE / F = 1/2 m v^2 / (I L B). For the given quantities we get `ds = 25 kg * (1.12 * 10^4 m/s)^2 / (2 * 2000 amps * .50 Tesla * .5 meters) = 3.2 * 10^6 meters, or about 3200 km. **
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RESPONSE --> ok looks good
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16:24:14 query 28.66 u quark + 2/3 e and d quark -1/3 e counterclockwise, clockwise in neutron (r = 1.20 * 10^-15 m) What are the current and the magnetic moment produced by the u quark?
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RESPONSE --> If r is the radius of the orbit and v the velocity then the frequency of an orbit is f = v / (2 pi r). The frequency tells you how many times the charge passes a given point per unit of time. If the charge is q then the current must therefore be }I = q f = q v / (2 pi r). Half the magnetic moment is due to the u quark, which carries charge equal and opposite to the combined charge of both d quarks, the other half to the d quarks (which circulate, according to this model, in the opposite direction with the same radius so that the two d quarks contribute current equal to, and of the same sign, as the u quark). The area enclosed by the path is pi r^2, so that the magnetic moment of a quark is I A = q v / (2 pi r) * pi r^2 = q v r / 2. The total magnetic moment is therefore 2/3 e * v r / 2 + 2 ( 1/3 e * v r / 2) = 4/3 e v r / 2 = 2/3 e v r.. Setting this equal to the observed magnetic moment mu we have 2/3 e v r = mu so that v = 3/2 mu / (e r) = 3/2 * 9.66 * 10^-27 A m^2 / (1.6 * 10^-19 C * 1.20 * 10^-15 m) = 7.5 * 10^7 m/s, approx.. Note that units are A m^2 / (C m) = C / s * m^2 / (C m) = m / s
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16:24:19 ** If r is the radius of the orbit and v the velocity then the frequency of an orbit is f = v / (2 pi r). The frequency tells you how many times the charge passes a given point per unit of time. If the charge is q then the current must therefore be }I = q f = q v / (2 pi r). Half the magnetic moment is due to the u quark, which carries charge equal and opposite to the combined charge of both d quarks, the other half to the d quarks (which circulate, according to this model, in the opposite direction with the same radius so that the two d quarks contribute current equal to, and of the same sign, as the u quark). The area enclosed by the path is pi r^2, so that the magnetic moment of a quark is I A = q v / (2 pi r) * pi r^2 = q v r / 2. The total magnetic moment is therefore 2/3 e * v r / 2 + 2 ( 1/3 e * v r / 2) = 4/3 e v r / 2 = 2/3 e v r.. Setting this equal to the observed magnetic moment mu we have 2/3 e v r = mu so that v = 3/2 mu / (e r) = 3/2 * 9.66 * 10^-27 A m^2 / (1.6 * 10^-19 C * 1.20 * 10^-15 m) = 7.5 * 10^7 m/s, approx.. Note that units are A m^2 / (C m) = C / s * m^2 / (C m) = m / s. **
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RESPONSE --> ok
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16:24:32 query univ 28.68 (29.56 10th edition) infinite L-shaped conductor toward left and downward. Point a units to right of L along line of current from left. Current I. What is the magnetic field at the specified point?
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RESPONSE --> dont do.
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16:24:37 STUDENT RESPONSE FOLLOWED BY SOLUTION: I could not figure out the magnetic field affecting point P. the current is cursing ** I assume you mean 'coursing', though the slip is understandable ** toward P then suddnely turns down at a right angle. If I assume that the magnetic field of a thin wire is radial in all directions perpendicular to the wire, then it is possible that at least one field line would be a straight line from the wire to point P. It seems to me that from that field line,down the to the lower length of the wire, would affect at P. SOLUTION: The r vector from any segment along the horizontal section of the wire would be parallel to the current segment, so sin(theta) would be 0 and the contribution `dB = k ' I `dL / r^2 sin(theta) would be zero. So the horizontal section contributes no current at the point. Let the y axis be directed upward with its origin at the 'bend'. Then a segment of length `dy at position y will lie at distance r = sqrt(y^2 + a^2) from the point and the sine of the angle from the r vector to the point is a / sqrt(y^2 + a^2). The field resulting from this segment is therefore `dB = k ' I `dy / (a^2 + y^2) * a / sqrt(a^2 + y^2). Crossing the I `dy vector with the r vector tells us the `dB is coming at us out of the paper (fingers extended along neg y axis, ready to 'turn' toward r results in thumb pointing up toward us away from the paper). This is the direction for all `dB contributions so B will have the same direction. Summing all contributions we have sum(k ' I `dy / (a^2 + y^2) * a / sqrt(a^2 + y^2), y from 0 to -infinity). Taking the limit as `dy -> 0 we get the integral of k ' I a / (a^2 + y^2)^(3/2) with respect to y, with y from 0 to -infinity. This integral is -k ' I / a. So the field is B = - k ' I / a, directed upward out of the page. **
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RESPONSE --> ok
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