course Phys 202 r}пDassignment #020
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14:28:37 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 --> You take the two points and subtract them from each other and htis is the distance. r=sqrt(dx^2+dy^2) Then using |F|=k*|q1*q2|/r^2 plug in the numbers to find the magnitude. HTe direction is given by if the charges are alike, they go in the same direction. If the charges are differant, then it goes in the opposite direction.
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14:30:46 ** 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 the charges mixed up with the q and Q. i do understand that. I had not realized the arctan(y/x) was used. I must have overlooked it and I will go back and look at the problems.
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14:42:40 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 -->
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14:42:48 ** 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 -->
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q{lͨ assignment #021 蚊ьU΅y Physics II 08-01-2006
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15:20:46 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 --> You first have to find differance between the 2 points. Then using the charge and work, use W=Q*dV to find the change in volts.
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15:21:40 ** 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 --> I got a similar answer but did not go into great detail. I do understand the answer.
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15:53:33 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 --> the magnitude is given by F = k q1 * Q / r^2 and by kowing the direction of the vector and determining the quadrant it is in can tell you how the two signs differ. Plugging the numbers into the equation gives the r.
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16:00:42 ** 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 --> I was completely offbase but that makes it much simpler to figure it out.
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16:15:55 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 --> Work=Force *displacement force=charge*electric field Using these equations, the force cab be determined by F=W/d Then the Force put into E=F/Q
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16:16:14 ** 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 same answer.
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16:22:02 In your own words explain the meaning of the electric field.
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RESPONSE --> The electric field is the current of electricity working
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16:22:23 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 understand the answer.
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16:22:42 In your own words explain the meaning of voltage.
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RESPONSE --> Voltage is the intensity of elctricity.
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16:23:26 ** 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 W/unit charge fromt point A to B. I understand the definition
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bw}ep assignment #022 蚊ьU΅y Physics II 08-01-2006
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16:40:09 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 --> Using the displacement and the Force, work can be determined W=F*d hten knowing that only work is the opposiosing force against the KE so the final KE will be the original KE minus the work. KEf= KE0-W
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16:40:56 ** 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 --> i have hte first part right but was a little off on the last part.
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16:54:07 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 --> having P provided, the equation: P=dW/dt dW=F*d F=Q*E
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16:55:22 ** 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 --> V=W/Q I understand that know. I knew how to find the work but forgot about V=W/Q
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17:06:06 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 --> As long as the distance r does not change, then the change in the charge changes the Electric field uniformly. A doubled charge would meandouble the flux divided by the same area and therefore double the field area. I fthe distance changes, hten it causes the area to increase or decrease causing it to either decrease the field if the distance is increased or increase the field if the distanc eis decreased.
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17:06:59 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 --> i wrote it in words rather then equations. Flux=4*pi*k*Q is the equation, correct?
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17:40:49 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 --> E=flux/area=4pikQ/(2pi*r*L)
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17:44:28 ** 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 --> I got the equation but I didn't really understand how it was fiqured out.
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}ӯlG assignment #023 蚊ьU΅y Physics II 08-01-2006
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18:02:57 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 --> Using the drift velocity to determine the distance in one second, we can determine the length of the wire it would move. HTen using that, we use the number of electrons to determine the current.
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18:03:54 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 --> I understand the thought process.
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18:37:33 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 --> E=dV/dx E is higher in the short wire so the velocity is higher which inturn means a higher current.
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18:37:51 ** 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 --> That is what I was saying.
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18:46:17 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 --> The thicker wire has more cross-sectional area and therefore more electrons can move through it.
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18:47:00 ** 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 --> why is it (d2/d1)^2?
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19:15:59 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 --> THe average net force is found by determining the work and divide it by the distance. W=F*d
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19:20:53 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 --> The higher the resistance, the greater the voltage which therefore means a greater current. I=c*V/R the current is inversely proportional to the resisitance. P=(V/R)*V
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19:22:01 ** 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 --> P=dW/dT I =V/R V=dW/Q
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19:55:02 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 --> the magnetic field causes a force on the current. The force goes in right angles of the magnetic field.
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RESPONSE -->
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19:58:04 ** 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 --> I understand and read about the right-hand rule.
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