send 20-26

course Phy 202

here are the homework assignments I have so far.

Good work overall. Be sure you completely understand the situations and solutions given in the Introductory Problem Sets.

Let me know if there's anything you need me to explain further.

éyéËÒíáöîO½–ûÅÁƒ›Ì³aÉG´îÉœï Physics II 11-21-2005

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19:46:02 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 --> We must first assume a test charge of 1 C. We then take the coordinates of the location away from the origin. We then use the squt((x2-x1)^2 + (y2-y1)^2) to get the distance between the charges. We then use the formula (k * (charge 1 * charge 2))/ the distance calculated above. This is the magnitude. We then calculate the forces in the x and y direction by the formulas (y2-y1)/Dist*Mag (x2-x1)/Dist*mag we then determine the Magnitude of the of the force by sqrt(yforce^2+xforce^2) The direction is arctan(y2-y1)/(x2-x1)

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19:49:56 ** 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, 180 deg more or less than this direction. **

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RESPONSE --> I forgot the square on the r in the magnitude. I did not state to add the 180 if x is negative. I did not know that this direction was oppisit if q1 and Q are unlike charges.

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19:51:14 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 --> This is done exactly the same as the previous example, however, I believe the direction represents the field direction.

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19:54:12 11-21-2005 19:54:12 ** 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 deg more or less than this direction. **

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NOTES -------> I was somewhat right, I see that if q1 is positive the field is away from the origin to the point, and if q1 is negative the field is from the origin oppisit the point.

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assignment #021 éyéËÒíáöîO½–ûÅÁƒ›Ì³aÉG´îÉœï Physics II 11-21-2005

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20:36:26 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 --> We find the potential difference in volts between two points on the x axis by dividing the work by the charge to get the potential difference

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20:42:07 11-21-2005 20:42:07 ** 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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NOTES -------> I forgot to calculate the force and multiply it by the distance. This is a very simple calculation, by finding the force at each point.

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20:45:43 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 --> I am not real sure how to do this as of right now.

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20:48:03 ** 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 still do not understand this problem

This is covered in the Introductory Problem Sets. Be sure you have mastered those problems. Let me know if you have questions on any of them.

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20:50:16 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 --> We would divide the work by the distance between the points, to get the force, we would then divide the force by the charge to get the field.

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20:50:27 ** You get ave force from work and distance: Fave = `dW / `ds. You get ave electric field from work and charge: Eave = F / q. An alternative: Find potential difference `dV = `dW / q. Ave electric field is Eave = `dV / `ds **

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RESPONSE --> I was correct

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20:50:48 In your own words explain the meaning of the electric field.

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RESPONSE --> An electric field is a directon of travel of an electric charge.

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20:51:46 11-21-2005 20:51:46 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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NOTES -------> I think that I was correct, I just did not state that it measures the magnatude and direction at a point.

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20:57:31 In your own words explain the meaning of voltage.

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RESPONSE --> Voltage is a consistant charge.

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20:58:13 ** Voltage is the work done per unit of charge in moving charge from one point to another. **

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RESPONSE --> I did not exactly know that voltage was work done per unit of charge in moving charge from one point to another.

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20:59:25 Query experiment 16 current flow and energy You should have submitted your writeup of this experiment. However, answer the following questions. First question: Why is it that when the circuit formed has the least resistance the generator is hardest to crank?

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RESPONSE --> Because the current is moving back through the generator and this causes more work to be done to make the current travel.

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21:00:24 11-21-2005 21:00:24 ** Less resistance implies more current, meaning more charge per unit of time and therefore more work per unit of time. To perform this work at a constant cranking rate (hence a constant voltage, and also a constant distance per unit of time) requires more force (since the distance in a unit of time is constant, in order to do work at a greater rate the force must be greater). **

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NOTES -------> I think my answers to this were correct in my write up.

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21:01:30 Second question: Why do you think it is that the generator is harder to crank with two bulbs in parallel then with the same to bulbs and series?

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RESPONSE --> Because there is more voltage traveling in the same amount of distance and thus requires more work to achieve the same voltage in the series circut.

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21:01:54 11-21-2005 21:01:54 ** In a parallel circuit the full voltage of the generator is applied to both branches of the circuit, since both are directly connected to the generator. In a series circuit, the circuit doesn't split and the voltage is divided (not equally unless the bulbs have equal resistance) between the two bulbs. So in the parallel circuit both bulbs experience greater voltage, and hence greater current, than in the series circuit. **

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NOTES -------> Looks good to me.

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éâ¾^áÖüœºûëwÒÎè®ëÑÂ¥õúš assignment #022 éyéËÒíáöîO½–ûÅÁƒ›Ì³aÉG´îÉœï Physics II 11-28-2005

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19:32:57 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 --> Because the only force acting is due to electric field, thus the Final energy will be the initial energy less the work done by moving the charge the distance. We find this work by multiplying the electric field by the charge we get the force, we can multiply this force by the given displacement to get the work and subtract this work from the initial KE

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19:33:37 ** 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 was right I believe.

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19:35:37 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 --> Given the power, charge and time we can use the expression P = q 'dv/'dt and solve for dv, which is the potential difference.

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19:36:07 ** 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 --> I was right again.

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19:39:05 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 --> We can find the flux through the formula 4 pi kq this flux will be spred over a spher of radius r. We can find this area by 4 pi r^2. We then divide the flux by the area to get the electric field.

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19:39:25 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 was correct.

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19:42:16 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 --> we can find the flux through the formula 4 pi kq and we assume that the dist. is much less than L we can assume a cylinder of radius r over distance L to get the surface area, we then divide this flux by the surface area to get the electric field.

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19:42:37 ** 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 was correct again.

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19:44:04 Why does the charge on the initially uncharged capacitor build more and more slowly if we continue cranking the generator at a constant rate?

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RESPONSE --> Because the capacitor starts to reach it's capasity and starts to discharge some of the energy, because the energy in the capacitor starts to be greater than the energy being created by the generatorl.

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19:45:55 ** The generator supplies a constant voltage; the charge flowing through the circuit builds on the capacitor, which builds voltage opposed to that of the generator. This results in less voltage across the rest of the circuit and therefore less flow of current. Less current requires less energy so the cranking is easier. UNIV PHY NOTE: Think of charge building on the capacitor, not current. The capacitor actually stores up the integral of the current with respect to time; since current is charge / time this integral has units of charge. **

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RESPONSE --> I think I was correct, but I am not sure, I do see how this is thought. That the energy in the capasitor builds and is opposed to the energy being supplied, creating less of a flow of current.

The key characteristic of a capacitor is that it stores charge separated into positive and negative regions. This requires energy and the discharge of the capacitor (which allows the separated charges to rejoin) releases energy.

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³ß†ôÎœ|w›`–Ææ±ý踃ÏèÑ©GºÊ assignment #023 éyéËÒíáöîO½–ûÅÁƒ›Ì³aÉG´îÉœï Physics II 11-28-2005

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20:28:46 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 --> We must first divide the drift velocity by the total length of the wire to find the length of the wire traveled in that second. We then multiply this number by the number of electrons in the wire to get the number in the second traveled. We then multiply the charge of the neutrons by this last number to get the current.

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20:29:04 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 mulitply 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 was correct

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20:34:43 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 --> There is more room for the current to spread and travel over the longer wire, thus produceing less current over the wire because the current carrying neutrons have more room to travel.

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20:36:08 ** 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 --> I knew it was something like this, but i did not know exactly why. I see that the potential gradient causes the acceleration and thus the greater potential gradient the greater the current.

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20:38:26 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 --> This will be similar to the last question in that the thicker wire will have a greater potential gradient because of the greater radius and larger area produces a greater current.

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20:40:48 ** 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 --> I missed this one. I should have remembered that the greater diamater over a given equal distance will have more electrons.

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20:43:02 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 --> We would begin by finding the work by multiplying the potential difference by the charge of an electron. We would then divide the work by the distance to get the force.

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20:43:19 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 --> I was correct

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20:45:57 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 --> we begin by finding the current through the formula I = V/R which will give us the current I. We can then divide the current by the volts(in equal units) to get the work in watts.

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20:46:22 ** 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 --> I was correct.

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20:49:15 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 either attracts or repels the current depending on the direction of current. If the magnetic field is changed the direction of the movement of the current is changed. In my experience, a sustained current experiences a sustained force.

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20:50:22 11-28-2005 20:50:22 ** 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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NOTES -------> I did not explain in these terms. I also did not know the I*L*B rule for force on a current.

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Ó|ê­¢w§«fÝéÃÒâŸõ±û˼Üý assignment #024 éyéËÒíáöîO½–ûÅÁƒ›Ì³aÉG´îÉœï Physics II 11-29-2005

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12:36:36 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 --> Because it is in a series the resistances are added together. We will then divide the voltage by this sum of resistances to get the current. We multiply this current by each resistance to get the voltage across the resistor.

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12:36:50 ** 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 --> I was correct.

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12:39:31 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 --> to find the current across each resistor we divide the voltage by the resistance of that specific resistor. We do this for both resistors. we would then multiply this current by the voltage to get the voltage across each resistor.

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12:40:08 ** 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 --> I missed the voltage. I see that because it is a parallel circut they will have the same voltage across the resistor.

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12:43:05 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 --> We must first find the potential difference which is the charge on the capacitor/capacitance. We then subtract this number from the total voltage being applied. We then divide this new voltage by the resistance to find the current.

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12:43:24 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 --> I was correct.

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12:46:39 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 --> Again we divide the charge by the capacitance to get the potential difference. The capacitor will be the source of power so this will be the voltage. We divide this voltage by the resistance to find the current. We take the 1% of the charge held by the capacitor and divide it by the current to get the time required to discharge this 1%.

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12:47:27 ** 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 --> I was correct, however I did not note that the current will be decreasing over this time, so the time will be a little quicker than what I solved.

The time will be a little longer, not quicker.

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áD‰¾ÒÁÔyyƒ¼x|á|e¥âz÷€~S¾ assignment #025 éyéËÒíáöîO½–ûÅÁƒ›Ì³aÉG´îÉœï Physics II 11-29-2005

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16:08:20 Query Gen Phy 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 --> we must solve for all the charges reacting on the given charge. Because of location all forces will be positive. we use the formula k*(Q1*Q2)/r^2. this gives the force in that direction. Two of the forces will already be in either the x or y direction. The only one we have to calculate x and y for is the diagonal and we use a 45 degree angle with sine and cosine to determine the x and y for this. We then add the x forces and y forces to get the total x and total y. we use these in the formula Magnatude = (x^2+y^2)^.5 and arctan (y/x) for direction. The magnatude is .624 for this question the direction is 45 degrees, because both the x and y are positive which puts it in the 1st quadrant.

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16:11:06 ** 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 --> I was taking the repulsive force leaving the point being repeled, not from the origin. This caused me to miss this problem.

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16:11:15 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 --> ok

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16:11:19 ** 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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16:11:23 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 --> ok

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16:11:26 ** 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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16:11:31 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 --> ok

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16:11:34 ** 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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ߊwŒyN›ëêÒº³›ª¾¡±§ÛèÞ¤eòßÑÒŽ assignment #026 éyéËÒíáöîO½–ûÅÁƒ›Ì³aÉG´îÉœï Physics II 11-29-2005

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17:10:39 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 --> We could use the formula E = k Q/r^2 since the point is directly in the middle of the charges their magnitude will be the same, only with different signs. 745 = (9.0*10^9(c))/.08^2 C= 5.3*10^-10

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17:12:15 ** The charges are each 8 cm from the field point. 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) = 372.5 N/C. Thus E = 372.5 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 = 372.5 N/C * (.08 m)^2 / (9 * 10^9 N m^2 / C^2) = 372.5 N/C * .0064 m^2 / (9 * 10^9 N m^2 / C^2) = 2.6 * 10^-10 C, approx. **

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RESPONSE --> I forgot to divide the in half before calculating the magnitude.

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17:14:58 If the charges are represented by Q and -Q, what is the electric field at the midpoint?

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RESPONSE --> E=k(Q1+Q2)/r^2

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17:16:18 ** 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 --> I was solving for a general solution, not the problem we just did. I see where the 2 came from because Q1 and Q2 are equal and thus 2Q.

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17:17:52 Query electrostatics. Explain why a charge enclosed in an aluminum cylinder is not affected by a charge outside the aluminum cylinder, when the charges would clearly affect one another in the absence of the cylinder.

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RESPONSE --> The aluminum has electrons that react with the inside charge and the outside charge seperately, thus there is no reaction between the charge inside the aluminum foil and the charge outside the foil.

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17:18:44 STUDENT RESPONSE AND INSTRUCTOR COMMENT: The electrons in a conductor will freely move and redidtribute them selves so that the field can not penetrate the cylinder. INSTRUCTOR COMMENT: Good explanation. See the next paragraph for a more detailed explanation: the aluminum cylinder is full of free charges that can migrate freely anywhere on the cylinder. The outside charge attracts opposite charges, which build to greatest density on regions of the aluminum cylinder nearest it. This process must continue until there is no electric field in the aluminum, because if there is an electric field in the aluminum then charges will move in response to it. Movement of charges must continue until the field is eliminated. This effectively 'shields' all point inside the cylinder from the effect of external charges.

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RESPONSE --> I understand how this works.

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17:19:20 Explain why bringing a charged plastic rod near the far end of an aluminum rod will have a direct affect on a charge close to the near end of the rod, whereas the same charge and plastic rod will exhibit no measurable interaction in the absence of the aluminum rod.

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RESPONSE --> I am not real sure why this happens, I was unable to get this to work in my lab.

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17:21:01 ** This question concerns the situation where a charged object is located near one end of the metal rod and the charged plastic rod is brought near the other end of the metal rod. Nothing ever touches anything else so there is no migration of charge from one object to the other. When the charged plastic rod is brought near there is however a redistribution of charge on the conducting rod and the force on the charged object will change. The redistribution of charges causes the end of the pipe near the rod to take on a charge opposite to that of the rod; this occurs by displacement of charge from the other end of the rod, which therefore ends up with the same type of charge as the plastic rod. Thus the charge near that end will experience a force in the same direction as if the rod was near. If the metal rod wasn't there, the distance of the plastic rod would cause the effect on that charge to be minimal. **

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RESPONSE --> I see how this works, it creates somewhat of a magnet where the oppisit charge is drawn and the same charge is repelled which attracts the other charged object at the far end.

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17:21:05 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 -->

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17:21:06 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 -->

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17:21:09 ** Flux is not a vector quantity so your flux values will not be multiples of the i, j and k vectors. The normal vectors 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 -->

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