Question 1
1. Students perform an experiment to determine the value of vacuum permittivity \(\varepsilon _{o}\) . Sphere 1 is nonconducting with charge +q and is attached to an insulating rod. Sphere 2 is nonconducting with charge +Q and has mass M . Sphere 2 is hung from a string of negligible mass and length L. Sphere 1 is brought near, without touching, Sphere 2, as shown. Equilibrium is established when the centers of the two spheres have the same vertical position, are a horizontal distance d apart, and the string is at an angle q from the vertical.
(a) On the following dot that represents Sphere 2 at the position shown in the previous figure, draw and label the forces (not components) that act on Sphere 2. Each force must be represented by a distinct arrow starting on, and pointing away from, the dot.
(b) Derive the relationship between the distance d and the angle q to show that \(d=\sqrt{\frac{Qq}{4\pi \varepsilon _{o}Mg \ tan\Theta }}\) .
(c) These values are collected in one trial: \(Q=q=6.0\times 10^{-8}C,\Theta =12^{\circ}and\ d=0.057m\) . Calculate the expected force of tension exerted on Sphere 2 by the string
(d) The students vary d and measure q after equilibrium is reached. The students use the collected data to plot the following graph of \(d^{2}vs. \frac{1}{tan\Theta }\).
i. Draw the best-fit line for the data.
ii. Using the best-fit line, calculate an experimental value for the vacuum permittivity \(\varepsilon _{o}\) when M = 0.0050 kg and Q = q = 6.0 ¥ 10−8 C.
(e) The students modify the experiment by replacing Sphere 1 with a conducting Sphere 3 that has the same size and charge +q. The experiment is repeated.
i. The circle in the following figure represents Sphere 3 when spheres 2 and 3 are at equilibrium. On the circle, draw a single “+” sign to represent the location of highest concentration of the excess positive charges.
ii. Briefly explain your reasoning for the sketch drawn in part (e)(i).
iii. In the original experiment, when the centers of the two spheres are a horizontal distance \(d _{1}\) apart, the string makes an angle \(\Theta _{1}\) from the vertical. In the modified experiment, when the centers of the two spheres are a horizontal distance \(d _{1}\) apart, the string makes an angle \(\Theta _{2}\) from the vertical. Is \(\Theta _{2}\) greater than, less than, or equal to \(\Theta _{1}\) ?
_____ \(\Theta _{2}\) > \(\Theta _{1}\) _____ \(\Theta _{2}\) < \(\Theta _{1}\) _____ \(\Theta _{2}\)= \(\Theta _{1}\)
Briefly justify your answer.
▶️Answer/Explanation
1 (a) Example Response
(b) Example Response
(c) Example Response
(d) (i) Example Response
(d) (ii) Example Response
(e) (i) Example Response
(e) (ii) Example Response
The negative charges on Sphere 3 move to the right due to the attractive forces from the positive charges on Sphere 2 , leaving a net positive charge on the left side of Sphere 3 .
(e) (iii) Example Response
Excess charges on Sphere 3 are now free to move, so excess like charges will be concentrated on the far ends of Sphere 3 when the spheres are in static equilibrium. The excess like charges, located on opposite sides of Sphere 3 , repel with less force than if the excess charges were located at the centers of Sphere 3 . Thus, the downward force due to gravity on Sphere 2 causes the center of Sphere 2 to hang closer to the center of Sphere 3 .
Question 2
2. Two horizontal, parallel, conducting rails are separated by distance L = 0.40 m. A resistor of resistance R = 0.30 Ω connects the rails. A horizontal ideal spring is located between the rails. The right end of the spring is free to move and the left end is fixed in place. A conducting bar of mass m = 0.23 kg is placed on the rails and is in contact with the spring, which is initially compressed. Frictional forces and the resistance of the bar and rails are negligible.
• At time t = 0, the bar is released from rest and is pushed to the right by the spring.
• At time \(t_{1} \) , the bar loses contact with the spring and slides to the right.
• At time \(t_{2} \), the bar enters and travels through a uniform magnetic field of magnitude B = 0.50 T that is directed into the page, as shown.
• At time \(t_{1} \), the bar enters a region where the magnitude of the uniform magnetic field is still B = 0.50 T but is directed out of the page.
• At time \(t_{1} \), the bar enters a region with no magnetic field.
Consider time \(t_{B} \) such that \(t_{2} \) < \(t_{B} \) < \(t_{3} \).
(a) On the following diagram of the bar, draw an arrow indicating the direction of the net force \(F_{net} \) exerted on the bar at time \(t_{B} \). If the net force is zero, write \(F_{net} \) = 0.
(b) At time \(t_{B} \), the speed of the bar is v = 2.5 m/s.
i. Calculate the magnitude of the current in the bar at time \(t_{B} \).
ii. Calculate the magnitude of the net force \(F_{net} \) exerted on the bar at time \(t_{B} \).
(c) On the following axes, sketch a graph of the speed v of the bar as a function of time t between t = 0 and \(t_{4} \).
(d) The scenario is repeated but an additional resistor of resistance R = 0.30 Ω is connected, as shown.
i. Determine the total resistance \(R_{total}\) of the closed circuit for the new scenario.
ii. In the original scenario, the magnitude of the acceleration of the bar immediately after the bar enters the first uniform magnetic field is \(a_{original}\). In the new scenario, the magnitude of the acceleration of the bar immediately after the bar enters the first uniform magnetic field is \(a_{new}\). Is \(a_{new}\) greater than, less than, or equal to \(a_{original}\) ? Justify your answer.
(e) Describe a modification to m, B, or L that will result in a smaller induced potential difference across the original resistor immediately after the bar enters the first uniform magnetic field. Justify your answer.
▶️Answer/Explanation
2 (a) Example Response
(b) (i) Example Response
(b) (ii) Example Response
(c) Example Response
(d) (i) Example Response
(d) (ii) Example Response
Since there is less resistance in the new circuit, there will be more current in the new circuit, so a larger force on the bar. Thus, since the force on the bar is larger, the new acceleration is greater than the original acceleration.
(e) Example Response
The potential difference due to the induced emf across the original resistor is described by the equation ε = −BLv . Induced potential difference ε is proportional to B . Therefore, if the magnitude of the magnetic field is smaller than B = 0.5 T in the new scenario compared to the original scenario, ε would be smaller.
OR
The potential difference due to the induced emf across the original resistor is described by the equation ε = −BLv .The induced potential difference ε is proportional to L , which represents the distance the conducting rails are separated. Therefore, if L is smaller than L = 0.4 m , ε would be smaller.
OR
The potential difference due to the induced emf across the original resistor is described by the equation ε = −BLv . If the mass of the bar is greater, the velocity entering the magnetic field is less. The induced potential difference ε is proportional to v . Therefore, a smaller v due to a greater mass will induce a smaller ε .
Question 3
3. The circuit shown consists of a battery of emf Ɛ, resistors 1 and 2 each with resistance R, capacitors 1 and 2 with capacitances C and 2C, respectively, and a switch. The switch is initially open and both capacitors are uncharged.
At time t = 0, the switch is closed to Position A.
(a) Write, but do NOT solve, a differential equation that can be used to determine the charge Q on the positive plate of Capacitor 1 as a function of time t after the switch is closed to Position A. Express your answer in terms of Ɛ, R, C, Q, t, and fundamental constants, as appropriate.
(b) On the axes shown, sketch graphs of the surface charge density \(\sigma \)on the positive plate of Capacitor 1 and the total power P dissipated by the resistors as functions of time t from time t = 0 until steady-state conditions are nearly reached.
A long time after the switch is closed to Position A, the charge on the positive plate of Capacitor 1 is \(Q_{o}\) and Capacitor 2 is uncharged.
(c) At time \(t_{1}\), the switch is closed to Position B.
- Immediately after time \(t_{1}\) , is the direction of the current in the switch directed toward the left, directed toward the right, or is there no current? Briefly justify your answer.
- Determine an expression for the total charge on the positive plate of Capacitor 2 a long time after \(\(t_{1}\). Express your answer in terms of \(Q_{o}\) and fundamental constants, as appropriate.
- Derive an expression for the total energy \(E_{R}\) dissipated by resistors 1 and 2 from immediately after time \(t_{1}\) until new steady-state conditions have been reached. Express your answer in terms of C, \(Q_{O}, and fundamental constants, as appropriate. With the switch still closed to Position B, the parallel plates of Capacitor 2 are moved so that the separation distance increases by a factor of 2.
(d) Determine the ratio \(\frac{U_{2}}{U_{1}}\) of the energy \(U_{2}\) stored in Capacitor 2 to the energy \(U_{2}\) stored in Capacitor 1 a long time after the plates of Capacitor 2 have been moved. Briefly justify your answer. With the capacitors still charged as in part (d), the switch is now closed to Position A.
(e) Express your answers to part (e)(i) and part (e)(ii) in terms of R, C, \(Q_{O}\), and fundamental constants, as appropriate.
i. Derive an expression for the current \(I_{O}\) from the battery immediately after the switch is closed to Position A.
ii. Determine the current \(I_{\infty }\) from the battery a long time after the switch is closed to Position .
▶️Answer/Explanation
3 (a) Example Response
(b) Example Response
(c) (i) Example Response
The current is directed towards the right because the top plate of Capacitor 1 is positively charged, meaning conventional current will flow clockwise.
OR
Toward the right. Current flows from high to low potential so it will flow from the top plate up and right through the switch.
(c) (ii) Example Response
The potential difference across Capacitor 1 is equal to the potential difference across Capacitor 2 . Capacitor 2 has twice the capacitance of Capacitor 1. Therefore, Capacitor 2 stores twice the charge that is stored on Capacitor 1. Due to conservation of charge, Capacitor 2 stores an amount of charge equal to \(\frac{2}{3}Q_{o}\).
(c) (iii) Example Response
(d) Example Response
After steady state conditions are reached, both capacitors have the same potential difference. The new capacitance of Capacitor 2 is equal to the capacitance of Capacitor 1 because the capacitance of a capacitor is inversely related to the distance between the plates of a capacitor. Therefore, since \( U_{c} = \frac{1}{2}C(\Delta V)^{2}, \frac{U_{2}}{U_{1}} = 1\)
(e) (i) Example Response
(e) (ii) For indicating that the current is zero