Electric Fields AP Physics 2 FRQ – Exam Style Questions etc.
Electric Fields AP Physics 2 FRQ
Unit 10: Electric Force, Field, and Potential
Weightage : 15–18%
Exam Style Practice Questions, Electric Fields AP Physics 2 FRQ
Question
The figure above shows two metal spheres that are far apart compared to their size and that are held in place. The spheres are connected by wires to either side of switch S. Initially, the switch is open. Sphere 1 has mass \(m_{1}\), radius \(r_{1}\), and a net positive charge \(+Q_{0}\). Sphere 2 has mass \(m_{2}\) and radius \(r_{2}\) < \(r_{1}\) and is initially uncharged. The switch is then closed. Afterward, sphere 1 has a charge \(Q_{1}\) and is at potential \(V_{1}\), and the electric field strength just outside its surface is \(E_{1}\). The corresponding values for sphere 2 are \(Q_{2}\), \(V_{2}\), and \(E_{2}\). Neglect air resistance and gravitational interactions.
(a) i. Indicate whether \(V_{1}\) is larger than, smaller than, or equal to \(V_{2}\). Briefly explain your reasoning using appropriate physics principles and/or mathematical models.
ii. Indicate whether \(Q_{1}\) is larger than, smaller than, or equal to \(Q_{2}\). Briefly explain your reasoning using appropriate physics principles and/or mathematical models.
iii. Indicate whether \(E_{1}\) is larger than, smaller than, or equal to \(E_{2}\). Show how you arrived at your answer using appropriate physics principles and/or mathematical models.
(b) The distance between the centers of sphere 1 and sphere 2 is D. The switch is now opened, the wires are disconnected from the spheres, and the spheres are released, all without changing the charges on the spheres. Write but do NOT solve equations that could be used to determine the velocities \(v_{1}\) and \(v_{2}\) of the spheres a long time after they are released, in terms of \(m_{1}\), \(m_{2}\), \(Q_{1}\), \(Q_{2}\), D, and physical constants, as appropriate.
(c) The spheres are now returned to their original locations. Sphere 1 once again has initial net charge \(+Q_{0}\), and sphere 2 is initially uncharged. The switch is again closed and then reopened. Sphere 3, an uncharged metal sphere of radius \(r_{3}\)> \(r_{1}\) > \(r_{2}\) on an insulating handle, is now brought into contact with sphere 2. Sphere 3 is then moved away.
i. Indicate the sign of the final charge on each sphere.
ii. Rank the absolute value of the final charge on each of the three spheres. Explain how you arrived at this answer.
Answer/Explanation
Ans:
(a) i) Indicating that \(V_{1}\) = \(V_{2}\) and giving a correct explanation
Examples:
- When the switch is closed, charge flows to eliminate the initial potential difference between the spheres.
- There is no source of potential besides the spheres, and they are now effectively one conductor.
ii) Using the fact that \(V_{1}\) = \(V_{2}\) (or whatever relationship was indicated in part i)
Equating the expressions for potential, \(KQ_{1}/r_{1} = kQ_{2}/r_{2}\), and indicating that since \(r_{1}\) > \(r_{2}\), \(Q_{1}\) > \(Q_{2}\) (or logic consistent with answer to part i)
iii) Combining \(E=\frac{kQ}{r^{2}}\) and \(V= kQ/r \) and using \(V_{1}\) = \(V_{2}\) and \(r_{1}\) > \(r_{2}\) to show that\(E_{1}\) < \(E_{2}\) (or logic consistent with answer to part i)
(b) Any indication that when the spheres are released, the electric potential energy is converted into kinetic energy of the spheres
Any indication that momentum of the spheres must be conserved
Both correct equations \(\frac{k Q_{1}Q_{2}}{ D}= \frac{1}{2} m_{1}(v_{1})^{2} + \frac{1}{2} m_{2}(v_{2})^{2}\)
\(0= m_{1}\underset {v_{2}}\rightarrow\) or \(m_{1}v_{1}=m_{2}v_{2}\)
(c) i) Spheres 1 and 3 have a positive charge, which is always true. If spheres 1 and 2 are well separated, all three spheres will ultimately have a positive charge. If the student assumes spheres 1 and 2 are close enough that charging by induction can occur, sphere 2 might end up with positive, zero, or negative charge depending on the size and placement of sphere 3 relative to sphere 2.
ii) Charge on sphere 2 is split between spheres 2 and 3 when they come into contact, which implies that \(Q_{1}\) > \(Q_{3}\) since it was shown that \(Q_{1}\) > \(Q_{2}\). Sphere 3 has more charge than sphere 2 because it is larger, which justifies the ranking of \(Q_{1}\) > \(Q_{3}\) > \(Q_{2}\) (or some correct reasoning relating the charges on spheres 1 and 3 based on the answer to part (a) and part (c)i
Question: (10 points – suggested time 20 minutes)
The apparatus shown in the figure above consists of two oppositely charged parallel conducting plates, each with area A = 0.25 m2 , separated by a distance d = 0.010 m . Each plate has a hole at its center through which electrons can pass. High velocity electrons produced by an electron source enter the top plate with speed v0 = 5.40 ×106 m/s, take 1.49 ns to travel between the plates, and leave the bottom plate with speed vf = 8.02 ×106 m/s .
(a) Which of the plates, top or bottom, is negatively charged? Support your answer with a reference to the direction of the electric field between the plates.
(b) Calculate the magnitude of the electric field between the plates.
(c) Calculate the magnitude of the charge on each plate.
(d) The electrons leave the bottom plate and enter the region inside the dashed box shown below, which contains a uniform magnetic field of magnitude B that is perpendicular to the page. The electrons then leave the magnetic field at point X.
i. On the figure above, sketch the path of the electrons from the bottom plate to point X. Explain why the path has the shape that you sketched.
ii. Indicate whether the magnetic field is directed into the page or out of the page. Briefly explain your choice.
Answer/Explanation
Ans:
The top plate is negatively charged. The electrons increase their speed, s their acceleration is downward. This means by F = ma, the electric force points downward. Since these are electrons, the force is in the opposite direction of the field lines, which thus point up. Field lines point towards the negative plate, which is the top one
W = ΔKE
Vg = ΔKE
Ed = \(\frac{\Delta KE}{g}\)
\(E = \frac{1}{2}\left ( \frac{1}{q} \Delta KE\right )\)
\(E = \left ( \frac{1}{0.010m} \right )\left ( \frac{1}{1.6\times 10^{-19}C} \right )\left ( \frac{1}{2} \right )\left ( 9.11\times 10^{-31}Kg \right )\left ( (8.02\times 10^{6}m/s)^{2}-(5.40\times 10^{6}m/s)^{2} \right )=1.0\times 10^{4}N/C\)
(c)
\(C = \frac{\varepsilon .A}{d}\)
\(\frac{Q}{V} = \frac{\varepsilon .A}{d}\)
\(Q = \frac{\varepsilon 0EdA}{d} = \varepsilon 0EA\)
= (8.85 × 10-12 C2/Nm2) (1.0×104N/C) (0.25m2) = 2.2×10-8C
(d)
i.
The magnetic force is always perpendicular to the direction of velocity. Thus, the force acts as a centripetal force, so the electrons undergo circular motion.
ii.
The magnetic field is directed out of the page. By the right hand rule (force to the right, velocity down,) the magnetic field is into the page. Since the electrons are negatively charged, we switch the direction so the field is out of the page (the right hand rule applies to positively charged particles.)