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
(10 points, suggested time 20 minutes)
Particles A and B each have positive charge +Q and are held fixed at two vertices of an equilateral triangle of side length d, as shown. Point P is located equidistant from each vertex of the triangle.
Students Y and Z discuss the electric field and the electric potential at Point P after a third charged particle is placed at the bottom-right vertex. The students make the following statements.
Student Y: “If a particle with positive charge +2Q is placed at the bottom-right vertex, the magnitude of the electric field will be zero at Point P.”
Student Z: “To make the value of the electric potential zero at Point P, a particle with negative charge −Q should be placed at the bottom-right vertex.”
(a) In a coherent, paragraph-length response, evaluate the accuracy of each student’s statement. If any aspect of either student’s statement is inaccurate, explain how to correct the student’s statement. Support your evaluations using appropriate physics principles.
(b) Particles A and B are once again held in place at two vertices of the equilateral triangle. The students want to represent the electric potential energy of a system of particles when a third charged particle is brought from very far away to the bottom-right vertex. Scenarios 1 and 2 are considered.
i. In Scenario 1, a third particle with positive charge +Q is moved from very far away to the bottom-right vertex and then held in place. A bar is shown on the following chart that represents the electric potential energy \(U_{i1}\) of the system consisting of all three particles when the third particle with positive charge is very far away from the other particles. In the grid provided, complete the bar chart.
• Draw a bar to represent the work \(W_{1}\) required to move the third particle with positive charge from very far away to the bottom-right vertex.
• Draw another bar to represent the electric potential energy \(U_{f1}\) of the system consisting of all three particles when the third particle with positive charge is held in place at the bottom-right vertex.
The height of each bar should be proportional to the energy represented. If the quantity is zero, write a “0” in that column.
ii. In Scenario 2, a particle with negative charge −Q is moved from very far away to the bottom-right vertex and then held in place. A bar is shown on the following chart that represents the electric potential energy \(U_{i2}\) of the system consisting of all three particles when the particle with negative charge is very far away from the other particles. In the grid provided, complete the bar chart.
• Draw a bar to represent the work \(@_{2}\) required to move the particle with negative charge from very far away to the bottom-right vertex.
• Draw another bar to represent the electric potential energy \(U_{f2}\) of the system consisting of all three particles when the particle with negative charge is held in place at the bottom-right vertex.
The height of each bar should be proportional to the energy represented. If the quantity is zero, write a “0” in that column.
▶️Answer/Explanation
1(a) Example Response
Student Y is incorrect. Before the third particle is placed at the bottom-right vertex, the electric field from particles A and B at Point P is down and to the right. The electric field from a positively charged particle placed at the bottom-right vertex is up and to the left. The third particle needs to have charge +Q , rather than +2Q , in order to have the correct magnitude to make the resultant field zero at Point P .
Student Z is incorrect. Before the third particle is placed at the bottom-right vertex, the value of the electric potential at Point P is positive. Because Point P is equidistant from all three particles, the electric potential at Point P is proportional to the total charge of the system. If the total charge of the system is zero, the electric potential at Point P will be zero. This requires the third particle to have charge −2Q .
OR
Student Y is incorrect that a particle with charge +2Q placed at the bottom-right vertex will result in no electric field at Point P . The horizontal component of the electric field from Particle A is less than the horizontal component of the electric field from the particle with charge +2Q . The sum of the vertical components of the fields from particles A and B is less than the vertical component of the field from the particle with charge +2Q . Therefore, the resulting electric field at Point P is nonzero and points in a direction between particles A and B.
Student Z is incorrect. Before the third particle is placed at the bottom-right vertex, the value of the electric potential at Point P is positive. Electric potential is a scalar quantity, so if the third particle has charge −2Q rather than −Q , the electric potential at Point P will be zero.
(b)(i) Example Response
(b)(ii) Example Response
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.)