AP® Physics C: E&M - Conservation of Electric Energy - FRQs
Question
1. A nonconducting rod of uniform positive linear charge density is near a sphere with charge −2.0 nC. The rod and sphere are held at rest on the x-axis, as shown in Figure 1. Equipotential lines and positions A, B, C, D, and E are labeled. Adjacent tick marks on the x-axis and the y-axis are 0.40 m apart.
(a) Calculate the absolute value of the electric flux through the Gaussian surface whose cross section is the −20.0 V equipotential line.
A positive test charge (not shown) is placed and held at rest at Position C. An external force is applied to the test charge to move the test charge to different positions in the order of C→ →E D→A. The test charge is momentarily held at rest at each position.
(b) The bar shown in Figure 2 represents the absolute value of the work \( W_{CE}\) done by the external force on the test charge to move the test charge from Position C to Position E.
i. Complete the following tasks on Figure 2.
• Draw a bar to represent the relative absolute value of the work \( W_{CD}\) done by the external force on the test charge to move the test charge from Position E to Position D.
• Draw a bar to represent the relative absolute value of the work \( W_{DA}\) done by the external force on the test charge to move the test charge from Position D to Position A.
• The height of each bar should be proportional to the value of \( W_{CE}\). If \( W_{ED}\) = and/or \( W_{DA}\) = 0, write a “0” in the corresponding columns, as appropriate.
ii. Calculate the approximate magnitude of the x-component of the electric field at Position B.
The positive test charge is placed at Position D. The test charge is then released from rest.
(c) Indicate the direction (not components) of the net electric force exerted on the test charge immediately after the test charge is released from rest.
_____ +x _____ +y _____ Directly away from the sphere
_____ −x _____ −y _____ Directly toward the sphere
Without using equations, justify your answer using physics principles.
The sphere and the test charge are removed. The rod has length 4L and uniform positive linear charge density \(+\lambda \). The rod is held at rest on the x-axis in the orientation shown in Figure 3. Position P (not shown) is located on the x-axis a distance \(x_{p}\) from the origin, where \(x_{p}\) > 4L.
(d) The electric potential \(V_{p}\) at \(x_{p}\) is \(V_{p}\) = \(K\lambda \) In\(\left ( \frac{x_{p}}{x_{p} – 4L} \right )\).
i. Using integral calculus, derive the expression for \(V_{p}\) provided.
ii. On Figure 4, sketch a graph of the x-component \(E_{x}\) of the electric field from the rod as a function of x in the region 4L < x < 12L.
▶️Answer/Explanation
1(a) Example Solution
1(b) (i) Example Response
1(b) (ii) Example Solution
1(c) Example Response
+y. The direction of an electric field vector is perpendicular to an equipotential line. Because the test charge has a positive charge, the test charge would move from a position of higher electric potential to a position of lower electric potential when an electric force is exerted on the test charge. Therefore, at Position D, the electric force is upward because that is the direction that is perpendicular to the equipotential line and in the direction of decreasing electric potential.
1(d) (i) Example Solutions
OR
1(d) (ii) Example Response
Question
1. A nonconducting rod of uniform positive linear charge density is near a sphere with charge −2.0 nC. The rod and sphere are held at rest on the x-axis, as shown in Figure 1. Equipotential lines and positions A, B, C, D, and E are labeled. Adjacent tick marks on the x-axis and the y-axis are 0.40 m apart.
(a) Calculate the absolute value of the electric flux through the Gaussian surface whose cross section is the −20.0 V equipotential line.
A positive test charge (not shown) is placed and held at rest at Position C. An external force is applied to the test charge to move the test charge to different positions in the order of C→ →E D→A. The test charge is momentarily held at rest at each position.
(b) The bar shown in Figure 2 represents the absolute value of the work \( W_{CE}\) done by the external force on the test charge to move the test charge from Position C to Position E.
i. Complete the following tasks on Figure 2.
• Draw a bar to represent the relative absolute value of the work \( W_{CD}\) done by the external force on the test charge to move the test charge from Position E to Position D.
• Draw a bar to represent the relative absolute value of the work \( W_{DA}\) done by the external force on the test charge to move the test charge from Position D to Position A.
• The height of each bar should be proportional to the value of \( W_{CE}\). If \( W_{ED}\) = and/or \( W_{DA}\) = 0, write a “0” in the corresponding columns, as appropriate.
ii. Calculate the approximate magnitude of the x-component of the electric field at Position B.
The positive test charge is placed at Position D. The test charge is then released from rest.
(c) Indicate the direction (not components) of the net electric force exerted on the test charge immediately after the test charge is released from rest.
_____ +x _____ +y _____ Directly away from the sphere
_____ −x _____ −y _____ Directly toward the sphere
Without using equations, justify your answer using physics principles.
The sphere and the test charge are removed. The rod has length 4L and uniform positive linear charge density \(+\lambda \). The rod is held at rest on the x-axis in the orientation shown in Figure 3. Position P (not shown) is located on the x-axis a distance \(x_{p}\) from the origin, where \(x_{p}\) > 4L.
(d) The electric potential \(V_{p}\) at \(x_{p}\) is \(V_{p}\) = \(K\lambda \) In\(\left ( \frac{x_{p}}{x_{p} – 4L} \right )\).
i. Using integral calculus, derive the expression for \(V_{p}\) provided.
ii. On Figure 4, sketch a graph of the x-component \(E_{x}\) of the electric field from the rod as a function of x in the region 4L < x < 12L.
▶️Answer/Explanation
1(a) Example Solution
1(b) (i) Example Response
1(b) (ii) Example Solution
1(c) Example Response
+y. The direction of an electric field vector is perpendicular to an equipotential line. Because the test charge has a positive charge, the test charge would move from a position of higher electric potential to a position of lower electric potential when an electric force is exerted on the test charge. Therefore, at Position D, the electric force is upward because that is the direction that is perpendicular to the equipotential line and in the direction of decreasing electric potential.
1(d) (i) Example Solutions
OR
1(d) (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
Object I, shown above, has a charge of + 3 x 10–6 coulomb and a mass of 0.0025 kilogram.
a. What is the electric potential at point P, 0.30 meter from object I?
Object II, of the same mass as object I, but having a charge of + 1 x 10–6 coulomb, is brought from infinity to point P, as shown above.
b. How much work must be done to bring the object II from infinity to point P?
c. What is the magnitude of the electric force between the two objects when they are 0.30 meter apart?
d. What are the magnitude and direction of the electric field at the point midway between the two objects?
The two objects are then released simultaneously and move apart due to the electric force between them. No other forces act on the objects.
e. What is the speed of object I when the objects are very far apart?
Answer/Explanation
Ans:
a. V = kQ/r = 9 × 104 V
b. W = qΔV (where V at infinity is zero) = 0.09 J
c. F = kqQ/r2 = 0.3 N
d. Between the two charges, the fields from each charge point in opposite directions, making the resultant field the difference between the magnitudes of the individual fields.
E = kQ/r2 gives EI = 1.2 × 106 N/C to the right and EII = 0.4 × 106 N/C to the left
The resultant field is therefore E = EI – EII = 8 × 105 N/C to the right
e. From conservation of momentum mIvI = mIIvII and since the masses are equal we have vI = vII. Conservation of energy gives U = K = 2(½ mv2) = 0.09 J giving v = 6 m