AP Physics C E and M Unit 8 Electric Fields

[Maximum number: 15]

Students perform an experiment to study the force between two charged objects using the apparatus shown above, which contains two identical conducting spheres. The upper sphere is attached to an insulating string, which can be used to move the sphere downward. The lower sphere sits on an insulating rod, which is on an electronic balance. The electronic balance is zeroed before the lower sphere and insulating rod are in place.
For the first trial, a charge of Q is placed on each sphere and then the upper sphere is slowly moved downward. The students measure the distance d between the centers of the spheres and the magnitude F of the force that appears on the electronic balance. The recorded data are shown on the graph of F as a function of $\frac{1}{d^{2}}$ shown below.

Figure for Question 8 Electric Charges, Fields, and Gauss’s Law question 1 — AP Physics C: Electricity & Magnetism

(a)

Draw a line that represents the best fit to the points shown.

Use the graph to calculate the charge Q.

iii. On the graph on the previous page, draw a circle around the data point that was taken when the distance between the centers of the spheres was the least.
iv. Determine the distance between the centers of the spheres for the data point indicated above.
v. What physical quantity does the vertical intercept represent?

Justify your answer.

Figure for Question (a) — AP Physics C: Electricity & Magnetism

The experiment is extended by collecting additional data points, which appear on the right side of the graph shown above. The new data points do not follow the linear pattern seen with the first points. The group of students tries to explain this discrepancy.

[ 8 ]

i. For drawing an appropriate best-fit line 1 point
Example response for part (a)(i)

Figure for Answer to Question (a) — AP Physics C: Electricity & Magnetism

For calculating the slope using two points from the best-fit line 1 point

\text { slope }=m=\frac{\Delta y}{\Delta x}=\frac{(12,000-10,500) \mu \mathrm{N}}{(72-40) \mathrm{m}^{-2}}=4.7 \times 10^{-5} \mathrm{~N} \cdot \mathrm{~m}^{2}

For correctly relating the slope to the charge 1 point

F=\frac{k Q^{2}}{r^{2}} \therefore \text { slope }=k Q^{2}

For correctly calculating the charge consistent with the slope from the best-fit line 1 point

Q=\sqrt{\frac{\text { slope }}{k}}=\sqrt{\frac{\left(4.7 \times 10^{-5} \mathrm{~N} \cdot \mathrm{~m}^{2}\right)}{\left(9 \times 10^{9} \mathrm{~N} \cdot \mathrm{~m}^{2} / \mathrm{C}^{2}\right)}}=7.2 \times 10^{-8} \mathrm{C}

iii. For drawing a circle around the appropriate data point 1 point

For a correct answer 1 point
From the graph, $\frac{1}{d^{2}}=82.5 \mathrm{~m}^{-2}$ $d=0.11 \mathrm{~m}$

For indicating that the y-intercept should be the weight of the conducting sphere 1 point and insulating rod
For a correct justification 1 point
Example responses for part (a)(v).
When the spheres are infinitely far apart, there is no electric force between them.
The only force recorded by the balance must be the weight of the lower sphere and insulating rod.
OR
The scale is zeroed before the insulating rod, and lower sphere are placed on it, so the nonzero y-intercept is their weight.

Total for part (a)

8 points

(b)

One student suspects that charge is slowly leaking off the top sphere. Could this explain the discrepancy? Yes No

Justify your answer.

[ 1 ]

(c)

A second student suspects that the excess charges have rearranged themselves, polarizing the spheres.

On the circles representing the spheres below, use a single "+" sign on each sphere to represent the locations of highest concentration of the excess positive charges.

Explain how this rearrangement could be responsible for the discrepancy.

[ 3 ]

(d)

A third student suggests that the experiment be modified so that the top sphere is given a negative charge that is equal in magnitude to the positive charge given to the bottom sphere.

On the circles representing the spheres below, use a single "+" sign on the bottom sphere to represent the location of highest concentration of the excess positive charges. Use a single "-" sign on the top sphere to represent the location of the highest concentration of the excess negative charges.

For a separation distance equal to that of the data point indicated in part (a)(iii), would the magnitude of the force reading with spheres of opposite charges be greater than, less than, or equal to the magnitude of the force reading with spheres of the same charges?

Greater than Less than Equal to
Justify your answer.

Figure for Question (d) — AP Physics C: Electricity & Magnetism

[ 3 ]

[Maximum number: 9]

A very long nonconducting cylinder is surrounded by a thin concentric conducting cylindrical shell, as shown in the cutout view. A segment of length L of the inner cylinder has a net charge of +Q uniformly distributed throughout its volume. A segment of length L of the outer shell has a net charge of +4 Q. The radii of the inner cylinder and outer shell are R and 3 R, respectively, as shown in the cross-section view.

(a)

Determine the charge on the outer surface of the cylindrical shell within length L.

[ 1 ]

For correctly determining the charge on the outer surface of the shell 1 point

Example Response

\begin{aligned} & q_{\text {net }}=q_{\text {inner }}+q_{\text {outer }} \\ & q_{\text {outer }}=q_{\text {net }}-q_{\text {inner }} \\ & q_{\text {outer }}=+4 Q-(-Q) \\ & q_{\text {outer }}=+5 Q \end{aligned}

Scoring Note: A correct response may earn a point even if no work is shown.

Total for part (a)

1 point

(b)

Using Gauss's law, derive an expression for the electric field a distance r from the center of the inner cylinder for r<R. Express your answers in terms of Q, R, r, L, and physical constants, as appropriate.

[ 3 ]

For using Gauss's law by substituting for either the area or the enclosed charge

Example Response
$\frac{q_{\text {enc }}}{\varepsilon_{0}}=\oint E \cdot d A=E(2 \pi r L)$

1 point

For using a correct expression for $q_{\text {enc }}$ as a function of r

Example Response
$q_{\text {enc }}=\rho V=\left(\frac{Q}{\pi R^{2} L}\right)\left(\pi r^{2} L\right)=Q \frac{r^{2}}{R^{2}}$

1 point

For using the correct area

Example Response
$\begin{aligned} & E=q_{\text {enc }} \frac{1}{\varepsilon_{0}(2 \pi r L)}=\left(Q \frac{r^{2}}{R^{2}}\right) \frac{1}{\varepsilon_{0}(2 \pi r L)} & E=Q \frac{r}{2 \pi \varepsilon_{0} R^{2} L} \end{aligned}$

1 point

Total for part (b) 3 points

(c)

The magnitude of the electric field at r=R is $12 \mathrm{~N} / \mathrm{C}$ . Calculate the value of the electric field at r=2 R.

[ 2 ]

For recognizing that the E-field varies as the reciprocal of the separation distance from the 1 point center of the nonconducting cylinder
Example Response

E \propto \frac{1}{r}

For the correct value, including units, of electric field at 2 R1 point
Example Response

\begin{aligned} & E_{\text {new }}=\frac{E_{\text {old }}}{2}=\frac{12 \mathrm{~N} / \mathrm{C}}{2} \\ & E_{\text {new }}=6 \mathrm{~N} / \mathrm{C} \end{aligned}

Total for part (c) 2 points

(d)

i. On the following axes that include regions I, II, and III, sketch the graph of the electric field E as a function of the distance r from the axis of the inner cylinder.

[ 3 ]

[Maximum number: 5]

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.

[ 2 ]

For using a correct equation for electric flux 1 point
Example Response

\Phi_{E}=\frac{Q}{\varepsilon_{0}}

For the correct numerical answer 1 point
Scoring Note: This point can be earned if a negative sign is included in the final answer or if units are missing and/or the units are incorrect.
Example Response

\left|\Phi_{E}\right|=226 \frac{\mathrm{~N} \cdot \mathrm{~m}^{2}}{\mathrm{C}}

Example Solution

\begin{aligned} & \oint \vec{E} \cdot d \vec{A}=\frac{Q}{\varepsilon_{0}} \\ & \Phi_{E}=\frac{Q}{\varepsilon_{0}} \\ & \Phi_{E}=\frac{-2.0 \times 10^{-9} \mathrm{C}}{8.85 \times 10^{-12} \frac{\mathrm{C}^{2}}{\mathrm{~N} \cdot \mathrm{~m}^{2}}} \\ & \left|\Phi_{E}\right|=226 \frac{\mathrm{~N} \cdot \mathrm{~m}^{2}}{\mathrm{C}} \end{aligned}

Total for part (a)

2 points

(b)

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.

Figure 3

The sphere and the test charge are removed. The rod has length 4 L 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 $x_{\mathrm{P}}$ from the origin, where $x_{\mathrm{P}}>4 L$ .

[ 3 ]

[Maximum number: 4]

A nonconducting rod of uniform negative linear charge density is near a sphere with charge +1.0 nC. The rod and sphere are held at rest on the y-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 on 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 0.0 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 held momentarily at rest at each position.

[ 2 ]

For using a correct equation for electric flux 1 point
Example Response

\Phi_{E}=\frac{Q}{\varepsilon_{0}}

For the correct numerical answer 1 point
Scoring Note: Correct units are not required to earn this point.
Example Response

\Phi_{E}=113 \frac{\mathrm{~N} \cdot \mathrm{~m}^{2}}{\mathrm{C}}

Example Solution

\begin{aligned} & \oint \vec{E} \cdot d \vec{A}=\frac{Q}{\varepsilon_{0}} \\ & \Phi_{E}=\frac{Q}{\varepsilon_{0}} \\ & \Phi_{E}=\frac{1.0 \times 10^{-9} \mathrm{C}}{8.85 \times 10^{-12} \frac{\mathrm{C}^{2}}{\mathrm{~N} \cdot \mathrm{~m}^{2}}} \\ & \Phi_{E}=113 \frac{\mathrm{~N} \cdot \mathrm{~m}^{2}}{\mathrm{C}} \end{aligned}

Total for part (a)

2 points

(b)

The electric potential $V_{\mathrm{P}}$ at $y_{\mathrm{P}}$ is $V_{\mathrm{P}}=-k \lambda \ln \left(\frac{y_{\mathrm{P}}}{y_{\mathrm{P}}-2 L}\right)$ .

[ 2 ]

(i)

ii. On Figure 4, sketch a graph of the y-component $E_{y}$ of the electric field resulting from the rod as a function of y in the region 2 L<y<12 L.

Figure 4

Figure 1

[ 2 ]

AP Physics C E and M Unit 8: Electric Charges and Fields

Covered syllabus points

Covered syllabus points

8 Electric Charges, Fields, and Gauss’s Law question 1

Question (a)

Question (b)

Question (c)

Question (d)

8 Electric Charges, Fields, and Gauss’s Law question 2

Question (a)

Question (b)

Question (c)

Question (d)

8 Electric Charges, Fields, and Gauss’s Law question 3

Question (a)

Question (b)

8 Electric Charges, Fields, and Gauss’s Law question 4

Question (a)

Question (b)

Question (i)