AP Physics C E and M Unit 9 Electric Potential

[Maximum number: 1]

Questions 31-33

Figure for Question 9 Electric Potential question 1 — AP Physics C: Electricity & Magnetism

Two small spheres have charges +Q and -Q and are located at the bottom corners of an equilateral triangle, as shown in the figure above. The equilateral triangle has sides of length a, and point P is at the top corner of the triangle.

The potential energy stored in the charge configuration shown is

A

$\frac{Q^{2}}{4 \pi \varepsilon_{0} a}$

B

$\frac{Q}{4 \pi \varepsilon_{0} a^{2}}$

C

zero

D

$-\frac{Q}{4 \pi \varepsilon_{0} a^{2}}$

E

$-\frac{Q^{2}}{4 \pi \varepsilon_{0} a}$

[Maximum number: 11]

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)

The bar shown in Figure 2 represents the absolute value of the work $W_{\mathrm{CE}}$ done by the external force on the test charge to move the test charge from Position C to Position E.

[ 4 ]

(i)

Complete the following tasks on Figure 2.

- Draw a bar to represent the relative absolute value of the work $W_{\mathrm{ED}}$ 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 $W_{\mathrm{DA}}$ 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 $W_{\mathrm{CE}}$ . If $W_{\mathrm{ED}}=0$ and/or $W_{\mathrm{DA}}=0$ , write a "0" in the corresponding columns, as appropriate.

Figure 2

[ 2 ]

For indicating that $W_{\mathrm{ED}}=0$

1 point

For drawing a bar representing $W_{\mathrm{DA}}$ that has a height of approximately one and a half units

1 point

Example Response

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

(ii)

Calculate the approximate magnitude of the x-component of the electric field at Position B.

The positive test charge is placed at Position C. The test charge is then released from rest.

[ 2 ]

(ii) For using an equation that relates the electric field to potential difference 1 point
Scoring Note: This point can be earned if the response begins with a correct relationship between electric field and potential difference in which numerical values are already substituted.
Example Responses

E_{y}=-\frac{d V}{d y} \quad \mathbf{O R} \quad\left|E_{y}\right|=\left|-\frac{d V}{d y}\right| \quad \mathbf{O R} \quad\left|E_{y}\right|=\left|-\frac{\Delta V}{\Delta y}\right| \quad \mathbf{O R} \quad \Delta V=-\int \vec{E} \cdot d \vec{r}

For correct substitutions of values of electric potential and the distance between equipotential 1 point lines that can be used to calculate the approximate magnitude of the electric field at Position B
Example Response

\left|E_{y}\right|=\left|-\frac{-20.0 \mathrm{~V}-(-10.0 \mathrm{~V})}{1.4 \mathrm{~m}}\right|

Example Solution

\begin{aligned} & E_{y}=-\frac{d V}{d y} \\ & \left|E_{y}\right|=\left|-\frac{d V}{d y}\right| \\ & \left|E_{y}\right|=\left|-\frac{\Delta V}{\Delta y}\right| \\ & \left|E_{y}\right|=\left|-\frac{-20.0 \mathrm{~V}-(-10.0 \mathrm{~V})}{1.4 \mathrm{~m}}\right| \\ & \left|E_{y}\right|=7.1 \frac{\mathrm{~V}}{\mathrm{~m}} \end{aligned}

Total for part (b)

4 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 2 L and uniform negative linear charge density $-\lambda$ . The rod is held at rest on the y-axis in the orientation shown in Figure 3. Position P (not shown) is located on the y-axis a distance $y_{\mathrm{P}}$ from the origin, where $y_{\mathrm{P}}>2 L$ .

[ 3 ]

(c)

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

[ 4 ]

(i)

Using integral calculus, derive the expression for $V_{\mathrm{P}}$ provided.

[ 4 ]

(i) For using an appropriate equation for determining the electric potential from a line of uniform charge
Example Responses
$V=\frac{1}{4 \pi \varepsilon_{0}} \sum \frac{q_{i}}{r_{i}} \quad$ OR $\quad V=\frac{1}{4 \pi \varepsilon_{0}} \int \frac{d q}{r} \quad$ OR $\quad \Delta V=-\int \vec{E} \cdot d \vec{r}$
For a correct determination of r, the distance between Point P and a point on the line of uniform charge
Example Responses
$V_{\mathrm{P}}=k \sum \frac{Q}{y_{\mathrm{P}}-y} \quad$ OR $\quad V_{\mathrm{P}}=k \int\left(\frac{1}{y_{\mathrm{P}}-y}\right) d q$
For a correct integral with $\lambda d y$ substituted for d q
Example Response
$V_{\mathrm{P}}=-k \lambda \int\left(\frac{1}{y_{\mathrm{P}}-y}\right) d y$
For the correct limits of integration
Example Response
$V_{\mathrm{P}}=-k \lambda \int_{0}^{2 L}\left(\frac{1}{y_{\mathrm{P}}-y}\right) d y$

Example Solutions

\begin{aligned} & V=\frac{1}{4 \pi \varepsilon_{0}} \sum \frac{q_{i}}{r_{i}} \\ & V_{\mathrm{P}}=k \int\left(\frac{1}{y_{\mathrm{P}}-y}\right) d q \\ & d q=-\lambda d y \\ & V_{\mathrm{P}}=-k \lambda \int\left(\frac{1}{y_{\mathrm{P}}-y}\right) d y \\ & V_{\mathrm{P}}=-k \lambda \int_{0}^{2 L}\left(\frac{1}{y_{\mathrm{P}}-y}\right) d y \\ & V_{\mathrm{P}}=\left.k \lambda \ln \left(y_{\mathrm{P}}-y\right)\right|_{0} ^{2 L}=-\left.k \lambda \ln \left(y_{\mathrm{P}}-y\right)\right|_{2 L} ^{0} \\ & V_{\mathrm{P}}=-k \lambda \ln \left(\frac{y_{\mathrm{P}}}{y_{\mathrm{P}}-2 L}\right) \end{aligned}

OR

\begin{aligned} & \Delta V=-\int \vec{E} \cdot d \vec{r} \\ & E(y)=\int \frac{k}{r^{2}} d q \\ & E(y)=\int_{0}^{2 L} \frac{k \lambda}{\left(y-y^{\prime}\right)^{2}} d y^{\prime}=k \lambda\left(\frac{1}{y-2 L}-\frac{1}{y}\right) \\ & V_{\mathrm{P}}=-\int_{\infty}^{y_{\mathrm{P}}} E(y) d y=-k \lambda \int_{\infty}^{y_{\mathrm{P}}}\left(\frac{1}{y-2 L}-\frac{1}{y}\right) d y \\ & V_{\mathrm{P}}=k \lambda \ln \left(\frac{y_{\mathrm{P}}}{y_{\mathrm{P}}-2 L}\right) \end{aligned}

[Maximum number: 10]

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)

The bar shown in Figure 2 represents the absolute value of the work $W_{\mathrm{CE}}$ done by the external force on the test charge to move the test charge from Position C to Position E.

Complete the following tasks on Figure 2.

- Draw a bar to represent the relative absolute value of the work $W_{\mathrm{ED}}$ 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 $W_{\mathrm{DA}}$ 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 $W_{\mathrm{CE}}$ . If $W_{\mathrm{ED}}=0$ and/or $W_{\mathrm{DA}}=0$ , write a "0" in the corresponding columns, as appropriate.

Figure 2

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.

[ 4 ]

For indicating that $W_{\mathrm{ED}}=0$

1 point

For drawing a bar representing $W_{\mathrm{DA}}$ that has a height of six units

1 point
Example Response

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

Scoring Note: This point can be earned if the response begins with a correct relationship between electric field and potential difference in which numerical values are already substituted.

Example Responses

E_{x}=-\frac{d V}{d x} \quad \mathbf{O R} \quad\left|E_{x}\right|=\left|-\frac{d V}{d x}\right| \quad \mathbf{O R} \quad\left|E_{x}\right|=\left|-\frac{\Delta V}{\Delta x}\right| \quad \mathbf{O R} \quad \Delta V=-\int \vec{E} \cdot d \vec{r}

For correct substitutions of values of electric potential and the distance between equipotential 1 point lines that can be used to calculate the approximate magnitude of the electric field at Position B

Example Response

\left|E_{x}\right|=\left|-\frac{20.0 \mathrm{~V}-0.0 \mathrm{~V}}{0.65 \mathrm{~m}}\right|

Example Solution

\begin{aligned} & E_{x}=-\frac{d V}{d x} \\ & \left|E_{x}\right|=\left|-\frac{d V}{d x}\right| \\ & \left|E_{x}\right|=\left|-\frac{\Delta V}{\Delta x}\right| \\ & \left|E_{x}\right|=\left|-\frac{20.0 \mathrm{~V}-0.0 \mathrm{~V}}{0.65 \mathrm{~m}}\right| \\ & \left|E_{x}\right|=31 \frac{\mathrm{~V}}{\mathrm{~m}} \end{aligned}

Total for part (b) 4 points

(b)

The electric potential $V_{\mathrm{P}}$ at $x_{\mathrm{P}}$ is $V_{\mathrm{P}}=k \lambda \ln \left(\frac{x_{\mathrm{P}}}{x_{\mathrm{P}}-4 L}\right)$ .

Using integral calculus, derive the expression for $V_{\mathrm{P}}$ 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 4 L<x<12 L.

Figure 4

Figure 1

[ 6 ]

(i) For using an appropriate equation for determining the electric potential from a line of uniform 1 point charge charge
Example Responses
$V=\frac{1}{4 \pi \varepsilon_{0}} \sum \frac{q_{i}}{r_{i}} \quad$ OR $\quad V=\frac{1}{4 \pi \varepsilon_{0}} \int \frac{d q}{r} \quad$ OR $\quad \Delta V=-\int \vec{E} \cdot d \vec{r}$
For a correct determination of r, the distance between Point P and a point on the line of 1 point uniform charge

Example Responses
$V_{\mathrm{P}}=k \sum \frac{Q}{x_{\mathrm{P}}-x} \quad$ OR $\quad V_{\mathrm{P}}=k \int\left(\frac{1}{x_{\mathrm{P}}-x}\right) d q$
For a correct integral with $\lambda d x$ substituted for d q 1 point

Example Response
$V_{\mathrm{P}}=k \lambda \int\left(\frac{1}{x_{\mathrm{P}}-x}\right) d x$
For the correct limits of integration 1 point

Example Response
$V_{\mathrm{P}}=k \lambda \int_{0}^{4 L}\left(\frac{1}{x_{\mathrm{P}}-x}\right) d x$

Example Solutions

\begin{aligned} & V=\frac{1}{4 \pi \varepsilon_{0}} \sum \frac{q_{i}}{r_{i}} \\ & V_{\mathrm{P}}=k \int\left(\frac{1}{x_{\mathrm{P}}-x}\right) d q \\ & d q=\lambda d x \\ & V_{\mathrm{P}}=k \lambda \int\left(\frac{1}{x_{\mathrm{P}}-x}\right) d x \\ & V_{\mathrm{P}}=k \lambda \int_{0}^{4 L}\left(\frac{1}{x_{\mathrm{P}}-x}\right) d x \\ & V_{\mathrm{P}}=-\left.k \lambda \ln \left(x_{\mathrm{P}}-x\right)\right|_{0} ^{4 L}=\left.k \lambda \ln \left(x_{\mathrm{P}}-x\right)\right|_{4 L} ^{0} \\ & V_{\mathrm{P}}=k \lambda \ln \left(\frac{x_{\mathrm{P}}}{x_{\mathrm{P}}-4 L}\right) \end{aligned}

OR

\begin{aligned} & \Delta V=-\int \vec{E} \cdot d \vec{r} \\ & E(x)=\int \frac{k}{r^{2}} d q \\ & E(x)=\int_{0}^{4 L} \frac{k \lambda}{\left(x-x^{\prime}\right)^{2}} d x^{\prime}=k \lambda\left(\frac{1}{x-4 L}-\frac{1}{x}\right) \\ & V_{\mathrm{P}}=-\int_{\infty}^{x_{\mathrm{P}}} E(x) d x=-k \lambda \int_{\infty}^{x_{\mathrm{P}}}\left(\frac{1}{x-4 L}-\frac{1}{x}\right) d x \\ & V_{\mathrm{P}}=k \lambda \ln \left(\frac{x_{\mathrm{P}}}{x_{\mathrm{P}}-4 L}\right) \end{aligned}

(ii) For sketching a curve or line that continually approaches the horizontal axis as position 1 point increases
For sketching a concave up curve that is always positive 1 point
Example Response

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

Total for part (d)

for question 115 points

[Maximum number: 2]

An isolated, air-filled, charged capacitor consists of two conducting, coaxial, cylindrical shells

that each have length L. The inner shell has radius $R_{1}$ and the outer shell has radius $R_{2}$ , as

shown in Figure 1, where $R_{1}<R_{2} \ll L$ . The surface charge densities (amounts of charge per unit

area) of the inner and outer shells are $+\sigma_{1}$ and $-\sigma_{2}$ , respectively. The absolute values of the

total charges on the shells are equal.

Figure 1

Note: Figures not drawn to scale.

A.

Derive an expression for the absolute value $|\Delta V|$ of the potential difference between

the outer and inner shells in terms of $R_{1}, R_{2}, \sigma_{1}$ , and physical constants, as appropriate.

Begin your derivation by writing a fundamental physics principle or an equation from

the reference information.

(ii) For substituting the expression for E from part A (i) into $\Delta V=-\int_{a}^{b} \vec{E} \cdot d \vec{r}$

Point A4

Scoring Notes:

- Vector notation is not required for this point to be earned.

- The sign of $\Delta V$ is not considered for this point to be earned.

For an attempt to solve the integral for $|\Delta V|$ that includes correct limits

Point A5

(e.g., $|\Delta V|=\frac{\sigma_{1} R_{1}}{\varepsilon_{0}} \int_{R_{1}}^{R_{2}} \frac{1}{r} d r$ )

Scoring Note: This point may be earned regardless of the order of the limits of

integration; for example, $\frac{\sigma_{1} R_{1}}{\varepsilon_{0}} \int_{R_{2}}^{R_{1}} \frac{1}{r} d r$ .

Example Response

$\begin{aligned} & |\Delta V|=\left|-\int_{a}^{b} \vec{E} \cdot d \vec{r}\right| \\ & |\Delta V|=\int_{R_{1}}^{R_{2}} \frac{\sigma_{1} R_{1}}{r \varepsilon_{0}} d r=\frac{\sigma_{1} R_{1}}{\varepsilon_{0}} \int_{R_{1}}^{R_{2}} \frac{1}{r} d r \\ & |\Delta V|=\left.\frac{\sigma_{1} R_{1}}{\varepsilon_{0}} \ln (r)\right|_{R_{1}} ^{R_{2}}=\frac{\sigma_{1} R_{1}}{\varepsilon_{0}} \ln \left(\frac{R_{2}}{R_{1}}\right) \end{aligned}$

AP Physics C E and M Unit 9: Electric Potential

Covered syllabus points

Covered syllabus points

9 Electric Potential question 1

9 Electric Potential question 2

Question (a)

Question (i)

Question (ii)

Question (b)

Question (c)

Question (i)

9 Electric Potential question 3

Question (a)

Question (b)

9 Electric Potential question 4