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AP Physics C Electricity : AP Physics C

Study concepts, example questions & explanations for AP Physics C Electricity

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Example Questions

Example Question #4 : Calculating Electric Potential

A proton moves in a straight line for a distance of . Along this path, the electric field is uniform with a value of $2*10^7 \frac{V}{m}$ . Find the work done on the proton by the electric field.

The charge of a proton is $1.61*10^{-19}\text{C}$ .

Possible Answers:

$1.61*10^{-11}J$

$6.44*10^{-11}J$

$0.8*10^{-11}J$

$3.22*10^{-11}J$

Correct answer:

$1.61*10^{-11}J$

Explanation:

Work done by an electric field is given by the product of the charge of the particle, the electric field strength, and the distance travelled.

W=qEd

We are given the charge ( $1.61*10^{-19}\text{C}$ ), the distance (), and the field strength ( $2*10^7 \frac{V}{m}$ ), allowing us to calculate the work.

$W=(1.61*10^{-19}C)(2*10^{7}\frac{V}{m})(5m)$

$W=(3.22*10^{-12}\frac{J}{m})(5m)$

$W=1.61*10^{-11}J$

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Example Question #4 : Calculating Electric Potential

A negative charge of magnitude $9.0 \mu C$ is placed in a uniform electric field of $3.0 * 10^4 \frac{N}{C}$ , directed upwards. If the charge is moved 1.0m upwards, how much work is done on the charge by the electric field in this process?

Possible Answers:

$2.7 * 10^{4}J$

$-3.0 * 10^{-13} J$

$-2.7 * 10^{-4}J$

$3.0 * 10^{-13} J$

$-3.3 * 10^{-6}J$

Correct answer:

$-2.7 * 10^{-4}J$

Explanation:

Relevant equations:

$\Delta V = E * d$

$W = q \Delta V$

Given:

$E = 3.0 * 10^4 \frac{N}{C}$

d = 1.0 m

$q = -9.0 \mu C = -9.0 * 10 ^ {-9}C$

First, find the potential difference between the initial and final positions:

$\Delta V = (3.0 * 10^4 \frac{N}{C})(1.0m)= 3.0 * 10^4 \frac{N\cdot m}{C}$

2. Plug this potential difference into the work equation to solve for W:

$W = (-9.0 * 10^{-9}C)(3.0 * 10^4 \frac{N\cdot m}{C})$

$W = -2.7 * 10^{-4}J$

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Example Question #1 : Calculating Electric Potential

Three point charges are arranged around the origin, as shown.

Ps0_threechargepotential

$\begin{matrix} &\text{Charge} & \text{Location}\\ Q_1& +3C&(-2cm,0) \\ Q_2&-5C &(4cm,0) \\ Q_3&+1C & (0,5cm) \end{matrix}$

Calculate the total electric potential at the origin due to the three point charges.

$k=9*10^9\frac{Nm^2}{C^2}$

Possible Answers:

$3.41*10^{11}V$

$4.05*10^{11}V$

$2.65*10^{12}V$

$1.05*10^{12}V$

Correct answer:

$4.05*10^{11}V$

Explanation:

Electric potential is a scalar quantity given by the equation:

$V = \frac{kq_{i}}{r_{i}}$

To find the total potential at the origin due to the three charges, add the potentials of each charge.

$V_{total}= V_{1}+V_{2}+V_{3} = \frac{kQ_{1}}{r_{1}}+\frac{kQ_{2}}{r_{2}}+\frac{kQ_{3}}{r_{3}}$

$V_{total}=\frac{(9*10^9)(3C)}{0.02m}+\frac{(9*10^9)(-5C)}{0.04m}+\frac{(9*10^9)(1C)}{0.05m}$

$V_{total}=1.35*10^{12}V+(-1.125*10^{12}V)+1.8*10^{11}V$

$V_{total}=4.05*10^{11}V$

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Example Question #5 : Calculating Electric Potential

Three identical point charges with $q=1\mu C$ are placed so that they form an equilateral triangle as shown in the figure. Find the electric potential at the center point (black dot) of that equilateral triangle, where this point is at a equal distance, d=3mm , away from the three charges.

3_charges

$k=8.99*10^9\frac{Nm^2}{C^2}$

Possible Answers:

5.43*10^8V

7.25*10^5V

8.99*10^6V

8.99*10^3V

1.90*10^6V

Correct answer:

8.99*10^6V

Explanation:

The electric potential from point charges is $V=\frac{kq}{d}$ .

Knowing that all three charges are identical, and knowing that the center point at which we are calculating the electric potential is equal distance from the charges, we can multiply the electric potential equation by three.

$V=3\frac{kq}{d}$

Plug in the given values and solve for .

$V=3\frac{(8.99*10^9\ \frac{\text{Nm}^2}{\text{C}^2})(1* 10^{-6}\ \text{C})}{3*10^{-3}\ \text{m}}$

$V=8.99* 10^6\ \text{V}$

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Example Question #21 : Electricity

Ap physics c e m potential problems 2 6 16 1

Eight point charges of equal magnitude are located at the vertices of a cube of side length . Calculate the potential at the center of the cube.

Possible Answers:

$\frac{8Q}{3\pi\epsilon_0a}$

$\frac{2\sqrt{2}Q}{\pi\epsilon_0a}$

$\frac{2Q}{\pi\epsilon_0a\sqrt{3}}$

$\frac{4Q}{\pi\epsilon_0a\sqrt{3}}$

$\frac{\sqrt{3}Q}{6\pi\epsilon_0a}$

Correct answer:

$\frac{4Q}{\pi\epsilon_0a\sqrt{3}}$

Explanation:

By the Pythagorean theorem, each charge is a distance

$d = \sqrt{\left (\frac{a}{2} \right )^2 + \left (\frac{a}{2} \right )^2+\left (\frac{a}{2} \right )^2} = \frac{\sqrt{3}}{2}a$

from the center of the cube, so the potential is

$V=8\times\left (\frac{Q}{4\pi\epsilon_0\left (\frac{\sqrt{3}}{2}a \right )} \right )$

$=\frac{4Q}{\pi\epsilon_0a\sqrt{3}}$

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Example Question #1 : Calculating Electric Potential

Ap physics c e m potential problems 2 6 16 1 4

An infinite plane has a nonuniform charge density given by $\sigma(r)=\frac{qa}{2\pi(r^2+a^2)^{3/2}}$ . Calculate the potential at a distance above the origin.

You may wish to use the integral:

$\int\frac{x}{(x^2+a^2)^2}dx=\frac{1}{2a^2}\left(\frac{x^2}{x^2+a^2} \right )+C$

Possible Answers:

$\frac{q}{8\pi\epsilon_0a}$

$\frac{q}{\pi\epsilon_0a}$

$\frac{q}{16\pi\epsilon_0a}$

$\frac{q}{2\pi\epsilon_0a}$

$\frac{q}{4\pi\epsilon_0a}$

Correct answer:

$\frac{q}{8\pi\epsilon_0a}$

Explanation:

Use polar coordinates with the given surface charge density, $\sigma(r)=\frac{qa}{2\pi(r^2+a^2)^{3/2}}$ and area element $rd\theta d\phi$ . Noting that a point from the origin is a distance $\sqrt{r^2+a^2}$ from the point of interest, we calculate the potential as follows, integrating with respect to from to $\infty$ .

$V=\frac{1}{4\pi\epsilon_0} \int_{0}^{2\pi} \int_{0}^{\infty}\left (\frac{qa}{2\pi(r^2+a^2)^{3/2}} \right ) \frac{rdrd\theta}{\sqrt{r^2+a^2}}$

$=\frac{1}{4\pi\epsilon_0} (2\pi) \left(\frac{qa}{2\pi} \right )\int_{0}^{\infty}\left (\frac{1}{(r^2+a^2)^{3/2}} \right ) \frac{rdr}{\sqrt{r^2+a^2}}$

$=\frac{qa}{4\pi\epsilon_0} \int_{0}^{\infty}\frac{r}{(r^2+a^2)^2} dr$

$=\frac{qa}{4\pi\epsilon_0} \left [\frac{1}{2a^2}\left (\frac{x^2}{x^2+a^2} \right ) \right ]_{0}^{\infty}$

$=\frac{q}{8\pi\epsilon_0a} (1-0)$ (by limit)

$=\frac{q}{8\pi\epsilon_0a}$

Remark: This is exactly the charge distribution that would be induced on an infinite sheet of (grounded) metal if a negative charge were held a distance above it.

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Example Question #2 : Calculating Electric Potential

Ap physics c e m potential problems 2 6 16 1 3

A nonuniformly charged hemispherical shell of radius (shown above) has surface charge density $\sigma(\theta)=b\cos^3\theta$ . Calculate the potential at the center of the opening of the hemisphere (the origin).

Possible Answers:

$\frac{Rb}{3\epsilon_0}$

$\frac{3Rb}{8\epsilon_0}$

$\frac{Rb}{2\epsilon_0}$

$\frac{Rb}{15\epsilon_0}$

$\frac{Rb}{8\epsilon_0}$

Correct answer:

$\frac{Rb}{8\epsilon_0}$

Explanation:

Use spherical coordinates with the given surface charge density $\sigma(\theta)=b\cos^3\theta$ , and area element $R^2\sin\theta d\theta d\phi$ . Every point on the hemispherical shell is a distance from the origin, so we calculate the potential as follows, noting the limits of integration for $\theta$ range from to $\pi/2$ .

$V=\frac{1}{4\pi\epsilon_0}\int_{0}^{2\pi}\int_{0}^{\pi/2}(b\cos^3\theta)\left (\frac{1}{R} \right )R^2\sin\theta d\theta d\phi$

$=\frac{Rb}{4\pi\epsilon_0} (2\pi) \int_{0}^{\pi/2}\cos^3\theta\sin\theta d\theta$

$=\frac{Rb}{2\epsilon_0} \left[\frac{-\cos^4\theta}{4} \right]_{0}^{\pi/2}$

$=\frac{Rb}{8\epsilon_0}$

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Example Question #11 : Calculating Electric Potential

Ap physics c e m potential problems 2 6 16 1 2

Three equal point charges are placed at the vertices of an equilateral triangle of side length . Calculate the potential at the center of the triangle, labeled P.

Possible Answers:

$\frac{\sqrt{3}Q}{4\pi\epsilon_0a}$

$\frac{3\sqrt{3}Q}{2\pi\epsilon_0a}$

$\frac{3\sqrt{3}Q}{4\pi\epsilon_0a}$

$\frac{3\sqrt{3}Q}{8\pi\epsilon_0a}$

$\frac{\sqrt{3}Q}{2\pi\epsilon_0a}$

Correct answer:

$\frac{3\sqrt{3}Q}{4\pi\epsilon_0a}$

Explanation:

Draw a line from the center perpendicular to any side of the triangle. This line divides the side into two equal pieces of length $\frac{a}{2}$ . From the center, draw another line to one of the vertices at the end of this side. This produces a 30-60-90 triangle with longer leg $\frac{a}{2}$ , so the hypotenuse (the distance from the vertex to the center) is $\frac{a}{\sqrt{3}}$ . The potential at the center is due to three of these charges, so it must be

$V=3\times \left ( \frac{Q}{4\pi\epsilon_0\left(\frac{a}{\sqrt{3}} \right )} \right)$

$=\frac{3\sqrt{3}Q}{4\pi\epsilon_0a}$

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Example Question #171 : Ap Physics C

Ap physics c e m potential problems 2 6 16 1 1

A uniformly charged hollow disk has inner radius R_1 and outer radius R_2 , and carries a total charge . Calculate the potential a distance from the center, on the axis of the disk.

Possible Answers:

$\frac{1}{8\pi\epsilon_0}\frac{Q}{R_2^2-R_1^2}\left(\sqrt{R_2^2+a^2}-\sqrt{R_1^2+a^2} \right )$

$\frac{1}{4\pi\epsilon_0}\frac{Q}{R_2^2}\left(\sqrt{R_2^2+a^2}-\sqrt{R_1^2+a^2} \right )$

$\frac{1}{4\pi\epsilon_0}\frac{Q}{R_2^2-R_1^2}\left(\sqrt{R_2^2+a^2}-\sqrt{R_1^2+a^2} \right )$

$\frac{1}{2\pi\epsilon_0}\frac{Q}{R_2^2-R_1^2}\left(\sqrt{R_2^2+a^2}-\sqrt{R_1^2+a^2} \right )$

$\frac{1}{2\pi\epsilon_0}\frac{Q}{R_2^2}\left(\sqrt{R_2^2+a^2}-\sqrt{R_1^2+a^2} \right )$

Correct answer:

$\frac{1}{2\pi\epsilon_0}\frac{Q}{R_2^2-R_1^2}\left(\sqrt{R_2^2+a^2}-\sqrt{R_1^2+a^2} \right )$

Explanation:

Use a polar coordinate system with surface charge density $\frac{Q}{\pi(R_2^2-R_1^2)}$ and area element $Rdrd\theta$ . The distance from the point of interest to a point a distance from the center is $\sqrt{r^2+a^2}$ , so the potential is

$V=\frac{1}{4\pi\epsilon_0}\int_{0}^{2\pi}\int_{R_1}^{R_2}\left(\frac{Q}{\pi(R_2^2-R_1^2)}\right )\frac{rdrd\theta}{\sqrt{r^2+a^2}}$

$=\frac{1}{4\pi\epsilon_0}(2\pi)\left(\frac{Q}{\pi(R_2^2-R_1^2)}\right )\int_{R_1}^{R_2}\frac{r}{\sqrt{r^2+a^2}}dr$

$=\frac{1}{2\pi\epsilon_0}\frac{Q}{R_2^2-R_1^2}\left[\sqrt{r^2+a^2} \right ]_{R_1}^{R_2}$

$=\frac{1}{2\pi\epsilon_0}\frac{Q}{R_2^2-R_1^2}\left(\sqrt{R_2^2+a^2} -\sqrt{R_1^2+a^2}\right )$

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Example Question #21 : Electricity

Ap physics c e m potential problems 2 6 16 1

Two point charges and are separated by a distance . Calculate the potential at point P, a distance from charge in the direction perpendicular to the line connecting the two charges.

Possible Answers:

$\frac{-Q}{48\pi\epsilon_0a}$

$\frac{Q}{24\pi\epsilon_0a}$

$\frac{-Q}{30\pi\epsilon_0a}$

$\frac{Q}{60\pi\epsilon_0a}$

$\frac{11Q}{60\pi\epsilon_0a}$

Correct answer:

$\frac{Q}{60\pi\epsilon_0a}$

Explanation:

By the Pythagorean theorem, the distance from point P to charge is . Because point P is also from charge , it follows that the potential is

$V=\frac{1}{4\pi\epsilon_0}\left( \frac{2Q}{5a}-\frac{Q}{3a}\right )$

$=\frac{Q}{60\pi\epsilon_0a}$

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All AP Physics C Electricity Resources

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AP Physics C Electricity : AP Physics C

Study concepts, example questions & explanations for AP Physics C Electricity

All AP Physics C Electricity Resources

Example Questions

Example Question #4 : Calculating Electric Potential

Example Question #4 : Calculating Electric Potential

Example Question #1 : Calculating Electric Potential

Example Question #5 : Calculating Electric Potential

Example Question #21 : Electricity

Example Question #1 : Calculating Electric Potential

Example Question #2 : Calculating Electric Potential

Example Question #11 : Calculating Electric Potential

Example Question #171 : Ap Physics C

Example Question #21 : Electricity

All AP Physics C Electricity Resources

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