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AP Physics C Electricity : Calculating Electric Potential

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Example Questions

Example Question #171 : Ap Physics C

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:

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

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

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

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

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

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 #172 : 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:

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

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

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

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

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

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 #173 : Ap Physics C

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:

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

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

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

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

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

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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Example Question #181 : Ap Physics C

Ap physics c e m potential problems 2 6 16 5

A nonuniformly charged ring of radius carries a linear charge density of $\lambda(\theta) = b\sin^2\theta$ . Calculate the potential at the center of the ring.

Possible Answers:

$\frac{b}{4\pi\epsilon_0}$

$\frac{b}{4\epsilon_0R}$

$\frac{b}{2\epsilon_0}$

$\frac{b}{4\epsilon_0}$

$\frac{b}{2\pi\epsilon_0}$

Correct answer:

$\frac{b}{4\epsilon_0}$

Explanation:

Use a polar coordinate system, the given linear charge density $\lambda(\theta)=b\sin^2\theta$ , and length element $Rd\theta$ . Since every point on the ring is the same distance from the center, we calculate the potential as

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

$=\frac{b}{4\pi\epsilon_0}\int_{0}^{2\pi}\sin^2\theta d\theta$

$=\frac{b}{4\pi\epsilon_0}\int_{0}^{2\pi}\frac{1-\cos2\theta}{2} d\theta$

$=\frac{b}{4\pi\epsilon_0}\left [\frac{\theta}{2}-\frac{\sin2\theta}{4} \right ]_{0}^{2\pi}$

$=\frac{b}{4\epsilon_0}$

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Example Question #182 : Ap Physics C

Ap physics c e m potential problems 2 6 16 6

In this model of a dipole, two charges and are separated by a distance as shown in the figure, where the charges lie on the x-axis at $+\frac{a}{2}$ and $-\frac{a}{2}$ respectively. Calculate the exact potential a distance from the origin at angle $\theta$ from the axis of the dipole.

Possible Answers:

$\frac{q}{4\pi\epsilon_0} \left ( \frac{1}{\sqrt{R^2 + \left (\frac{a}{2} \right)^2 +2Ra \cos\theta}} - \frac{1}{\sqrt{R^2 + \left (\frac{a}{2} \right)^2 - 2Ra\cos\theta}} \right )$

$\frac{q}{4\pi\epsilon_0} \left ( \frac{1}{\sqrt{R^2 + \left (\frac{a}{2} \right)^2 -Ra \cos\theta}} - \frac{1}{\sqrt{R^2 + \left (\frac{a}{2} \right)^2 + Ra\cos\theta}} \right )$

$\frac{q}{4\pi\epsilon_0} \left ( \frac{1}{\sqrt{R^2 + \left (\frac{a}{2} \right)^2 -2Ra \cos\theta}} - \frac{1}{\sqrt{R^2 + \left (\frac{a}{2} \right)^2 + 2Ra\cos\theta}} \right )$

$\frac{q}{4\pi\epsilon_0} \left ( \frac{1}{\sqrt{R^2 + \left (\frac{a}{2} \right)^2 +Ra \cos\theta}} - \frac{1}{\sqrt{R^2 + \left (\frac{a}{2} \right)^2 - Ra\cos\theta}} \right )$

$\frac{q}{4\pi\epsilon_0} \left ( \frac{1}{\sqrt{R^2 + a^2 -2Ra \cos\theta}} - \frac{1}{\sqrt{R^2 + a^2 + 2Ra\cos\theta}} \right )$

Correct answer:

$\frac{q}{4\pi\epsilon_0} \left ( \frac{1}{\sqrt{R^2 + \left (\frac{a}{2} \right)^2 -Ra \cos\theta}} - \frac{1}{\sqrt{R^2 + \left (\frac{a}{2} \right)^2 + Ra\cos\theta}} \right )$

Explanation:

By the law of cosines, the distance from the point to charge is

$\sqrt{R^2+\left(\frac{a}{2} \right )^2 - 2R\left(\frac{a}{2} \right )\cos\theta}$ .

The distance to charge can be found by using the law of cosines using the supplementary angle $\pi-\theta$ , for which $\cos(\pi-\theta) = \cos(\pi)\cos\theta - \sin(\pi)\sin\theta = -\cos\theta$ . Therefore the distance to is

$\sqrt{R^2+\left(\frac{a}{2} \right )^2 - 2R\left(\frac{a}{2} \right )(-\cos\theta)}$ .

Lastly, the exact potential is given by

$\frac{q}{4\pi\epsilon_0} \left ( \frac{1}{\sqrt{R^2 + \left (\frac{a}{2} \right)^2 -Ra \cos\theta}} - \frac{1}{\sqrt{R^2 + \left (\frac{a}{2} \right)^2 + Ra\cos\theta}} \right )$ .

Remark: Far from the dipole (approximating $R\gg a$ ) gives the much simpler equation for the potential of an ideal electric dipole $\frac{qa\cos\theta}{4\pi\epsilon_0 R^2}$ .

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Example Question #183 : Ap Physics C

Ap physics c e m potential problems 2 6 16 4

A uniformly charged hollow spherical shell of radius carries a total charge . Calculate the potential a distance (where a>R ) from the center of the sphere.

Possible Answers:

$\frac{Q}{8\pi\epsilon_0R}$

$\frac{Q}{4\pi\epsilon_0R}$

$\frac{Q}{4\pi\epsilon_0a^2}$

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

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

Correct answer:

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

Explanation:

Use a spherical coordinate system and place the point of interest a distance from the center on the z-axis. By the law of cosines, the distance from this point to any point on the sphere is $\sqrt{R^2 + a^2 -2Ra\cos\theta}$ . Using surface charge density $\frac{Q}{4\pi R^2}$ and area element $R^2\sin\theta d\theta d\phi$ , we evaluate the potential as:

$V =\frac{1}{4\pi\epsilon_0} \int_{0}^{2\pi}\int_{0}^{\pi}\left (\frac{Q}{4\pi R^2} \right )\frac{R^2\sin\theta d\theta d\phi}{\sqrt{R^2 + a^2-2Ra\cos\theta}}$

. $=\frac{1}{4\pi\epsilon_0} \left (2\pi \right )\left (\frac{Q}{4\pi} \right )\int_{0}^{\pi}\frac{\sin\theta d\theta d\phi}{\sqrt{R^2 + a^2-2Ra\cos\theta}}$

$=\frac{1}{4\pi\epsilon_0} \left (\frac{Q}{2} \right ) \left [\frac{1}{Ra}\sqrt{R^2 + a^2-2Ra\cos\theta} \right ]_{0}^{\pi}$

$=\frac{1}{4\pi\epsilon_0} \left (\frac{Q}{2} \right ) \frac{1}{Ra}\left (|R+a|-|R-a| \right )$

$=\frac{1}{4\pi\epsilon_0} \left (\frac{Q}{2} \right ) \frac{1}{Ra}(2R)$

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

Remarkably, this is the same potential that would exist a distance from a point charge.

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Example Question #31 : Electricity And Magnetism Exam

Ap physics c e m potential problems 2 6 16 3

A thin bar of length L lies in the xy plane and carries linear charge density $\lambda(x) = bx$ , where ranges from 0 to . Calculate the potential at the point (0, a) on the y-axis.

Possible Answers:

$\frac{b}{4\pi\epsilon_0}\sqrt{L^2+a^2}$

$\frac{b}{8\pi\epsilon_0}\sqrt{L^2+a^2}$

$\frac{b}{2\pi\epsilon_0}\left (\sqrt{L^2+a^2}-a \right )$

$\frac{b}{4\pi\epsilon_0}\left (\sqrt{L^2+a^2}-a \right )$

$\frac{b}{8\pi\epsilon_0}\left (\sqrt{L^2+a^2}-a \right )$

Correct answer:

$\frac{b}{4\pi\epsilon_0}\left (\sqrt{L^2+a^2}-a \right )$

Explanation:

Use the linear charge density $\lambda(x)=bx$ and length element , where each point is $\sqrt{x^2+a^2}$ from the point (0,a) . The potential is therefore

$V=\frac{1}{4\pi\epsilon_0}\int_{0}^{L}\frac{bx}{\sqrt{x^2+a^2}}dx$

$= \frac{b}{4\pi\epsilon_0}\left [ \sqrt{x^2+a^2}\right ]_{0}^{L}$

$=\frac{b}{4\pi\epsilon_0}\left (\sqrt{L^2+a^2}-a \right )$

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Example Question #32 : Electricity And Magnetism Exam

Ap physics c e m potential problems 2 6 16 2

A uniformly charged ring of radius carries a total charge . Calculate the potential a distance from the center, on the axis of the ring.

Possible Answers:

$\frac{Q}{4\pi\epsilon_0R\sqrt{R^2 + a^2}}$

$\frac{Qa}{4\pi\epsilon_0(R^2+a^2)^{\frac{3}{2}}}$

$\frac{Q}{4\pi\epsilon_0(R^2 + a^2)}$

$\frac{Q}{4\pi\epsilon_0\sqrt{R^2 + a^2}}$

$\frac{Qa}{4\pi\epsilon_0(R^2 + a^2)}$

Correct answer:

$\frac{Q}{4\pi\epsilon_0\sqrt{R^2 + a^2}}$

Explanation:

Use the linear charge density $\frac{Q}{2\pi R}$ and length element $Rd\theta$ . The distance from each point on the ring to the point on the axis is $\sqrt{R^2+a^2}$ . Lastly, integrate over $\theta$ from to $2\pi$ to obtain

$V=\frac{1}{4\pi\epsilon_0}\int_{0}^{2\pi}\left (\frac{Q}{2\pi R} \right )\frac{1}{\sqrt{R^2+a^2}}Rd\theta$

$=\frac{Q}{4\pi\epsilon_0\sqrt{R^2+a^2}}$

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Example Question #33 : Electricity And Magnetism Exam

Ap physics c e m potential problems 2 6 16

A uniformly charged square frame of side length carries a total charge . Calculate the potential at the center of the square.

You may wish to use the integral:

$\int{\frac{dx}{\sqrt{x^2+b^2}}} = \ln {(\sqrt{x^2+b^2} + x)} + C$

Possible Answers:

$\frac{Q}{\pi\epsilon_{0}a}\ln{(1+\sqrt{2})}$

$\frac{2Q}{\pi\epsilon_{0}a}\ln{(1+\sqrt{2})}$

$\frac{Q}{4\pi\epsilon_{0}a}\ln{(1+\sqrt{2})}$

$\frac{Q}{2\pi\epsilon_{0}a}\ln{(1+\sqrt{2})}$

$\frac{Q}{8\pi\epsilon_{0}a}\ln{(1+\sqrt{2})}$

Correct answer:

$\frac{Q}{2\pi\epsilon_{0}a}\ln{(1+\sqrt{2})}$

Explanation:

Calculate the potential due to one side of the bar, and then multiply this by to get the total potential from all four sides. Orient the bar along the x-axis such that its endpoints are at $\pm\frac{a}{2}$ , and use the linear charge density $\frac{Q}{4a}$ . The potential is therefore

$V=4\times\left (\frac{1}{4\pi\epsilon_{0}}\int_{-a/2}^{a/2} \left (\frac{Q}{4a} \right )\frac{1}{\sqrt{x^2+\frac{a^2}{4}}}dx\right )$

$=\frac{Q}{4\pi\epsilon_{0}a}\int_{-a/2}^{a/2} \frac{1}{\sqrt{x^2+\frac{a^2}{4}}}dx$

$=\frac{Q}{4\pi\epsilon_{0}a}\left[ \ln\left (\sqrt{x^2+\frac{a^2}{4}} +x\right )\right]_{-a/2}^{a/2}$

$=\frac{Q}{4\pi\epsilon_0a}\ln\left (\frac{\sqrt{2}+1}{\sqrt{2}-1} \right )$

$=\frac{Q}{2\pi\epsilon_0a}\ln(1+\sqrt{2})$

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AP Physics C Electricity : Calculating Electric Potential

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

All AP Physics C Electricity Resources

Example Questions

Example Question #171 : Ap Physics C

Example Question #172 : Ap Physics C

Example Question #173 : Ap Physics C

Example Question #181 : Ap Physics C

Example Question #182 : Ap Physics C

Example Question #183 : Ap Physics C

Example Question #31 : Electricity And Magnetism Exam

Example Question #32 : Electricity And Magnetism Exam

Example Question #33 : Electricity And Magnetism Exam

All AP Physics C Electricity Resources

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