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AP Physics C: Mechanics : Electricity and Magnetism Exam

Study concepts, example questions & explanations for AP Physics C: Mechanics

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2 Diagnostic Tests 92 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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Example Question #1 : 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:

8.99*10^6V

8.99*10^3V

7.25*10^5V

1.90*10^6V

5.43*10^8V

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 #1 : Calculating Electric Potential

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:

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

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

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

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

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

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:

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

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

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

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

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

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 #1 : 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:

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

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

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

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

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

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:

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

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

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

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

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

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 #11 : Calculating Electric Potential

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:

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

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

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

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

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

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 #12 : Calculating Electric Potential

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:

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

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

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

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

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

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:

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

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

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

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

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

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 #181 : 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:

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

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

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

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

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

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 #181 : 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:

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

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

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

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

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

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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All AP Physics C: Mechanics Resources

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AP Physics C: Mechanics : Electricity and Magnetism Exam

Study concepts, example questions & explanations for AP Physics C: Mechanics

All AP Physics C: Mechanics Resources

Example Questions

Example Question #1 : Calculating Electric Potential

Example Question #1 : Calculating Electric Potential

Example Question #1 : Calculating Electric Potential

Example Question #1 : Calculating Electric Potential

Example Question #11 : Calculating Electric Potential

Example Question #11 : Calculating Electric Potential

Example Question #12 : Calculating Electric Potential

Example Question #181 : Ap Physics C

Example Question #181 : Ap Physics C

Example Question #181 : Ap Physics C

All AP Physics C: Mechanics Resources

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