Electricity and Magnetism Exam - AP Physics C: Electricity and Magnetism

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Question

For a ring of charge with radius and total charge , the potential is given by $V=\frac{1}{4\pi\epsilon_0}\frac{q}{\sqrt{x^2+a^2}}$ .

Find the expression for electric field produced by the ring.

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Answer

We know that $E=-\frac{dV}{dx}$ .

Using the given formula, we can find the electric potential expression for the ring.

$E=-\frac{d}{dx}(\frac{1}{4\pi\epsilon_0}\frac{q}{\sqrt{x^2+a^2}})$

Take the derivative and simplify.

$E=-(\frac{q}{4\pi\epsilon_0}\frac{1}{(x^2+a^2)^{3/2}}-\frac{1}{2}2x)=\frac{1}{4\pi\epsilon_0}\frac{qx}{(x^2+a^2)^{3/2}}$

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Question

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, , away from the three charges.

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

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Answer

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 .

$V=3\frac{(8.9910^9\ \frac{\text{Nm}^2}{\text{C}^2})(1 10^{-6}\ \text{C})}{310^{-3}\ \text{m}}$

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

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Question

A proton moves in a straight line for a distance of . Along this path, the electric field is uniform with a value of $210^7 \frac{V}{m}$ . Find the work done on the proton by the electric field.

The charge of a proton is $1.6110^{-19}\text{C}$ .

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Answer

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.

We are given the charge ( $1.6110^{-19}\text{C}$ ), the distance (), and the field strength ( $210^7 \frac{V}{m}$ ), allowing us to calculate the work.

$W=(1.6110^{-19}C)(210^{7}\frac{V}{m})(5m)$

$W=(3.2210^{-12}\frac{J}{m})(5m)$

$W=1.6110^{-11}J$

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Question

A proton moves in a straight line for a distance of . Along this path, the electric field is uniform with a value of $210^7 \frac{V}{m}$ . Find the potential difference created by the movement.

The charge of a proton is $1.6110^{-19}\text{C}$ .

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Answer

Potential difference is given by the change in voltage

$V_a-V_b=\frac{W}{q}$

Work done by an electric field is equal to the product of the electric force and the distance travelled. Electric force is equal to the product of the charge and the electric field strength.

$\frac{W}{q}=\frac{Fd}{q}=\frac{qEd}{q}=Ed$

The charges cancel, and we are able to solve for the potential difference.

$\Delta V=Ed=(210^7\frac{V}{m})(5m)=110^8V$

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Question

The potential outside of a charged conducting cylinder with radius and charge per unit length is given by the below equation.

$V=\frac{l}{2\pi\epsilon_0}ln\frac{R}{r}$

What is the electric field at a point located at a distance from the surface of the cylinder?

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Answer

The radial electric field outside the cylinder can be found using the equation $E_r=-\frac{dV}{dr}$ .

Using the formula given in the question, we can expand this equation.

$E_r=-\frac{dV}{dr}=-\frac{d}{dr}(\frac{l}{2\pi\epsilon_0}(lnR-lnr))$

Now, we can take the derivative and simplify.

$E_r=-\frac{1}{2\pi \epsilon_0}(\frac{1}{R}-\frac{1}{r})$

$E_r=-(-\frac{l}{2\pi\epsilon_0r})=\frac{l}{2\pi\epsilon_0r}$

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Question

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 upwards, how much work is done on the charge by the electric field in this process?

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Answer

Relevant equations:

$\Delta V = E * d$

$W = q \Delta V$

Given:

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

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

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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Question

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

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

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

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

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Answer

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.

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

$V_{total}=\frac{(910^9)(3C)}{0.02m}+\frac{(910^9)(-5C)}{0.04m}+\frac{(910^9)(1C)}{0.05m}$

$V_{total}=1.3510^{12}V+(-1.12510^{12}V)+1.810^{11}V$

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

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Question

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:

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

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Answer

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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Question

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.

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Answer

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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Question

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

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Answer

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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Question

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 on the y-axis.

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Answer

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

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

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Answer

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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Question

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.

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Answer

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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Question

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.

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Answer

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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Question

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.

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Answer

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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Question

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

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Answer

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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Question

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.

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Answer

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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Question

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).

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Answer

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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Question

Two infinite parallel conducting sheets each have positive charge density $\sigma$ . What is the magnitude and direction of the electric field to the right of the right sheet?

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Answer

Relevant equations:

$E = \frac{\sigma }{2 \epsilon _o}$ (field due to single infinite plane)

Electric field is additive; in other words, the total electric field from the two planes is the sum of their individual fields:

$E_{total} = E_1 + E_2$

$E _{total} = \frac{\sigma}{2 \epsilon _o}+\frac{\sigma}{2 \epsilon _o}$

$E_{total}=\frac{\sigma}{\epsilon_o}$

The direction of the electric field is away from positive source charges. Thus, to the right of these positively charged planes, the field points away to the right.

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Question

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:

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

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Answer

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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