AP Physics C: Electricity and Magnetism - 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 potential difference created by the movement.

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

$1*10^{10}V$

Explanation

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.

$\Delta V=Ed=(2*10^7\frac{V}{m})(5m)=1*10^8V$

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?

$\frac{l}{2\pi\epsilon_0r}$

$\frac{l}{2\pi\epsilon_0r^2}$

$\frac{l}{2\pi\epsilon_0r^3 }$

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

Explanation

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.

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

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

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 $1.61*10^{-19}\text{C}$ .

$1.61*10^{-11}J$

$3.22*10^{-11}J$

$6.44*10^{-11}J$

$0.8*10^{-11}J$

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.

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.

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

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

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

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).

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

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

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

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

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

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 $\pi/2$ .

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

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

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

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

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.

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

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

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

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

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

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

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

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

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:

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

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

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

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

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

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

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

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

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

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

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

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

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 ) from the center of the sphere.

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

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

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

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

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

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.

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.

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

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

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

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

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

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

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

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

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 on the y-axis.

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

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

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

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

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

Explanation

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

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

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

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

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.

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

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

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

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

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

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