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AP Physics C: Mechanics : AP Physics C

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Example Questions

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:

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

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

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

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

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

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 #11 : Calculating Electric Potential

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:

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

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

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

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

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

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 #21 : Electricity And Magnetism Exam

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:

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

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

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

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

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

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

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:

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

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

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

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

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

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

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:

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

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

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

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

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

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 #12 : Calculating Electric Potential

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:

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

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

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

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

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

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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Example Question #1 : Understanding Charge Distributions On Objects

Consider a spherical shell with radius and charge . What is the magnitude of the electric field at the center of this spherical shell?

Possible Answers:

$\frac{4}{3}Q\pi r^3$

$4kQ\pi r^3$

Zero

$4Q\pi r^3$

$\frac{4}{3}kQ\pi r^3$

Correct answer:

Zero

Explanation:

According to the shell theorem, the total electric field at the center point of a charged spherical shell is always zero. At this point, any electric field lines will result in symmetry, canceling each other out and creating a net field of zero at that point.

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Example Question #1 : Interpreting Electric Charge Diagrams

A charge of unknown value is held in place far from other charges. Its electric field lines and some lines of electric equipotential (V₁ and V₂) are shown in the diagram.

Onecharge_negative_fieldandpotentiallines

A second point charge, known to be negative, is placed at point A in the diagram. In which direction will the second, negative charge freely move?

Possible Answers:

Toward the original charge Q

Toward point C

Toward point A

It will remain stationary

Correct answer:

Toward point C

Explanation:

The original charge Q is negative, as indicated by the direction of the electric field lines in the diagram. A negative point charge of any value placed at point A will cause both charges to feel a mutually repelling force on each other. Therefore, the second charge will be repelled by the original negative charge with a force pointing radially along a line connecting the center of Q and the point A, resulting in free movement toward C. Additionally, point B is on a line of equipotential to point A; negative point charges will move from regions of lower electric potential to higher electric potential (against the direction of the electric field lines), so point B is not a viable answer.

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Example Question #182 : Ap Physics C

A proton traveling $600\frac{m}{s}$ enters a uniform magnetic field and experiences a magnetic force, causing it to travel in a circular path. Taking the magnetic field to be 0.0002T , what is the radius of this circular path (shown in red)?

Uniform_field

$m_{proton}=1.67*10^{-27}\ \text{kg}$

$q_{proton}=1.6*10^{-19}\ \text{C}$

Possible Answers:

18.5cm

2.5mm

5.9cm

2.3cm

3.1cm

Correct answer:

3.1cm

Explanation:

To calculate the magnetic force of a single charge, we use F=qvB , where is the charge of the proton, is its velocity, is the uniform magnetic field.

Since this magnetic force causes the proton to travel in a circular path, we set this magnetic force equation equal to the centripetal force equation.

$qvB=\frac{mv^2}{r}$

is the mass of the proton and is the radius of the circular path. Solve for .

$r=\frac{mv}{qB}$

Using the values given in the question, we can solve for the radius.

$r=\frac{(1.67*10^{-27}kg)(600\frac{m}{s})}{(1.6*10^{-19}C)(0.0002T)}$

$r=0.031\ \text{m}=3.1\ \text{cm}$

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Example Question #183 : Ap Physics C

Which of the following best describes the net magnetic flux through a closed sphere, in the presence of a magnet?

Possible Answers:

Zero regardless of the orientation of the magnet

More than one of the other options is true

Zero only if the magnet is completely enclosed within the surface

Negative only if the north pole of the magnet is within the surface

Positive only if the north pole of the magnet is within the surface

Correct answer:

Zero regardless of the orientation of the magnet

Explanation:

The net magnetic flux (or net field flowing in and out) through any closed surface must always be zero. This is because magnetic field lines have no starting or ending points, so any field line going into the surface must also come out. In other words, "there are no magnetic monopoles."

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All AP Physics C: Mechanics Resources

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AP Physics C: Mechanics : AP Physics C

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

All AP Physics C: Mechanics Resources

Example Questions

Example Question #181 : Ap Physics C

Example Question #11 : Calculating Electric Potential

Example Question #21 : Electricity And Magnetism Exam

Example Question #31 : Electricity

Example Question #32 : Electricity

Example Question #12 : Calculating Electric Potential

Example Question #1 : Understanding Charge Distributions On Objects

Example Question #1 : Interpreting Electric Charge Diagrams

Example Question #182 : Ap Physics C

Example Question #183 : Ap Physics C

All AP Physics C: Mechanics Resources

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