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AP Physics C: Mechanics : Electricity

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

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All AP Physics C: Mechanics Resources

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

Example Questions

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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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:

$4kQ\pi r^3$

Zero

$4Q\pi r^3$

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

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

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 point A

It will remain stationary

Toward the original charge Q

Toward point C

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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All AP Physics C: Mechanics Resources

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AP Physics C: Mechanics : Electricity

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

All AP Physics C: Mechanics Resources

Example Questions

Example Question #31 : Electricity And Magnetism Exam

Example Question #32 : Electricity And Magnetism Exam

Example Question #33 : Electricity And Magnetism Exam

Example Question #1 : Understanding Charge Distributions On Objects

Example Question #1 : Interpreting Electric Charge Diagrams

All AP Physics C: Mechanics Resources

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