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AP Physics C Electricity : Electricity

Study concepts, example questions & explanations for AP Physics C Electricity

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1 Diagnostic Test 46 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #161 : Ap Physics C

Two capacitors are in parallel, with capacitance values of 6mF $\displaystyle 6mF$ and 10mF $\displaystyle 10mF$. What is their equivalent capacitance?

Possible Answers:

4mF $\displaystyle 4mF$

16mF $\displaystyle 16mF$

2.67mF $\displaystyle 2.67mF$

12mF $\displaystyle 12mF$

Correct answer:

16mF $\displaystyle 16mF$

Explanation:

The equivalent capacitance for capacitors in parallel is the sum of the individual capacitance values.

$C_{eq}=C_1+C_2$ $\displaystyle C_{eq}=C_1+C_2$

Using the values given in the question, we can find the equivalent capacitance.

$C_{eq}=10mF+6mF=16mF$ $\displaystyle C_{eq}=10mF+6mF=16mF$

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Example Question #12 : Electricity And Magnetism Exam

A proton moves in a straight line for a distance of $\displaystyle 5m$. Along this path, the electric field is uniform with a value of $2*10^7 \frac{V}{m}$ $\displaystyle 2*10^7 \frac{V}{m}$. Find the force on the proton.

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

Possible Answers:

$1.61*10^{-12}N$ $\displaystyle 1.61*10^{-12}N$

$6.44*10^{-12}N$ $\displaystyle 6.44*10^{-12}N$

$0.8*10^{-12}N$ $\displaystyle 0.8*10^{-12}N$

$3.22*10^{-12}N$ $\displaystyle 3.22*10^{-12}N$

Correct answer:

$3.22*10^{-12}N$ $\displaystyle 3.22*10^{-12}N$

Explanation:

The force of an electric field is given by the equation F=qE $\displaystyle F=qE$, where $\displaystyle q$ is the charge of the particle and $\displaystyle E$ is the electric field strength. We can use the given values from the question to solve for the force.

$F=qE=(1.61*10^{-19}C)(2*10^{7}\frac{V}{m})=3.22*10^{-12}N$ $\displaystyle F=qE=(1.61*10^{-19}C)(2*10^{7}\frac{V}{m})=3.22*10^{-12}N$

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Example Question #13 : Electricity And Magnetism Exam

Two point charges, Q_1 $\displaystyle Q_1$ and Q_2 $\displaystyle Q_2$ are separated by a distance $\displaystyle d$.

Ps0_twochargeefield

The values of the charges are:

$Q_{1} = - \,2\;nC$ $\displaystyle Q_{1} = - \,2\;nC$

$Q_{2} = + \,5\;nC$ $\displaystyle Q_{2} = + \,5\;nC$

The distance is 4.0cm. The point $\displaystyle P$ lies 1.5cm away from Q_1 $\displaystyle Q_1$ on a line connecting the centers of the two charges.

What is the magnitude and direction of the net electric field at point $\displaystyle P$ due to the two charges?

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

Possible Answers:

$8000\ \frac{V}{m},\ \text{toward Q}_2$ $\displaystyle 8000\ \frac{V}{m},\ \text{toward Q}_2$

_{$\displaystyle 152000\ \frac{V}{m},\ \text{toward Q}_1$}

$8000\ \frac{V}{m},\ \text{toward Q}_1$ $\displaystyle 8000\ \frac{V}{m},\ \text{toward Q}_1$

$152000\ \frac{V}{m},\ \text{toward Q}_2$ $\displaystyle 152000\ \frac{V}{m},\ \text{toward Q}_2$

Correct answer:

_{$\displaystyle 152000\ \frac{V}{m},\ \text{toward Q}_1$}

Explanation:

At point $\displaystyle P$, the electric field due to Q_1 $\displaystyle Q_1$ points toward Q_1 $\displaystyle Q_1$ with a magnitude given by:

$E_{1}=\left| \frac{kQ_{1}}{r_{1}^{2}} \right| =\left| \frac{(9*10^9\frac{Nm^2}{C^2})(-2nC)}{(1.5cm)^{2}} \right| =80000 \;\frac{V}{m}$ $\displaystyle E_{1}=\left| \frac{kQ_{1}}{r_{1}^{2}} \right| =\left| \frac{(9*10^9\frac{Nm^2}{C^2})(-2nC)}{(1.5cm)^{2}} \right| =80000 \;\frac{V}{m}$

At point P, the electric field due to Q₂ points away from Q₂ with a magnitude given by

$E_{2}=\left| \frac{kQ_{2}}{r_{2}^{2}} \right|= \left| \frac{(9*10^9\frac{Nm^2}{C^2})(5nC)}{(4.0-1.5cm)^{2}} \right| =72000 \;\frac{V}{m}$ $\displaystyle E_{2}=\left| \frac{kQ_{2}}{r_{2}^{2}} \right|= \left| \frac{(9*10^9\frac{Nm^2}{C^2})(5nC)}{(4.0-1.5cm)^{2}} \right| =72000 \;\frac{V}{m}$

The addition of these two vectors, both pointing in the same direction, results in a net electric field vector of magnitude 152000 volts per meter, pointing toward Q_1 $\displaystyle Q_1$.

$80000\frac{V}{m}+72000\frac{V}{m}=152000\frac{V}{m}$ $\displaystyle 80000\frac{V}{m}+72000\frac{V}{m}=152000\frac{V}{m}$

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Example Question #12 : Electricity And Magnetism Exam

A charge, $\displaystyle Q$, is enclosed by two spherical surfaces of radii r_1 $\displaystyle r_1$ and $\displaystyle r_2$, with r_1<r_2 $\displaystyle r_1< r_2$. The cross-sectional side view is shown.

Ps0_gauss

Which is the correct relationship between the electric flux passing through the two spherical surfaces around the point charge?

Possible Answers:

$\phi _{1}<\phi _{2}$ $\displaystyle \phi _{1}< \phi _{2}$

$\phi _{1}>\phi _{2}$ $\displaystyle \phi _{1}>\phi _{2}$

$\phi _{1}=\phi _{2}=0$ $\displaystyle \phi _{1}=\phi _{2}=0$

$\phi _{1}=\phi _{2}$ $\displaystyle \phi _{1}=\phi _{2}$

Correct answer:

$\phi _{1}=\phi _{2}$ $\displaystyle \phi _{1}=\phi _{2}$

Explanation:

Electric flux is given by either side of the equation of Gauss's Law:

$\phi _{e} = \oint \vec{E}\cdot d\vec{A}=\frac{Q}{\varepsilon_{o} }$ $\displaystyle \phi _{e} = \oint \vec{E}\cdot d\vec{A}=\frac{Q}{\varepsilon_{o} }$

Since the charge is the same for both spherical surfaces, even though these surfaces are of different radii, the amounts of electric flux passing through each surface is the same.

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Example Question #15 : Electricity And Magnetism Exam

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

Possible Answers:

1*10^7V $\displaystyle 1*10^7V$

1*10^8V $\displaystyle 1*10^8V$

$1*10^{10}V$ $\displaystyle 1*10^{10}V$

1*10^9V $\displaystyle 1*10^9V$

Correct answer:

1*10^8V $\displaystyle 1*10^8V$

Explanation:

Potential difference is given by the change in voltage

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

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.

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

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

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

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Example Question #16 : Electricity And Magnetism Exam

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

Find the expression for electric field produced by the ring.

Possible Answers:

$\frac{1}{4\pi\epsilon_0}\frac{q}{(x^2+a^2)^{1/2}}}$ $\displaystyle \frac{1}{4\pi\epsilon_0}\frac{q}{(x^2+a^2)^{1/2}}}$

$\frac{1}{4\pi\epsilon_0}\frac{qx}{(x^2+a^2)^{3/2}}}$ $\displaystyle \frac{1}{4\pi\epsilon_0}\frac{qx}{(x^2+a^2)^{3/2}}}$

$\frac{1}{4\pi\epsilon_0}\frac{q}{(x^2+a^2)^{3/2}}}$ $\displaystyle \frac{1}{4\pi\epsilon_0}\frac{q}{(x^2+a^2)^{3/2}}}$

$\frac{1}{4\pi\epsilon_0}\frac{qx}{(x^2+a^2)^{1/2}}}$ $\displaystyle \frac{1}{4\pi\epsilon_0}\frac{qx}{(x^2+a^2)^{1/2}}}$

Correct answer:

$\frac{1}{4\pi\epsilon_0}\frac{qx}{(x^2+a^2)^{3/2}}}$ $\displaystyle \frac{1}{4\pi\epsilon_0}\frac{qx}{(x^2+a^2)^{3/2}}}$

Explanation:

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

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

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

Take the derivative and simplify.

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

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Example Question #17 : Electricity And Magnetism Exam

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

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

What is the electric field at a point located at a distance $\displaystyle r$ from the surface of the cylinder?

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

Now, we can take the derivative and simplify.

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

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

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Example Question #18 : Electricity And Magnetism Exam

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

Possible Answers:

$1.61*10^{-11}J$ $\displaystyle 1.61*10^{-11}J$

$6.44*10^{-11}J$ $\displaystyle 6.44*10^{-11}J$

$0.8*10^{-11}J$ $\displaystyle 0.8*10^{-11}J$

$3.22*10^{-11}J$ $\displaystyle 3.22*10^{-11}J$

Correct answer:

$1.61*10^{-11}J$ $\displaystyle 1.61*10^{-11}J$

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.

W=qEd $\displaystyle W=qEd$

We are given the charge ( $1.61*10^{-19}\text{C}$ $\displaystyle 1.61*10^{-19}\text{C}$), the distance ($\displaystyle 5m$), and the field strength ( $2*10^7 \frac{V}{m}$ $\displaystyle 2*10^7 \frac{V}{m}$), allowing us to calculate the work.

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

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

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

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Example Question #1 : Calculating Electric Potential

A negative charge of magnitude $9.0 \mu C$ $\displaystyle 9.0 \mu C$ is placed in a uniform electric field of $3.0 * 10^4 \frac{N}{C}$ $\displaystyle 3.0 * 10^4 \frac{N}{C}$, directed upwards. If the charge is moved 1.0m $\displaystyle 1.0m$ upwards, how much work is done on the charge by the electric field in this process?

Possible Answers:

$-3.3 * 10^{-6}J$ $\displaystyle -3.3 * 10^{-6}J$

$3.0 * 10^{-13} J$ $\displaystyle 3.0 * 10^{-13} J$

$-3.0 * 10^{-13} J$ $\displaystyle -3.0 * 10^{-13} J$

$-2.7 * 10^{-4}J$ $\displaystyle -2.7 * 10^{-4}J$

$2.7 * 10^{4}J$ $\displaystyle 2.7 * 10^{4}J$

Correct answer:

$-2.7 * 10^{-4}J$ $\displaystyle -2.7 * 10^{-4}J$

Explanation:

Relevant equations:

$\Delta V = E * d$ $\displaystyle \Delta V = E * d$

$W = q \Delta V$ $\displaystyle W = q \Delta V$

Given:

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

$\displaystyle d = 1.0 m$

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

First, find the potential difference between the initial and final positions:

$\Delta V = (3.0 * 10^4 \frac{N}{C})(1.0m)= 3.0 * 10^4 \frac{N\cdot m}{C}$ $\displaystyle \Delta V = (3.0 * 10^4 \frac{N}{C})(1.0m)= 3.0 * 10^4 \frac{N\cdot m}{C}$

2. Plug this potential difference into the work equation to solve for W:

$W = (-9.0 * 10^{-9}C)(3.0 * 10^4 \frac{N\cdot m}{C})$ $\displaystyle W = (-9.0 * 10^{-9}C)(3.0 * 10^4 \frac{N\cdot m}{C})$

$W = -2.7 * 10^{-4}J$ $\displaystyle W = -2.7 * 10^{-4}J$

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Example Question #2 : Calculating Electric Potential

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

Ps0_threechargepotential

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

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

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

Possible Answers:

$1.05*10^{12}V$ $\displaystyle 1.05*10^{12}V$

$4.05*10^{11}V$ $\displaystyle 4.05*10^{11}V$

$3.41*10^{11}V$ $\displaystyle 3.41*10^{11}V$

$2.65*10^{12}V$ $\displaystyle 2.65*10^{12}V$

Correct answer:

$4.05*10^{11}V$ $\displaystyle 4.05*10^{11}V$

Explanation:

Electric potential is a scalar quantity given by the equation:

$V = \frac{kq_{i}}{r_{i}}$ $\displaystyle V = \frac{kq_{i}}{r_{i}}$

To find the total potential at the origin due to the three charges, add the potentials of each charge.

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

$V_{total}=\frac{(9*10^9)(3C)}{0.02m}+\frac{(9*10^9)(-5C)}{0.04m}+\frac{(9*10^9)(1C)}{0.05m}$ $\displaystyle V_{total}=\frac{(9*10^9)(3C)}{0.02m}+\frac{(9*10^9)(-5C)}{0.04m}+\frac{(9*10^9)(1C)}{0.05m}$

$V_{total}=1.35*10^{12}V+(-1.125*10^{12}V)+1.8*10^{11}V$ $\displaystyle V_{total}=1.35*10^{12}V+(-1.125*10^{12}V)+1.8*10^{11}V$

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

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All AP Physics C Electricity Resources

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AP Physics C Electricity : Electricity

Study concepts, example questions & explanations for AP Physics C Electricity

All AP Physics C Electricity Resources

Example Questions

Example Question #161 : Ap Physics C

Example Question #12 : Electricity And Magnetism Exam

Example Question #13 : Electricity And Magnetism Exam

Example Question #12 : Electricity And Magnetism Exam

Example Question #15 : Electricity And Magnetism Exam

Example Question #16 : Electricity And Magnetism Exam

Example Question #17 : Electricity And Magnetism Exam

Example Question #18 : Electricity And Magnetism Exam

Example Question #1 : Calculating Electric Potential

Example Question #2 : Calculating Electric Potential

All AP Physics C Electricity Resources

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