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AP Physics C Electricity : AP Physics C

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

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All AP Physics C Electricity Resources

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

Example Questions

Example Question #4 : Electricity

A point charge of $q_1=10{nC}$ exerts a force of 235N on another charge with $q_2=2{nC}$ . How far apart are the two charges?

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

Possible Answers:

$1.2*10^{-2}m$

$2.77*10^{-5}m$

$4.9*10^{4}m$

$5.82*10^{-4}m$

$7.5*10^{-6}m$

Correct answer:

$2.77*10^{-5}m$

Explanation:

To find the distance between the two charges, use Coulomb's Law.

$F=\frac{kq_1q_2}{r^2}$

Since we want to find distance, , we solve for .

$r=\sqrt{\frac{kq_1q_2}{F}}$

We know the values of the force and the two charges.

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

$q_1=10{nC} = 10* 10^{-9}{C}$

$q_2=2{nC}=2* 10^{-9}{C}$

$F=235\ \text{N}$

We can plug in these values and solve for the distance.

$r=\sqrt{\frac{(8.99*10^9\frac{{Nm}^2}{{C}^2})(10* 10^{-9}{C})(2* 10^{-9}{C})}{235{N}}}$

$r=\sqrt{\frac{1.8*10^{-7}N*m^2}{235{N}}}$

$r=\sqrt{7.65*10^{-10}m^2}$

$r=2.77* 10^{-5}{m}$

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Example Question #5 : Electricity

What is the electric force between two charges, q_1=50nC and q_2=50nC , located 5cm apart?

$\epsilon_0=8.85* 10^{-12} \frac{F}{m}$

Possible Answers:

$9.0*10^{-3}N$

$1.8*10^{-3}N$

$4.5*10^{-3}N$

$6.7*10^{-3}N$

Correct answer:

$9.0*10^{-3}N$

Explanation:

The equation for finding the electric force between two charges is $F=\frac{kq_1q_2}{r^2}$ , where $k=\frac{1}{4\pi \epsilon_0}$ . Using this, we can rewrite the force equation.

$F=\frac{1}{4\pi \epsilon_0}\frac{q_1q_2}{r^2}$

Now, we can use the values given in the question to solve for the electric force between the two particles.

$F=\frac{1}{4\pi(8.85*10^{-12}\frac{F}{m})}\frac{(50*10^{-9}C)(50*10^{-9}C)}{(0.05m)^2}$

$F=\frac{1}{1.11*10^{-10}\frac{F}{m}}\frac{2.5*10^{-15}C^2}{0.0025m^2}$

$F=9*10^{9}\frac{m}{F}(1*10^{-12}\frac{C^2}{m^2})$

$F=9*10^{-3}N$

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Example Question #1 : Electricity

What is the magnitude of the electric field at a field point from a point charge of q=5nC ?

$\epsilon_0=8.85* 10^{-12} \frac{F}{m}$

Possible Answers:

$11.25\frac{N}{C}$

$180.0\frac{N}{C}$

$45.0\frac{N}{C}$

$2.25\frac{N}{C}$

Correct answer:

$45.0\frac{N}{C}$

Explanation:

The equation to find the strength of an electric field is $E=\frac{kq}{r^2}=\frac{1}{4\pi \epsilon_0}\frac{q}{r^2}$ .

We can use the given values to solve for the strength of the field at a distance of .

$E=\frac{1}{4\pi(8.85*10^{-12}\frac{F}{m})}\frac{5*10^{-9}C}{1m}$

$E=\frac{5*10^{-9}C}{1.11*10^{-10}\frac{F}{m}}$

$E=45\frac{N}{C}$

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Example Question #161 : Ap Physics C

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

Possible Answers:

2.67mF

4mF

16mF

12mF

Correct answer:

16mF

Explanation:

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

$C_{eq}=C_1+C_2$

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

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

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Example Question #162 : Ap Physics C

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 force on the proton.

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

Possible Answers:

$6.44*10^{-12}N$

$0.8*10^{-12}N$

$3.22*10^{-12}N$

$1.61*10^{-12}N$

Correct answer:

$3.22*10^{-12}N$

Explanation:

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

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

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Example Question #3 : Using Coulomb's Law

Two point charges, Q_1 and Q_2 are separated by a distance .

Ps0_twochargeefield

The values of the charges are:

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

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

The distance is 4.0cm. The point lies 1.5cm away from 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 due to the two charges?

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

Possible Answers:

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

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

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

Correct answer:

Explanation:

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

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

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

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

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 .

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

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Example Question #1 : Using Gauss's Law

A charge, , is enclosed by two spherical surfaces of radii r_1 and , with 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:

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

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

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

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

Correct answer:

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

Explanation:

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

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

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 #1 : 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 $1.61*10^{-19}\text{C}$ .

Possible Answers:

1*10^9V

1*10^7V

$1*10^{10}V$

1*10^8V

Correct answer:

1*10^8V

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.

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

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Example Question #2 : Calculating Electric Potential

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.

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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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Example Question #3 : Calculating Electric Potential

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?

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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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All AP Physics C Electricity Resources

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AP Physics C Electricity : AP Physics C

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

All AP Physics C Electricity Resources

Example Questions

Example Question #4 : Electricity

Example Question #5 : Electricity

Example Question #1 : Electricity

Example Question #161 : Ap Physics C

Example Question #162 : Ap Physics C

Example Question #3 : Using Coulomb's Law

Example Question #1 : Using Gauss's Law

Example Question #1 : Calculating Electric Potential

Example Question #2 : Calculating Electric Potential

Example Question #3 : Calculating Electric Potential

All AP Physics C Electricity Resources

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