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AP Physics C: Mechanics : Electricity

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

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

Example Questions

Example Question #1 : Electricity And Magnetism Exam

A hollow metal sphere with a diameter of 10cm has a net charge of $4\mu C$ distributed uniformly across its surface. What is the magnitude of the field a distance 2.0m from the center of the sphere?

$k = 9.0 * 10^9 \frac{N\cdot m^2}{C^2}$

Possible Answers:

$9.0 * 10^3 \frac{N}{C}$

$1.8 * 10^4 \frac{N}{C}$

$2.7 * 10^{4} \frac{N}{C}$

$3.6 * 10^4 \frac{N}{C}$

$8.6*10^3 \frac{N}{C}$

Correct answer:

$8.6*10^3 \frac{N}{C}$

Explanation:

Relevant equations:

$E = \frac{kq}{r^2}$ (electric field of point charge)

Anywhere outside the metal sphere, the electric field is the same as it would be for a point charge of the same magnitude, located at the center of the sphere. So, calculate the electric field of a point charge given:

$q = 4.0 * 10^{-6}C$

r = 2.0m+0.05m=2.05m

$k = 9.0 * 10^9 \frac{N\cdot m^2}{C^2}$

Plugging in gives:

$E = \frac{(9.0*10^9\frac{N\cdot m^2}{C^2})(4.0 * 10^{-6}C)}{(2.05m)^2}=\frac{36000}{4.2} \frac{N}{C}=8.6*10^3 \frac{N}{C}$

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Example Question #2 : Electricity And Magnetism Exam

Two infinite parallel conducting sheets each have positive charge density $\sigma$ . What is the magnitude and direction of the electric field to the right of the right sheet?

Possible Answers:

$\frac{\sigma}{\epsilon _o}$ , to the right

$\frac{\sigma}{2 \epsilon _o}$ , to the right

$\frac{2 \sigma}{\epsilon _o}$ , to the left

$\frac{2 \sigma}{\epsilon _o}$ , to the right

Correct answer:

$\frac{\sigma}{\epsilon _o}$ , to the right

Explanation:

Relevant equations:

$E = \frac{\sigma }{2 \epsilon _o}$ (field due to single infinite plane)

Electric field is additive; in other words, the total electric field from the two planes is the sum of their individual fields:

$E_{total} = E_1 + E_2$

$E _{total} = \frac{\sigma}{2 \epsilon _o}+\frac{\sigma}{2 \epsilon _o}$

$E_{total}=\frac{\sigma}{\epsilon_o}$

The direction of the electric field is away from positive source charges. Thus, to the right of these positively charged planes, the field points away to the right.

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Example Question #3 : Electricity And Magnetism Exam

Four particles, each of charge , make up the four corners of a square with equal side lengths of . For the charge in the top left corner of the square, in what direction is the net force that it experiences due to its interactions with the other three particles?

Possible Answers:

Directly to the left

$45^\circ$ upwards of left

Directly to the right

$45^\circ$ downwards of right

Correct answer:

$45^\circ$ upwards of left

Explanation:

The correct answer is 45 degrees upwards of left. Since all particles have charge , all forces will be repulsive (there will be no attracting forces). The particle in the top-right corner creates a repulsive force directly to the left, and the particle in the bottom-left corner creates a repulsive force directly upwards. These are equal in magnitude, since they are both at distance from the top left corner. The bottom-right corner also creates a repulsive force, but acts along the same direction as the vector sum of a leftwards and upwards force.

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Example Question #3 : Electricity And Magnetism Exam

Consider a spherical capacitor made of two nested spheres. The smaller sphere has a radius of R_1 and a charge of , and lies within a larger sphere with radius R_2 and a charge of .

Which of the following equations accurately describes the capacitance of this spherical capacitor?

Possible Answers:

$C = \frac{R_{1}R_{2}}{k(R_{2} - R_{1})}$

$C = \frac{R_{1}}{k(R_{2} - R_{1})}{}$

Due to symmetry, this scenario would not produce capacitance

$C = \frac{R_{2}}{k(R_{2} - R_{1})}{}$

$C = \frac{(R_{2} - R_{1})}{R_{1}R_{2}}$

Correct answer:

$C = \frac{R_{1}R_{2}}{k(R_{2} - R_{1})}$

Explanation:

To solve this problem, we will need to derive an equation.

We know that:

$C=\left | \frac{Q}{V}\right |$

$V=\oint E\cdot dr$

We can use Gauss's law to derive the electric field between the two circles yielding:

$E = \frac{Q}{\epsilon_{0}4\pi r^2}$

Doing our integration with respect to from R_1 to R_2 , we get:

$V = \frac{-Q(R_{2} - R_{1})}{\epsilon _{0}4\pi R_{1}R_{2}}$

We can plug this back into our equation for capacitance to get:

$C = \frac{R_{1}R_{2}}{k(R_{2} - R_{1})}$

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Example Question #1 : Using Coulomb's Law

We have a point charge of 100nC . Determine the electric field at a distance of $500\mu m$ away from that charge.

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

Possible Answers:

$8.5*10^2\frac{N}{C}$

$2.7*10^8\frac{N}{C}$

$3.6*10^6\frac{N}{C}$

$3.6*10^9\frac{N}{C}$

$6.3*10^5\frac{N}{C}$

Correct answer:

$3.6*10^9\frac{N}{C}$

Explanation:

Coulomb's law for the electric field from point charges is $E=\frac{kq}{r^2}$ , where we know the values of the following variables.

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

$q=100\ \text{nC}= 100*10^{-9}\ \text{C}$

$r=500\ \mu\text{m}= 500*10^{-6}\ \text{m}$

Using these values, we can solve for the electric field.

$E=\frac{(8.99* 10^9\ \frac{\text{Nm}^2}{\text{C}^2})(100*10^{-9}\ \text{C})}{(500*10^{-6}\ \text{m})^2}$

$E=3.6* 10^9 \frac{\text{N}}{\text{C}}$

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Example Question #2 : Using Coulomb's Law

Two positive point charges of q_1 and q_2 are place at a distance $1000\mu m$ away from each other, as shown below. If a positive test charge, , is placed in between, at what distance away from q_1 will this test charge experience zero net force?

Charges

Possible Answers:

$9.7*10^{-2}m$

$4.9*10^{-3}m$

$4.3*10^{-4}m$

$5.1*10^{-3}m$

$4.5*10^{-4}m$

Correct answer:

$4.5*10^{-4}m$

Explanation:

To find the location at which the test charge experience zero net force, write the net force equation as $F_{net}=F_1-F_2=0$ , where F_1 is the force on the test charge from q_1 , and F_2 is the force on the same test charge from q_2 . Using Coulomb's law, we can rewrite the force equation and set it equal to zero.

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

$F_{net}=\frac{kq_1q}{r^2}-\frac{kq_2q}{(d-r)^2}=0$

In this equation, the distance, , is how far away the test charge is from q_1 , while (d-r) represents how far away the test charge is from q_2 . Now, we simplify and solve for .

Cross-multiply.

kq_1q(d-r)^2=kq_2qr^2

We can cancel and . We do not need to know these values in order to solve the question.

q_1(d-r)^2=q_2r^2

$\sqrt{q_1}(d-r)=\sqrt{q_2}r$

$\sqrt{q_1}d-\sqrt{q_1}r=\sqrt{q_2}r$

$\sqrt{q_1}d=r(\sqrt{q_2}+\sqrt{q_1})$

$r=\frac{\sqrt{q_1}d}{\sqrt{q_2}+\sqrt{q_1}}$

Now that we have isolated , we can plug in the values given in the question and solve.

$r=\frac{\sqrt{21* 10^{-9}\ \text{C}}(1000*10^{-6}\ \text{m})}{\sqrt{32*10^{-9}\ \text{C}}+\sqrt{21* 10^{-9}\ \text{C}}}$

$r=\frac{(1.45*10^{-4}\ \text{C})(1000*10^{-6}\ \text{m})}{1.79*10^{-4}\ \text{C}+1.45*10^{-4}\ \text{C}}$

$r=4.5*10^{-4}\ \text{m}$

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Example Question #3 : Using Coulomb's Law

You are standing on top of a very large positively charged metal plate with a surface charge of $\sigma=9\times10^{-8}\ \frac{\text{C}}{\text{m}^2}$ .

Assuming that the plate is infinitely large and your mass is 67kg , how much charge does your body need to have in order for you to float?

$\epsilon_0=8.85*10^{-12}\frac{C^2}{Nm^2}$

Possible Answers:

0.13mC

0.13C

$5.8\mu C$

2.5C

3.1C

Correct answer:

0.13C

Explanation:

Consider the forces that are acting on you. There is the downward (negative direction) force of gravity, . In order for you to float, there has to be an upward (positive direction) force, and that upward force is coming from the metal plate, . To show that you would float, the net force equation is written as $F_{net}=qE-mg=0$ , where is the charge on you.

For plates that are charged, know that $E=\frac{\sigma}{2\epsilon_0}$ .

Knowing this, the force equation becomes $q\frac{\sigma}{2\epsilon_0}-mg=0$ .

Solve for .

$q=\frac{2\epsilon_0mg}{\sigma}$

Now we can plug in our given values, and solve for the charge.

$q=\frac{2(8.85*10^{-12}\ \frac{\text{C}^2}{\text{Nm}^2})(67\ \text{kg})(9.8\ \frac{\text{m}}{\text{s}^2})}{9*10^{-8}\ \frac{\text{C}}{\text{m}^2}}$

$q=\frac{1.2*10^{-8}\frac{C^2}{m^2}}{9*10^{-8}\frac{C}{m^2}}$

$q=0.13\ \text{C}$

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Example Question #4 : Electricity And Magnetism Exam

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:

$4.9*10^{4}m$

$2.77*10^{-5}m$

$5.82*10^{-4}m$

$7.5*10^{-6}m$

$1.2*10^{-2}m$

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 #1 : Using Coulomb's Law

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:

$9.0*10^{-3}N$

$4.5*10^{-3}N$

$6.7*10^{-3}N$

$1.8*10^{-3}N$

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 : Using Coulomb's Law

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:

$180.0\frac{N}{C}$

$2.25\frac{N}{C}$

$11.25\frac{N}{C}$

$45.0\frac{N}{C}$

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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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 #1 : Electricity And Magnetism Exam

Example Question #2 : Electricity And Magnetism Exam

Example Question #3 : Electricity And Magnetism Exam

Example Question #3 : Electricity And Magnetism Exam

Example Question #1 : Using Coulomb's Law

Example Question #2 : Using Coulomb's Law

Example Question #3 : Using Coulomb's Law

Example Question #4 : Electricity And Magnetism Exam

Example Question #1 : Using Coulomb's Law

Example Question #1 : Using Coulomb's Law

All AP Physics C: Mechanics Resources

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