AP Physics C Electricity and Magnetism Quiz: Redistribution Of Charge Between Conductors

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Redistribution Of Charge Between ConductorsQuestion 1 of 20

A conducting sphere of radius $R$ has an initial charge of $-6 \mu C$ . It is connected by a very long wire to a distant conducting sphere of radius $2R$ that has an initial charge of $+15 \mu C$ . After electrostatic equilibrium is reached, what is the final charge on the smaller sphere?

+3 \mu C

+4.5 \mu C

+6 \mu C

+9 \mu C

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AP Physics C Electricity and Magnetism Quiz

AP Physics C Electricity and Magnetism Quiz: Redistribution Of Charge Between Conductors

Practice Redistribution Of Charge Between Conductors in AP Physics C Electricity and Magnetism with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Redistribution Of Charge Between Conductors, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics C Electricity and Magnetism.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

$+3 \mu C$ (correct answer)
$+4.5 \mu C$
$+6 \mu C$
$+9 \mu C$

Explanation: The total charge is conserved:

Q_{total} = -6 \mu C + 15 \mu C = +9 \mu C

. When connected, their potentials are equal:

k q_1/R = k q_2/(2R)

, which means

q_2 = 2q_1

. The total charge is distributed such that

q_1 + q_2 = Q_{total}

. Substituting gives

q_1 + 2q_1 = 9 \mu C

, so

3q_1 = 9 \mu C

. Therefore, the final charge on the smaller sphere is

q_1 = +3 \mu C

Question 2

A neutral conducting sphere rests on an insulating stand. A positively charged rod is brought near, but not touching, the left side of the sphere. While the rod is held in place, the right side of the sphere is connected to ground. The ground connection is then removed, and finally the rod is moved far away.

What is the final state of the conducting sphere?

It is electrically neutral.
It has a net positive charge distributed uniformly.
It has a net negative charge distributed uniformly. (correct answer)
It is polarized with an excess of negative charge on the left and positive charge on the right.

Explanation: The positive rod attracts electrons to the left side of the sphere. When grounded, more electrons are attracted from the ground onto the sphere. Removing the ground connection traps this excess negative charge. Once the inducing rod is removed, this net negative charge spreads out uniformly over the surface of the sphere due to mutual repulsion.

Question 3

Two isolated conducting spheres with different initial charges and different radii are connected by a conducting wire. The system is allowed to reach electrostatic equilibrium.

Which of the following statements correctly describes the system in its final state?

The electric charge is equally distributed between the spheres, and their electric potentials are equal.
The total electric charge of the system is conserved, and the electric potential is the same for both spheres. (correct answer)
The total electric potential energy of the system is conserved, and the surface charge densities are equal.
The total electric charge of the system is conserved, and the electric fields at the surfaces are equal in magnitude.

Explanation: For an isolated system, total charge is conserved. When conductors are connected, charge flows between them until they reach the same electric potential. Charge is generally not distributed equally unless the conductors are identical. Potential energy is not conserved, as some energy is dissipated as heat during charge redistribution. Surface charge densities and fields are typically unequal.

Question 4

Two identical solid conducting spheres, A and B, are isolated from their surroundings. Sphere A has a net charge of +Q and sphere B is electrically neutral. The spheres are brought into contact and then separated. What is the final net charge on sphere B?

+Q/2 (correct answer)
+Q
0
-Q/2

Explanation: When two identical conducting spheres are brought into contact, the total net charge of the system distributes itself equally between the two spheres to reach electrostatic equilibrium, where both spheres are at the same potential. The total charge is Q + 0 = Q. Therefore, each sphere will have a final charge of +Q/2.

Question 5

A positively charged rod is brought near, but does not touch, a neutral, isolated conducting sphere. The sphere is then connected to the ground with a conducting wire. What is the net charge on the sphere after the grounding wire is disconnected, and then the rod is removed?

The sphere will be negatively charged. (correct answer)
The sphere will be positively charged.
The sphere will remain neutral.
The sphere will have a separated positive and negative charge.

Explanation: This is charging by induction. The positive rod attracts electrons in the sphere to the side near the rod, leaving the far side with a net positive charge. When grounded, electrons flow from the ground to the sphere, neutralizing the far side and giving the sphere a net negative charge. Disconnecting the ground first traps this excess negative charge. When the rod is removed, the negative charge spreads uniformly over the sphere.

Question 6

A small, isolated conducting sphere with a charge $+Q$ is brought into contact with a very large, isolated, and electrically neutral conducting sphere.

After the spheres are separated, the charge remaining on the small sphere is $q$ . Which statement is the best description of $q$ ?

$q$ is approximately equal to $+Q/2$ , as charge is shared.
$q$ is approximately equal to $+Q$ , as it is the original charged object.
$q$ is a very small fraction of $+Q$ , approaching zero. (correct answer)
$q$ is approximately equal to $-Q$ , due to induction effects.

Explanation: When in contact, the spheres reach the same potential. Let the radii be

r

and

R

, with

R \gg r

. Final charges

q

and

Q_{large}

satisfy

k q/r = k Q_{large}/R

and

q + Q_{large} = Q

. This leads to

q = Q / (1 + R/r)

. Since

R/r

is a very large number, the denominator is very large, making

q

a very small fraction of the original charge

Q

Question 7

A hollow, neutral conducting spherical shell surrounds a point charge $+Q$ located at its center. The outer surface of the shell is then grounded. What is the net charge on the outer surface of the shell?

$0$ (correct answer)
$-Q$
$+Q$
The charge cannot be determined without the shell's radius.

Explanation: The central charge

+Q

induces

-Q

on the inner surface of the shell. Without grounding,

+Q

would be induced on the outer surface. Grounding connects the outer surface to a reservoir at zero potential. The charge on the outer surface and the central charge determine the potential of the shell. To make the potential zero, the

+Q

that would have been on the outer surface flows to the ground. Thus, the final charge on the outer surface is zero.

Question 8

A conducting sphere with an initial charge $Q_1$ and potential $V_1$ is connected by a wire to a distant, uncharged conducting sphere with potential $V_2 = 0$ . Charge flows from the first sphere to the second. During this process, the total electrostatic potential energy of the system...

increases, as work is done to move the charges.
remains constant, because charge is conserved.
decreases, as energy is dissipated during charge redistribution. (correct answer)
becomes zero, as the final potential is lower than the initial potential.

Explanation: When charge flows through the connecting wire, which has some resistance, energy is dissipated as heat (

I^2R

loss). Also, the accelerating charges radiate electromagnetic energy. The final configuration has a lower total electrostatic potential energy than the initial configuration. The only case where energy would be constant is if the initial potentials were already equal, in which case no charge would flow.

Question 9

A solid conducting sphere A has an initial charge of $+Q$ . It is brought into contact with an identical, but neutral, solid conducting sphere B. After they are separated, sphere B is connected by a wire to ground.

What is the final net charge on sphere B after being grounded?

$+Q/4$
$+Q/2$
0 (correct answer)
$-Q/2$

Explanation: When sphere A touches identical sphere B, the total charge

+Q

is shared equally. Sphere B acquires a charge of

+Q/2

. When any conductor is connected to ground, charge flows until the conductor's potential becomes zero. For an isolated conductor, this means its net charge must become zero. Therefore, electrons flow from the ground to neutralize the

+Q/2

charge on sphere B.

Question 10

A conducting sphere of radius $R$ has an initial charge of $+3Q$ . A second conducting sphere of radius $2R$ has an initial charge of $-Q$ . The spheres are far apart and are then connected by a long conducting wire.

What is the final electric potential on the surface of the sphere of radius $R$ ?

$k \frac{2Q}{3R}$ (correct answer)
$k \frac{Q}{R}$
$k \frac{2Q}{R}$
$k \frac{4Q}{3R}$

Explanation: Total charge is conserved:

Q_{tot} = 3Q + (-Q) = 2Q

. After connecting, the potentials are equal:

V' = k \frac{Q_1'}{R} = k \frac{Q_2'}{2R}

, which implies

Q_2' = 2Q_1'

. Since

Q_1' + Q_2' = 2Q

, we have

Q_1' + 2Q_1' = 2Q

, so

3Q_1' = 2Q

and

Q_1' = \frac{2}{3}Q

. The final potential is

V' = k \frac{Q_1'}{R} = k \frac{(2/3)Q}{R} = k \frac{2Q}{3R}

Question 11

A neutral, solid conducting sphere is placed in a region of uniform external electric field pointing to the right. The sphere is then grounded on the right side. The ground connection is removed, and then the external field is turned off. What is the final state of the sphere?

The sphere has a net positive charge distributed uniformly. (correct answer)
The sphere has a net negative charge distributed uniformly.
The sphere is neutral.
The sphere is polarized with positive charge on the right and negative on the left.

Explanation: The external field pushes positive charge to the right and pulls negative charge (electrons) to the left side of the sphere. The sphere's potential is non-uniform before grounding. Grounding brings the point of contact to zero potential. Since the left side is at a negative potential relative to the point of contact, electrons flow from the left side of the sphere to the ground, leaving the sphere with a net positive charge. When the ground and then the field are removed, this net positive charge distributes uniformly.

Question 12

A neutral conducting sphere is briefly touched by a negatively charged rod. The rod is then removed. What is the process of charge transfer and the final state of the sphere?

Protons are transferred from the sphere to the rod, leaving the sphere negatively charged.
Electrons are transferred from the rod to the sphere, leaving the sphere negatively charged. (correct answer)
Protons are transferred from the rod to the sphere, leaving the sphere positively charged.
Electrons are transferred from the sphere to the rod, leaving the sphere positively charged.

Explanation: This process is charging by contact. In conductors, electrons are the mobile charge carriers. Since the rod is negatively charged, it has an excess of electrons. When it touches the neutral sphere, electrons are repelled from the rod and flow onto the sphere until the potential is equalized. This leaves the sphere with a net negative charge.

Question 13

Two spherical conductors have radii $r_1$ and $r_2$ and initial charges $q_1$ and $q_2$ . They are connected by a conducting wire. In which direction will charge flow?

From the sphere with the larger charge to the sphere with the smaller charge.
From the sphere with the higher potential to the sphere with the lower potential. (correct answer)
From the sphere with the larger radius to the sphere with the smaller radius.
From the sphere with the higher surface charge density to the one with lower density.

Explanation: Positive charge carriers, by convention, flow from a region of higher electric potential to a region of lower electric potential, similar to how mass flows from higher gravitational potential to lower gravitational potential. The flow continues until the potentials are equal. The amount of charge, radius, or charge density alone does not determine the direction of flow; the potential, which depends on both charge and radius (

V=kQ/R

), is the determining factor.

Question 14

A small conducting sphere of radius $r$ has a net charge of $+Q$ . A larger, electrically neutral conducting sphere has a radius of $2r$ . The spheres are far apart and are then connected by a long, thin conducting wire.

After the system reaches electrostatic equilibrium, how does the final charge on the larger sphere, $Q_{large}$ , compare to the final charge on the smaller sphere, $Q_{small}$ ?

$Q_{large} = Q_{small}$
$Q_{large} = 2 Q_{small}$ (correct answer)
$Q_{large} = 4 Q_{small}$
$Q_{large} = \frac{1}{2} Q_{small}$

Explanation: When connected by a conducting wire, the two spheres form a single equipotential surface, so their final electric potentials must be equal. The potential of a spherical conductor is given by

V = kQ/R

. Setting the potentials equal:

V_{small} = V_{large} \Rightarrow k\frac{Q_{small}}{r} = k\frac{Q_{large}}{2r}

. This simplifies to

Q_{large} = 2 Q_{small}

Question 15

A conducting sphere of radius $R_1 = 5.0 \text{ cm}$ has an initial charge of $Q_1 = +12 \text{ nC}$ . It is connected by a long, thin conducting wire to a distant, initially neutral conducting sphere of radius $R_2 = 10.0 \text{ cm}$ .

After electrostatic equilibrium is reached, what is the approximate final charge on the sphere of radius $R_2$ ?

+4.0 nC
+6.0 nC
+8.0 nC (correct answer)
+12 nC

Explanation: The total charge of the isolated system is conserved:

Q_{tot} = +12 \text{ nC}

. When connected, the spheres reach the same potential,

V'

. Let the final charges be

Q_1'

and

Q_2'

. We have

Q_1' + Q_2' = 12 \text{ nC}

and

k \frac{Q_1'}{R_1} = k \frac{Q_2'}{R_2}

. This gives

Q_1' = Q_2' (R_1/R_2) = Q_2' (5/10) = 0.5 Q_2'

. Substituting into the charge conservation equation:

0.5 Q_2' + Q_2' = 12 \text{ nC} \Rightarrow 1.5 Q_2' = 12 \text{ nC} \Rightarrow Q_2' = 8.0 \text{ nC}

Question 16

A solid conducting sphere carries a net positive charge $+Q$ . It is momentarily connected by a conducting wire to a large, neutral object that represents an electrical ground.

After the wire to ground is removed, what is the net charge on the sphere?

The charge remains $+Q$ due to charge conservation.
The charge becomes zero. (correct answer)
The charge becomes $-Q$ due to electron flow.
The charge becomes approximately $+Q/2$ .

Explanation: The ground is considered an infinite reservoir of charge at a potential of zero volts. The positively charged sphere is at a high positive potential. When connected to ground, electrons flow from the ground (lower potential) to the sphere (higher potential) until the sphere's potential is also reduced to zero. For a conductor to have zero potential, its net charge must be zero.

Question 17

A student wishes to give a neutral, isolated conducting sphere a net negative charge using the process of induction. A positively charged rod is brought near the sphere without touching it. The sphere is then momentarily grounded.

What is the correct next step in the process to ensure the sphere is left with a net negative charge?

The grounding wire should be removed, and then the positively charged rod should be taken away. (correct answer)
The positively charged rod should be taken away, and then the grounding wire should be removed.
The grounding wire and the positively charged rod should be removed simultaneously to trap the charge.
The sphere should be touched with the positively charged rod while it is still grounded to transfer charge.

Explanation: With the positive rod nearby, electrons are attracted from the ground onto the sphere. To trap this excess negative charge, the path to the ground must be broken first by removing the grounding wire. Only after the sphere is isolated again should the inducing rod be removed. If the rod were removed first, the excess electrons would simply flow back to the ground.

Question 18

A small conducting sphere with charge $+q$ is placed at the center of a thick, hollow, and electrically neutral conducting spherical shell. The small sphere is then moved to touch the inner surface of the shell.

After contact is made, what is the charge on the outer surface of the hollow shell?

0
$+q$ (correct answer)
$-q$
The charge depends on the relative radii of the sphere and the shell.

Explanation: When the inner sphere touches the inner surface of the shell, they effectively become a single conductor. For any conductor in electrostatic equilibrium, all of its net excess charge must reside on its outermost surface. The total charge of the system is

+q

, so this entire charge moves to the outer surface of the shell.

Question 19

A solid conducting sphere of radius $R_1$ and charge $+Q$ is enclosed by a concentric conducting shell of inner radius $R_2$ and outer radius $R_3$ . The shell initially has a net charge of $-2Q$ . The sphere and shell are then connected by a thin conducting wire.

What is the final charge on the outer surface of the shell (at radius $R_3$ )?

$-2Q$
$-Q$ (correct answer)
0
$+Q$

Explanation: Connecting the sphere and shell with a wire makes them a single continuous conductor. The total net charge of this conductor is the sum of the initial charges:

(+Q) + (-2Q) = -Q

. In electrostatic equilibrium, all net charge on a conductor resides on its outermost surface. Thus, the entire charge of

-Q

will be on the outer surface at radius

R_3

Question 20

A small conducting sphere of radius $r$ and a large conducting sphere of radius $R > r$ are connected by a long conducting wire. The entire system is given a net positive charge, which distributes between the spheres.

How does the surface charge density $\sigma_r$ on the small sphere compare to the surface charge density $\sigma_R$ on the large sphere?

$\sigma_r = \sigma_R$
$\sigma_r = (R/r) \sigma_R$ (correct answer)
$\sigma_r = (r/R) \sigma_R$
$\sigma_r = (R/r)^2 \sigma_R$

Explanation: The spheres are at the same potential

V

. For a sphere,

V = kQ/a

and

Q = \sigma A = \sigma(4\pi a^2)

. So,

V = k\sigma(4\pi a^2)/a = 4\pi k a \sigma

. Since

V

is the same for both spheres,

4\pi k r \sigma_r = 4\pi k R \sigma_R

. This simplifies to

r\sigma_r = R\sigma_R

, or

\sigma_r = (R/r)\sigma_R

. Since

R > r

, the surface charge density is greater on the smaller sphere.