Redistribution of Charge Between Conductors
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AP Physics C: Electricity and Magnetism › Redistribution of Charge Between Conductors
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?
$$+3 \mu C$$
$$+9 \mu C$$
$$+4.5 \mu C$$
$$+6 \mu C$$
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 $$q_1 = +3 \mu C$$.
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.
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.
A charged object $$X$$ is used to charge a conductor $$Y$$ by induction. Which of the following statements is true about the charge on $$Y$$?
The charge on $$Y$$ will have the opposite sign to the charge on $$X$$.
The charge on $$Y$$ will be neutral, but polarized.
The charge on $$Y$$ is zero unless physical contact is made with $$X$$.
The charge on $$Y$$ will have the same sign as the charge on $$X$$.
Explanation
Charging by induction involves using a charged object to polarize a conductor, which is then grounded. For instance, if object X is positive, it attracts electrons in Y to the near side. Grounding allows more electrons to flow onto Y. Removing the ground traps these electrons. When X is removed, Y is left with a net negative charge, opposite to X.
Which of the following statements correctly describes the system in its final state?
The total electric charge of the system is conserved, and the electric potential is the same for both spheres.
The total electric charge of the system is conserved, and the electric fields at the surfaces are equal in magnitude.
The electric charge is equally distributed between the spheres, and their electric potentials are equal.
The total electric potential energy of the system is conserved, and the surface charge densities are equal.
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.
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
+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.
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.
The sphere will remain neutral.
The sphere will have a separated positive and negative charge.
The sphere will be positively charged.
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.
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 $$+Q$$, as it is the original charged object.
$$q$$ is approximately equal to $$-Q$$, due to induction effects.
$$q$$ is a very small fraction of $$+Q$$, approaching zero.
$$q$$ is approximately equal to $$+Q/2$$, as charge is shared.
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 $$Q$$.
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.
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.
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?
$$0$$
The charge cannot be determined without the shell's radius.
$$-Q$$
$$+Q$$
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.
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...
decreases, as energy is dissipated during charge redistribution.
increases, as work is done to move the charges.
becomes zero, as the final potential is lower than the initial potential.
remains constant, because charge is conserved.
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.