Redistribution of Charge Between Conductors - AP Physics C Electricity and Magnetism

Which of the following describes the initial flow of charge once the wire is connected?
Answer: Positive charge flows from sphere A to sphere B. Conventional current (positive charge flow) moves from regions of higher electric potential to lower electric potential. Since sphere A is at +100 V and sphere B is at +50 V, positive charge flows from A to B until both spheres reach the same equilibrium potential.
Two isolated conducting spheres, Sphere 1 with radius $R$ and Sphere 2 with radius $2R$ , are far apart. They are connected by a thin conducting wire. If a total charge $+Q$ is placed on this system, how will the surface charge density $\sigma$ on Sphere 1 compare to that on Sphere 2 at equilibrium?
Answer: $\sigma_1 = 2\sigma_2$ At equilibrium, the potentials are equal: $V_1 = V_2$ . This means $k Q_1/R = k Q_2/(2R)$ , which simplifies to $Q_2 = 2Q_1$ . Surface charge density is $\sigma = Q/A = Q/(4\pi R^2)$ . So, $\sigma_1 = Q_1/(4\pi R^2)$ and $\sigma_2 = Q_2/(4\pi (2R)^2) = (2Q_1)/(16\pi R^2) = Q_1/(8\pi R^2)$ . Comparing them, we see that $\sigma_1 = 2\sigma_2$ .
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?
Answer: From the sphere with the higher potential to the sphere with the lower potential. 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.
What is the final net charge on sphere B after being grounded?
Answer: 0 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.
What is the final electric potential on the surface of the sphere of radius $R$ ?
Answer: $k \frac{2Q}{3R}$ 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 $V' = k \frac{Q_1'}{R} = k \frac{(2/3)Q}{R} = k \frac{2Q}{3R}$ .
How does the surface charge density $\sigma_r$ on the small sphere compare to the surface charge density $\sigma_R$ on the large sphere?
Answer: $\sigma_r = (R/r) \sigma_R$ 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.
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...
Answer: decreases, as energy is dissipated during charge redistribution. 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.
A solid conducting sphere of radius $R$ is given a net charge $+Q$ . A second, initially neutral solid conducting sphere of radius $3R$ is placed far away. If the spheres are connected by a thin conducting wire, which statement best describes the final electric field at the surface of each sphere?
Answer: The field is stronger at the surface of the smaller sphere. When connected, their potentials become equal: $V_s = V_l$ . The electric field at the surface of a sphere is $E = kQ/R^2$ , and its potential is $V = kQ/R$ . Thus, $E = V/R$ . Since $V_s = V_l = V$ , we have $E_s = V/R$ and $E_l = V/(3R)$ . Therefore, the electric field at the surface of the smaller sphere ( $E_s$ ) is three times stronger than the field at the surface of the larger sphere ( $E_l$ ).
After the grounding wire is removed, but while the sphere remains in the external field, what is the net charge on the sphere?
Answer: The sphere acquires a net charge that creates an internal opposing field. When placed in the external field, the sphere polarizes. Grounding it forces its potential to zero. To maintain zero potential throughout its volume while in the external field, charge must flow from the ground onto the sphere. This induced net charge creates its own electric field that, combined with the field from the polarized charges, perfectly cancels the external field inside the conductor. Removing the ground traps this net charge.
After the wire to ground is removed, what is the net charge on the sphere?
Answer: The charge becomes zero. 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.
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?
Answer: $+3 \mu C$ 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 $q_1 = +3 \mu C$ .
A charged object $X$ is used to charge a conductor $Y$ by induction. Which of the following statements is true about the charge on $Y$ ?
Answer: The charge on $Y$ will have the opposite sign to the charge on $X$ . 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.
What is the final charge on the outer surface of the shell (at radius $R_3$ )?
Answer: $-Q$ 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 $R_3$ .
A spherical conductor of radius $R$ with charge $Q$ is brought into contact with an identical uncharged sphere, and then separated. This process is repeated with $N-1$ other identical uncharged spheres, one by one. What is the charge remaining on the first sphere after it has touched all $N$ spheres (including itself)?
Answer: $Q/2^{N-1}$ Each time the charged sphere touches an identical uncharged sphere, its charge is halved. After touching the first uncharged sphere (the 2nd sphere in total), its charge is $Q/2$ . After touching the second uncharged sphere (the 3rd sphere in total), its charge becomes $(Q/2)/2 = Q/4 = Q/2^2$ . After touching the $(n-1)^{th}$ uncharged sphere, its charge will be $Q/2^{n-1}$ . So after touching $N-1$ other spheres, the charge is $Q/2^{N-1}$ .
After contact is made, what is the charge on the outer surface of the hollow shell?
Answer: $+q$ 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.
A large, neutral, hollow conducting box encloses a smaller, isolated conducting sphere with a charge $+q$ . If the small sphere is brought into contact with the inner surface of the box, what is the final charge on the outer surface of the box?
Answer: $+q$ When the charged sphere touches the inner surface of the conductor, the sphere and the box form a single conductor. In electrostatic equilibrium, all net charge resides on the outermost surface. Therefore, the entire charge $+q$ from the small sphere will transfer to the outer surface of the box. The small sphere will be left with no net charge.
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?
Answer: +Q/2 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.
Three identical conducting spheres, A, B, and C, have initial charges of $+6q$ , $-2q$ , and $+5q$ , respectively. Sphere A is touched to Sphere B and then they are separated. Then, Sphere A is touched to Sphere C and they are separated. What is the final charge on Sphere A?
Answer: $+3.5q$ Step 1: Sphere A ( $+6q$ ) touches Sphere B ( $-2q$ ). The total charge is $+6q - 2q = +4q$ . Since they are identical, they share this charge equally. After separation, A has $+2q$ and B has $+2q$ . Step 2: Sphere A ( $+2q$ ) touches Sphere C ( $+5q$ ). The total charge is $+2q + 5q = +7q$ . They share this charge equally. After separation, A has $+3.5q$ and C has $+3.5q$ .
After the spheres are separated, the charge remaining on the small sphere is $q$ . Which statement is the best description of $q$ ?
Answer: $q$ is a very small fraction of $+Q$ , approaching zero. 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 $Q$ .
What is the final charge distribution on the sphere?
Answer: A net positive charge distributed uniformly over the surface. The negative rod repels the free electrons in the sphere to the far (right) side. When grounded, these repelled electrons have a path to flow away from the sphere into the ground. After the ground is removed, the sphere is left with a deficit of electrons, resulting in a net positive charge. When the inducing rod is removed, this net positive charge distributes itself uniformly over the sphere's surface due to mutual repulsion.

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