Question

Three concentric conducting spherical shells of radii R, 2R and 3R carry charges Q, -2Q and 3Q respectively. The charges at the six surfaces (starting from the innermost ) are

• A. 0, Q, -Q, -Q, +Q, +2Q
• B. 0, Q, -2Q, +Q, +Q, +2Q
• C. 0, Q, 0, -2Q, 0, +3Q
• D. 0, Q, -2Q, 0, +Q, +2Q

Answer 1

Answer: (A)

A 0, Q, -Q, -Q, +Q, +2Q

Answer 2

The problem involves finding the charge distribution on the surfaces of three concentric conducting spherical shells. The radii are R, 2R, and 3R, and the total charges on the shells are Q, -2Q, and 3Q respectively. Let the shells be denoted as $S_1, S_2, S_3$ from innermost to outermost.

For a conductor in electrostatic equilibrium, the electric field inside the conductor is zero. Also, any net charge on a conductor resides on its surface. For a hollow conductor, the charge can be distributed on both the inner and outer surfaces.

Shell $S_1$ (radius R, total charge $Q_1 = Q$):
Let $q_{1,in}$ and $q_{1,out}$ be the charges on the inner and outer surfaces of $S_1$.
$q_{1,in} + q_{1,out} = Q_1 = Q$.
Consider a Gaussian surface within the material of $S_1$. The electric field is zero, so the total charge enclosed is zero. The only charge enclosed is $q_{1,in}$. Thus, $q_{1,in} = 0$.
Then, $q_{1,out} = Q - q_{1,in} = Q - 0 = Q$.
Charges on $S_1$: inner surface 0, outer surface Q.

Shell $S_2$ (radius 2R, total charge $Q_2 = -2Q$):
Let $q_{2,in}$ and $q_{2,out}$ be the charges on the inner and outer surfaces of $S_2$.
$q_{2,in} + q_{2,out} = Q_2 = -2Q$.
Consider a Gaussian surface within the material of $S_2$. The electric field is zero, so the total charge enclosed is zero. The charges enclosed are the total charge on $S_1$ and the charge on the inner surface of $S_2$.
Total charge enclosed = $Q_1 + q_{2,in} = Q + q_{2,in}$.
Since the total charge enclosed is zero, $Q + q_{2,in} = 0$, which gives $q_{2,in} = -Q$.
Then, $q_{2,out} = Q_2 - q_{2,in} = -2Q - (-Q) = -2Q + Q = -Q$.
Charges on $S_2$: inner surface -Q, outer surface -Q.

Shell $S_3$ (radius 3R, total charge $Q_3 = 3Q$):
Let $q_{3,in}$ and $q_{3,out}$ be the charges on the inner and outer surfaces of $S_3$.
$q_{3,in} + q_{3,out} = Q_3 = 3Q$.
Consider a Gaussian surface within the material of $S_3$. The electric field is zero, so the total charge enclosed is zero. The charges enclosed are the total charge on $S_1$, the total charge on $S_2$, and the charge on the inner surface of $S_3$.
Total charge enclosed = $Q_1 + Q_2 + q_{3,in} = Q + (-2Q) + q_{3,in} = -Q + q_{3,in}$.
Since the total charge enclosed is zero, $-Q + q_{3,in} = 0$, which gives $q_{3,in} = Q$.
Then, $q_{3,out} = Q_3 - q_{3,in} = 3Q - Q = 2Q$.
Charges on $S_3$: inner surface Q, outer surface 2Q.

The charges at the six surfaces, starting from the innermost, are:

1. Inner surface of shell with radius R: $q_{1,in} = 0$
2. Outer surface of shell with radius R: $q_{1,out} = Q$
3. Inner surface of shell with radius 2R: $q_{2,in} = -Q$
4. Outer surface of shell with radius 2R: $q_{2,out} = -Q$
5. Inner surface of shell with radius 3R: $q_{3,in} = Q$
6. Outer surface of shell with radius 3R: $q_{3,out} = 2Q$

The sequence of charges is 0, Q, -Q, -Q, Q, 2Q.

Comparing this sequence with the given options, option A matches our calculated charges.