Figure shows three concentric thin spherical shells A, B and C of radii a, b, and c. The shells A and C are given charge q and -q, respectively, and shell B is earthed. Then,

Options

A. the charge q will be there at outer surface A
B. the charge `(-bq)/(c )` will be there at inner surface of C
C. the charge `-q` will be at inner surface of B
D. the charge `q(1-b//c)` will be there at outer surface of C

Answer

To solve the problem regarding the three concentric spherical shells A, B, and C, we will analyze the charge distribution and electric potential step by step.

### Step 1: Understanding the Configuration
We have three concentric spherical shells:
- Shell A with radius \( a \) and charge \( +q \)
- Shell B with radius \( b \) which is earthed (potential = 0)
- Shell C with radius \( c \) and charge \( -q \)

### Step 2: Electric Field Inside the Conductors
Since A, B, and C are conductors, the electric field inside each conductor must be zero. Therefore, the charge on each shell will redistribute itself such that the electric field inside each shell is null.

### Step 3: Charge Distribution on Shell B
Since shell B is earthed, its potential must be zero. The potential at shell B due to the charges on shells A and C must balance out to maintain this condition.

### Step 4: Potential at Shell B
The potential \( V_B \) at shell B due to shell A and shell C can be expressed as:
\[ V_B = V_A + V_C \]
Where:
- \( V_A = \frac{q}{4\pi \epsilon_0 b} \) (due to shell A)
- \( V_C = \frac{-q}{4\pi \epsilon_0 b} + \frac{Q'}{4\pi \epsilon_0 b} \) (due to shell C and the induced charge \( Q' \) on shell B)

Setting \( V_B = 0 \):
\[ \frac{q}{4\pi \epsilon_0 b} + \left( \frac{-q + Q'}{4\pi \epsilon_0 b} \right) = 0 \]

### Step 5: Solving for Induced Charge on Shell B
From the equation above, we can simplify:
\[ \frac{q + Q' - q}{4\pi \epsilon_0 b} = 0 \]
This leads to:
\[ Q' = -\frac{q b}{c} \]
Thus, the charge induced on the inner surface of shell B is \( -Q' = \frac{q b}{c} \).

### Step 6: Charge on the Outer Surface of Shell C
The charge on the outer surface of shell C can be calculated as:
\[ Q_{outer\_C} = -q + Q' = -q + \frac{q b}{c} \]
This simplifies to:
\[ Q_{outer\_C} = q \left( \frac{b}{c} - 1 \right) \]

### Step 7: Charge on the Inner Surface of Shell C
The charge on the inner surface of shell C will be the negative of the charge induced on shell B:
\[ Q_{inner\_C} = -Q' = -\frac{q b}{c} \]

### Final Charge Distribution Summary
1. Charge on the outer surface of shell A: \( +q \)
2. Charge on the inner surface of shell B: \( -\frac{q b}{c} \)
3. Charge on the inner surface of shell C: \( -\frac{q b}{c} \)
4. Charge on the outer surface of shell C: \( q \left( \frac{b}{c} - 1 \right) \)

### Conclusion
Based on the calculations:
- The charge on the outer surface of shell A is \( +q \) (Correct).
- The charge on the inner surface of shell C is \( -\frac{q b}{c} \) (Correct).
- The charge on the inner surface of shell B is \( -q \) (Correct).
- The charge on the outer surface of shell C is \( q \left( \frac{b}{c} - 1 \right) \) (Incorrect).