Two concentric conducting shells A and B are of radii R and `2R`. A charge `+ q` is placed at the centre of the shells. Shell B is earthed and a charge q is given to shell A. Find the charge on outer surface of A and B.

Explanation

To solve the problem of finding the charge on the outer surfaces of the concentric conducting shells A and B, we will follow these steps:

### Step 1: Understand the System
We have two concentric conducting shells:
- Shell A has a radius \( R \) and is given a charge \( +q \).
- Shell B has a radius \( 2R \) and is earthed (connected to the ground).
- A charge \( +q \) is placed at the center of the shells.

### Step 2: Analyze the Charge Distribution on Shell A
Since there is a charge \( +q \) at the center, the inner surface of shell A will acquire a charge of \( -q \) to neutralize the electric field inside the conductor. This is due to the property of conductors that the electric field inside them must be zero.

- **Charge on inner surface of shell A**: \( Q_{\text{inner, A}} = -q \)

Now, since the total charge on shell A is \( +q \), the charge on the outer surface of shell A can be calculated as follows:

\[ Q_{\text{outer, A}} = Q_{\text{total, A}} - Q_{\text{inner, A}} = +q - (-q) = +q + q = +2q \]

### Step 3: Analyze the Charge Distribution on Shell B
Shell B is earthed, which means its potential is zero. The charge on the inner surface of shell B will be influenced by the charge on the outer surface of shell A.

The charge on the inner surface of shell B must be equal and opposite to the charge on the outer surface of shell A to maintain the electric field inside shell B as zero:

- **Charge on inner surface of shell B**: \( Q_{\text{inner, B}} = -Q_{\text{outer, A}} = -2q \)

Since shell B is earthed, the total charge on shell B must also ensure that the potential is zero. The charge on the outer surface of shell B can be calculated as follows:

Let \( Q_{\text{outer, B}} \) be the charge on the outer surface of shell B. The total charge on shell B is zero (because it is earthed), so:

\[ Q_{\text{total, B}} = Q_{\text{inner, B}} + Q_{\text{outer, B}} = 0 \]

Substituting the value of \( Q_{\text{inner, B}} \):

\[ -2q + Q_{\text{outer, B}} = 0 \]

This gives:

\[ Q_{\text{outer, B}} = +2q \]

### Summary of Charges
- Charge on the outer surface of shell A: \( +2q \)
- Charge on the outer surface of shell B: \( 0 \)

### Final Answer
- Charge on the outer surface of shell A: \( +2q \)
- Charge on the outer surface of shell B: \( 0 \)