A spherical conductor `A` of radius `r` is placed concentrically inside a conducting shell `B` of radius `R(R gt r)`. A charge `Q` is given to `A` , and then `A` is joined to `B` by a metal wire. The charge flowing from `A` to `B` will be

Options

A. `Q((R)/(R + r))`
B. `R((Q)/(R+ r))`
C. Q
D. Zero

Explanation

To solve the problem step by step, let's analyze the situation involving the spherical conductor A and the conducting shell B.

### Step 1: Understand the Setup
We have two concentric spherical conductors:
- Conductor A (inner sphere) with radius \( r \) and charge \( Q \).
- Conducting shell B (outer sphere) with radius \( R \) (where \( R > r \)).

### Step 2: Determine the Initial Potential
The potential \( V \) of a charged conductor is given by the formula:
\[ V = \frac{kQ}{r} \]
where \( k \) is the electrostatic constant, \( Q \) is the charge, and \( r \) is the radius of the conductor.

- The potential at the surface of conductor A (denoted as \( V_A \)) is:
\[ V_A = \frac{kQ}{r} \]

- The potential at the surface of conductor B (denoted as \( V_B \)) is initially zero because it is a conducting shell and has no charge initially:
\[ V_B = 0 \]

### Step 3: Calculate the Potential Difference
The potential difference \( \Delta V \) between the two conductors is:
\[ \Delta V = V_B - V_A = 0 - \frac{kQ}{r} = -\frac{kQ}{r} \]

### Step 4: Joining A and B with a Wire
When a metal wire connects A and B, charge will flow from A to B until the potential difference becomes zero. The charge will flow until the potentials of both conductors equalize.

### Step 5: Final Charge Distribution
Since the potential of conductor B will be influenced by the charge that flows from A, we denote the charge that flows from A to B as \( Q' \).

- After the charge \( Q' \) flows from A to B, the new charge on A will be \( Q - Q' \) and the charge on B will be \( Q' \).

### Step 6: Set the Final Potentials Equal
At equilibrium, the potentials of both conductors must be equal:
\[ V_A' = V_B' \]
Substituting the expressions for the potentials:
\[ \frac{k(Q - Q')}{r} = \frac{kQ'}{R} \]

### Step 7: Solve for \( Q' \)
By simplifying the equation:
\[ \frac{Q - Q'}{r} = \frac{Q'}{R} \]
Cross-multiplying gives:
\[ R(Q - Q') = rQ' \]
Rearranging the equation:
\[ RQ - RQ' = rQ' \]
\[ RQ = RQ' + rQ' \]
\[ RQ = Q'(R + r) \]
Thus, we can solve for \( Q' \):
\[ Q' = \frac{RQ}{R + r} \]

### Conclusion
The charge flowing from conductor A to conductor B is:
\[ Q' = \frac{RQ}{R + r} \]