A charge .q. is distrubuted over two concertric hollow conducting sphere of radii r and R `(lt r)` such that their surface charge densite are equal. The potential at their common centre is

Options

A. A) Zero
B. B) `(q)/(4piepsilon_(0))((r+R))/((r^(2)+R^(2))^(2))`
C. C) `(q)/(4piin_(0))[(1)/(r)+(1)/(R)]`
D. D) `(q)/(4piin_(0))[(r+R)/((r^(2)+R^(2)))]`

Answer

To solve the problem, we need to find the potential at the common center of two concentric hollow conducting spheres with a total charge \( q \) distributed between them. The surface charge densities of the two spheres are equal, and we denote the radii of the inner and outer spheres as \( R \) and \( r \) respectively, where \( R < r \).

### Step-by-Step Solution:

1. **Understanding Surface Charge Densities**:
- Let \( q_a \) be the charge on the inner sphere and \( q_b \) be the charge on the outer sphere.
- The surface charge density \( \sigma \) for a sphere is given by:
\[ \sigma = \frac{q}{A} = \frac{q}{4\pi R^2} \]
- For the inner sphere:
\[ \sigma_a = \frac{q_a}{4\pi R^2} \]
- For the outer sphere:
\[ \sigma_b = \frac{q_b}{4\pi r^2} \]
- Given that \( \sigma_a = \sigma_b \), we can equate these two expressions:
\[ \frac{q_a}{4\pi R^2} = \frac{q_b}{4\pi r^2} \]
- This simplifies to:
\[ q_a \cdot r^2 = q_b \cdot R^2 \quad \text{(1)} \]

2. **Total Charge Distribution**:
- The total charge is given by:
\[ q_a + q_b = q \quad \text{(2)} \]

3. **Expressing \( q_a \) and \( q_b \)**:
- From equation (1), we can express \( q_b \) in terms of \( q_a \):
\[ q_b = \frac{q_a \cdot r^2}{R^2} \]
- Substituting this into equation (2):
\[ q_a + \frac{q_a \cdot r^2}{R^2} = q \]
- Factoring out \( q_a \):
\[ q_a \left(1 + \frac{r^2}{R^2}\right) = q \]
- Solving for \( q_a \):
\[ q_a = \frac{q}{1 + \frac{r^2}{R^2}} = \frac{q R^2}{R^2 + r^2} \]
- Now substituting back to find \( q_b \):
\[ q_b = q - q_a = q - \frac{q R^2}{R^2 + r^2} = \frac{q r^2}{R^2 + r^2} \]

4. **Calculating the Potential at the Center**:
- The potential \( V \) at the common center due to both spheres is given by:
\[ V = k \frac{q_a}{R} + k \frac{q_b}{r} \]
- Substituting the expressions for \( q_a \) and \( q_b \):
\[ V = k \left( \frac{\frac{q R^2}{R^2 + r^2}}{R} + \frac{\frac{q r^2}{R^2 + r^2}}{r} \right) \]
- Simplifying:
\[ V = k \left( \frac{q R}{R^2 + r^2} + \frac{q r}{R^2 + r^2} \right) \]
\[ V = k \frac{q (R + r)}{R^2 + r^2} \]

5. **Final Expression**:
- The potential at the common center of the two hollow conducting spheres is:
\[ V = k \frac{q (R + r)}{R^2 + r^2} \]
- Where \( k = \frac{1}{4 \pi \epsilon_0} \).