A charge Q is distributed over two concentric hollow spheres of radii r and `R (gt r)` such that the surface charge densities are equal. Find the potential at the common centre.

Explanation

To find the potential at the common center of two concentric hollow spheres with equal surface charge densities, we can follow these steps:

### Step 1: Understand the problem
We have two concentric hollow spheres with radii \( r \) and \( R \) (where \( R > r \)). The total charge \( Q \) is distributed over both spheres such that their surface charge densities are equal.

### Step 2: Define surface charge density
Let the surface charge density on both spheres be \( \sigma \). The surface area of the smaller sphere (radius \( r \)) is \( 4\pi r^2 \) and for the larger sphere (radius \( R \)) it is \( 4\pi R^2 \).

### Step 3: Relate total charge to surface charge density
The total charge \( Q \) can be expressed in terms of the surface charge density:
\[ Q = \sigma \cdot 4\pi r^2 + \sigma \cdot 4\pi R^2 \]
Factoring out \( \sigma \):
\[ Q = \sigma \cdot 4\pi (r^2 + R^2) \]

### Step 4: Solve for surface charge density
From the equation above, we can solve for \( \sigma \):
\[ \sigma = \frac{Q}{4\pi (r^2 + R^2)} \]

### Step 5: Calculate the potential at the center
The potential \( V \) at the center due to a charged sphere is given by the formula:
\[ V = \frac{Q}{4\pi \epsilon_0 r} \]
For both spheres, we need to calculate the potential at the center due to each sphere separately and then add them up.

#### Potential due to the smaller sphere:
Using the charge on the smaller sphere:
\[ V_a = \frac{\sigma \cdot 4\pi r^2}{4\pi \epsilon_0 r} = \frac{\sigma r}{\epsilon_0} \]

#### Potential due to the larger sphere:
Using the charge on the larger sphere:
\[ V_b = \frac{\sigma \cdot 4\pi R^2}{4\pi \epsilon_0 R} = \frac{\sigma R}{\epsilon_0} \]

### Step 6: Total potential at the center
The total potential \( V \) at the center is the sum of the potentials due to both spheres:
\[ V = V_a + V_b = \frac{\sigma r}{\epsilon_0} + \frac{\sigma R}{\epsilon_0} = \frac{\sigma (r + R)}{\epsilon_0} \]

### Step 7: Substitute for \( \sigma \)
Now substituting the value of \( \sigma \):
\[ V = \frac{1}{\epsilon_0} \cdot \frac{Q}{4\pi (r^2 + R^2)} (r + R) \]

### Final Expression for Potential
Thus, the potential at the common center of the two spheres is:
\[ V = \frac{Q (r + R)}{4\pi \epsilon_0 (r^2 + R^2)} \]