A charge 'Q' is distributed over two concentric hollow spheres of radii 'r' and 'R' (gtr) such that the surface 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: Define the Problem
Let the charge \( Q \) be distributed over two concentric hollow spheres of radii \( r \) (inner sphere) and \( R \) (outer sphere) such that the surface charge densities are equal.

### Step 2: Understand Surface Charge Density
The surface charge density \( \sigma \) is defined as:
\[ \sigma = \frac{Q}{A} \]
where \( A \) is the surface area of the sphere. For the inner sphere of radius \( r \):
\[ \sigma_1 = \frac{Q_1}{4\pi r^2} \]
For the outer sphere of radius \( R \):
\[ \sigma_2 = \frac{Q_2}{4\pi R^2} \]
Since the surface densities are equal, we have:
\[ \sigma_1 = \sigma_2 \implies \frac{Q_1}{4\pi r^2} = \frac{Q_2}{4\pi R^2} \]
This simplifies to:
\[ Q_1 R^2 = Q_2 r^2 \implies Q_2 = Q_1 \frac{R^2}{r^2} \]

### Step 3: Total Charge Relation
The total charge \( Q \) is given by:
\[ Q = Q_1 + Q_2 \]
Substituting \( Q_2 \):
\[ Q = Q_1 + Q_1 \frac{R^2}{r^2} = Q_1 \left(1 + \frac{R^2}{r^2}\right) \]
From this, we can express \( Q_1 \):
\[ Q_1 = \frac{Q}{1 + \frac{R^2}{r^2}} = \frac{Qr^2}{r^2 + R^2} \]

### Step 4: Calculate the Potential at the Center
The potential \( V \) at the center due to a spherical shell of charge is given by:
\[ V = \frac{1}{4\pi \epsilon_0} \left( \frac{Q_1}{r} + \frac{Q_2}{R} \right) \]
Substituting \( Q_2 \):
\[ V = \frac{1}{4\pi \epsilon_0} \left( \frac{Q_1}{r} + \frac{Q_1 \frac{R^2}{r^2}}{R} \right) \]
This simplifies to:
\[ V = \frac{1}{4\pi \epsilon_0} \left( \frac{Q_1}{r} + \frac{Q_1 R}{r^2} \right) \]
Factoring out \( Q_1 \):
\[ V = \frac{Q_1}{4\pi \epsilon_0} \left( \frac{1}{r} + \frac{R}{r^2} \right) \]

### Step 5: Substitute \( Q_1 \)
Now substituting \( Q_1 \):
\[ V = \frac{Qr^2}{r^2 + R^2} \cdot \frac{1}{4\pi \epsilon_0} \left( \frac{1}{r} + \frac{R}{r^2} \right) \]
This becomes:
\[ V = \frac{Q}{4\pi \epsilon_0} \cdot \frac{1}{r^2 + R^2} \left( r + R \right) \]

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