Two connectric spheres of radii `R` and `r` have similar charges with equal surface charge densities `(sigam)`. The electric potential at their common centre is

Options

A. `sigma//epsilon_(0)`
B. `(sigma)/(epsilon_(0)) (R-r)`
C. `(sigma)/(epsilon_(0)) (R+r)`
D. `(sigma)/(4pi epsilon_(0)) (R +r)`

Explanation

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

### Step 1: Understand the Problem
We have two concentric spheres:
- The larger sphere has a radius \( R \) and a surface charge density \( \sigma \).
- The smaller sphere has a radius \( r \) and the same surface charge density \( \sigma \).

### Step 2: Electric Potential Due to a Charged Sphere
The electric potential \( V \) at a distance \( d \) from the center of a uniformly charged sphere (outside the sphere) is given by:
\[ V = \frac{Q}{4 \pi \epsilon_0 d} \]
where \( Q \) is the total charge of the sphere and \( \epsilon_0 \) is the permittivity of free space.

### Step 3: Calculate the Charge on Each Sphere
The total charge \( Q \) on a sphere can be expressed in terms of the surface charge density \( \sigma \) and the surface area \( A \):
\[ Q = \sigma \cdot A \]
For a sphere, the surface area \( A \) is \( 4 \pi r^2 \). Thus, for the smaller sphere (radius \( r \)):
\[ Q_1 = \sigma \cdot 4 \pi r^2 \]
For the larger sphere (radius \( R \)):
\[ Q_2 = \sigma \cdot 4 \pi R^2 \]

### Step 4: Electric Potential at the Center from Each Sphere
1. **Potential from the smaller sphere**:
The potential \( V_1 \) at the center due to the smaller sphere is:
\[ V_1 = \frac{Q_1}{4 \pi \epsilon_0 r} = \frac{\sigma \cdot 4 \pi r^2}{4 \pi \epsilon_0 r} = \frac{\sigma r}{\epsilon_0} \]

2. **Potential from the larger sphere**:
The potential \( V_2 \) at the center due to the larger sphere is:
\[ V_2 = \frac{Q_2}{4 \pi \epsilon_0 R} = \frac{\sigma \cdot 4 \pi R^2}{4 \pi \epsilon_0 R} = \frac{\sigma R}{\epsilon_0} \]

### Step 5: Total Electric Potential at the Center
The total electric potential \( V \) at the common center of the two spheres is the sum of the potentials from both spheres:
\[ V = V_1 + V_2 = \frac{\sigma r}{\epsilon_0} + \frac{\sigma R}{\epsilon_0} \]
Factoring out \( \frac{\sigma}{\epsilon_0} \):
\[ V = \frac{\sigma}{\epsilon_0} (R + r) \]

### Final Answer
Thus, the electric potential at their common center is:
\[ V = \frac{\sigma}{\epsilon_0} (R + r) \]