Two concentric spheres kept in air have radii R and r. They have similar charge and equal surface charge density `sigma`. The electrical potential at their common centre is (where, `epsi_(0) =` permittivity of free space)

Options

A. `(sigma (R+r))/(epsi_(0))`
B. `(sigma(R-r))/epsi_(0)`
C. `(sigma(R+r))/(2 epsi_(0))`
D. `(sigma (R+r))/(4 epsi_(0))`

Answer

To solve the problem of finding the electrical potential at the common center of two concentric spheres with radii \( R \) and \( r \), and equal surface charge density \( \sigma \), we can follow these steps:

### Step 1: Understand the Configuration
We have two concentric spheres:
- The inner sphere has radius \( r \) and charge \( Q_1 \).
- The outer sphere has radius \( R \) and charge \( Q_2 \).
Both spheres have the same surface charge density \( \sigma \).

### Step 2: Relate Charge and Surface Charge Density
The surface charge density \( \sigma \) is defined as:
\[ \sigma = \frac{Q}{A} \]
where \( A \) is the surface area of the sphere. The surface area of a sphere is given by \( 4\pi R^2 \).

For the inner sphere:
\[ Q_1 = \sigma \cdot 4\pi r^2 \]

For the outer sphere:
\[ Q_2 = \sigma \cdot 4\pi R^2 \]

### Step 3: Calculate the Electric Potential
The electric potential \( V \) at a distance \( d \) from a charged sphere is given by:
\[ V = \frac{1}{4\pi \epsilon_0} \frac{Q}{d} \]

#### For the inner sphere (at the center):
Since we are at the center of the inner sphere, the potential due to the inner sphere is:
\[ V_r = \frac{1}{4\pi \epsilon_0} \frac{Q_1}{r} = \frac{1}{4\pi \epsilon_0} \cdot \frac{\sigma \cdot 4\pi r^2}{r} = \frac{\sigma r}{\epsilon_0} \]

#### For the outer sphere (at the center):
The potential due to the outer sphere at the center is:
\[ V_R = \frac{1}{4\pi \epsilon_0} \frac{Q_2}{R} = \frac{1}{4\pi \epsilon_0} \cdot \frac{\sigma \cdot 4\pi R^2}{R} = \frac{\sigma R}{\epsilon_0} \]

### Step 4: Total Potential at the Center
The total potential \( V_0 \) at the common center of the spheres is the sum of the potentials due to both spheres:
\[ V_0 = V_r + V_R = \frac{\sigma r}{\epsilon_0} + \frac{\sigma R}{\epsilon_0} \]

### Step 5: Simplifying the Expression
Combining the terms, we get:
\[ V_0 = \frac{\sigma (r + R)}{\epsilon_0} \]

### Final Answer
Thus, the electrical potential at their common center is:
\[ V_0 = \frac{\sigma (R + r)}{\epsilon_0} \]

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