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 find 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 electric potential due to a charged sphere
The electric potential \( V \) at a distance \( d \) from the center of a charged sphere (outside the sphere) is given by the formula:
\[ V = \frac{1}{4\pi \epsilon_0} \frac{Q}{d} \]
where \( Q \) is the total charge and \( d \) is the distance from the center of the sphere.

### Step 2: Calculate the total charge on each sphere
The total charge \( Q \) on a sphere can be calculated using the surface charge density \( \sigma \) and the surface area of the sphere:
\[ Q = \sigma \times \text{Surface Area} = \sigma \times 4\pi R^2 \quad \text{(for the larger sphere)} \]
\[ Q = \sigma \times 4\pi r^2 \quad \text{(for the smaller sphere)} \]

### Step 3: Calculate the potential at the center due to each sphere
1. **Potential due to the larger sphere (radius \( R \))**:
Since we are at the center of the sphere, the potential due to the larger sphere is:
\[ V_R = \frac{1}{4\pi \epsilon_0} \frac{Q_1}{0} = \frac{1}{4\pi \epsilon_0} \frac{\sigma \cdot 4\pi R^2}{R} = \frac{\sigma R}{\epsilon_0} \]

2. **Potential due to the smaller sphere (radius \( r \))**:
Similarly, the potential due to the smaller sphere is:
\[ V_r = \frac{1}{4\pi \epsilon_0} \frac{Q_2}{0} = \frac{1}{4\pi \epsilon_0} \frac{\sigma \cdot 4\pi r^2}{r} = \frac{\sigma r}{\epsilon_0} \]

### Step 4: Calculate the total potential at the center
The total potential \( V \) at the center due to both spheres is the sum of the potentials from each sphere:
\[ V = V_R + V_r = \frac{\sigma R}{\epsilon_0} + \frac{\sigma r}{\epsilon_0} \]
\[ V = \frac{\sigma (R + r)}{\epsilon_0} \]

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

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