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

Options

A. `sigma/epsi_(0) (R+r)`
B. `sigma/epsi_(0) (R-r)`
C. `sigma/epsi_(0) (1/R+1/r)`
D. `sigma/epsi_(0)(1/R)`

Answer

To find the electric potential at the common center of two concentric spheres with radii \( R \) and \( r \) and equal surface charge densities \( \sigma \), we can follow these steps:

### Step 1: Understand the Configuration
We have two concentric spheres:
- Inner sphere (A) with radius \( r \)
- Outer sphere (B) with radius \( R \)

Both spheres have the same surface charge density \( \sigma \).

### Step 2: Calculate the Charge on Each Sphere
The total charge \( q \) on a sphere can be calculated using the formula:
\[ q = \sigma \times A \]
where \( A \) is the surface area of the sphere given by \( A = 4\pi r^2 \).

For sphere A (radius \( r \)):
\[ q_A = \sigma \times 4\pi r^2 \]

For sphere B (radius \( R \)):
\[ q_B = \sigma \times 4\pi R^2 \]

### Step 3: Calculate the Electric Potential at the Center
The electric potential \( V \) at a point due to a charged sphere is given by:
\[ V = \frac{kq}{d} \]
where \( k \) is Coulomb's constant \( \left( k = \frac{1}{4\pi \epsilon_0} \right) \) and \( d \) is the distance from the center to the point where the potential is being calculated.

At the center \( O \):
- The potential due to sphere A:
\[ V_{AO} = \frac{k q_A}{r} = \frac{k \sigma \cdot 4\pi r^2}{r} = k \sigma \cdot 4\pi r \]

- The potential due to sphere B:
\[ V_{BO} = \frac{k q_B}{R} = \frac{k \sigma \cdot 4\pi R^2}{R} = k \sigma \cdot 4\pi R \]

### Step 4: Total Electric Potential at the Center
The total potential \( V_O \) at the center \( O \) is the sum of the potentials due to both spheres:
\[ V_O = V_{AO} + V_{BO} = k \sigma \cdot 4\pi r + k \sigma \cdot 4\pi R \]
Factoring out \( k \sigma \cdot 4\pi \):
\[ V_O = k \sigma \cdot 4\pi (r + R) \]

### Step 5: Substitute \( k \)
Substituting \( k = \frac{1}{4\pi \epsilon_0} \):
\[ V_O = \frac{1}{4\pi \epsilon_0} \sigma \cdot 4\pi (r + R) \]
The \( 4\pi \) cancels out:
\[ V_O = \frac{\sigma}{\epsilon_0} (r + R) \]

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

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