Two concentric sphere of radii `r_1 and r_2` carry charges `q_1 and q_2` respectively. If the surface charge density (`sigma`) is same for both spheres, the electric potential at the common centre will be

Options

A. `( sigma )/( in_0) (r_1)/( r_2)`
B. `(sigma)/( in_0) (r_2)/(r_1)`
C. `(sigma)/( in_0) (r_1 -r_2)`
D. `( sigma )/( in_0) ( r_1 +r_2 )`

Answer

To find the electric potential at the common center of two concentric spheres with given radii and charges, we can follow these steps:

### Step 1: Understand the Problem
We have two concentric spheres with radii \( r_1 \) and \( r_2 \), carrying charges \( q_1 \) and \( q_2 \) respectively. The surface charge density \( \sigma \) is the same for both spheres. We need to find the electric potential at their common center.

### Step 2: Relate Surface Charge Density to Charge
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 first sphere:
\[ \sigma = \frac{q_1}{4\pi r_1^2} \implies q_1 = 4\pi \sigma r_1^2 \]

For the second sphere:
\[ \sigma = \frac{q_2}{4\pi r_2^2} \implies q_2 = 4\pi \sigma r_2^2 \]

### Step 3: Calculate the Electric Potential at the Center
The electric potential \( V \) due to a point charge \( q \) at a distance \( r \) is given by:
\[ V = k \frac{q}{r} \]
where \( k \) is Coulomb's constant (\( k = \frac{1}{4\pi \epsilon_0} \)).

The total potential at the center due to both spheres is the sum of the potentials due to each sphere:
\[ V = V_1 + V_2 = k \frac{q_1}{r_1} + k \frac{q_2}{r_2} \]

### Step 4: Substitute the Values of \( q_1 \) and \( q_2 \)
Substituting the expressions for \( q_1 \) and \( q_2 \):
\[ V = k \frac{4\pi \sigma r_1^2}{r_1} + k \frac{4\pi \sigma r_2^2}{r_2} \]
This simplifies to:
\[ V = k \left( 4\pi \sigma r_1 + 4\pi \sigma r_2 \right) \]

### Step 5: Factor Out Common Terms
Factoring out \( 4\pi k \sigma \):
\[ V = 4\pi k \sigma (r_1 + r_2) \]

### Step 6: Substitute the Value of \( k \)
Substituting \( k = \frac{1}{4\pi \epsilon_0} \):
\[ V = 4\pi \left( \frac{1}{4\pi \epsilon_0} \right) \sigma (r_1 + r_2) \]
The \( 4\pi \) cancels out:
\[ V = \frac{\sigma}{\epsilon_0} (r_1 + r_2) \]

### Final Answer
The electric potential at the common center of the two concentric spheres is:
\[ V = \frac{\sigma}{\epsilon_0} (r_1 + r_2) \]