A charge `q` is distributed over two concentric hollow sphere of radii `a` and `b(bgtc)` such that the surface densities are equal.Find the potential at the common centre.is

Options

A. zero
B. `(q)/(4piepsilon_(0))((a+b))/((a^(2)+b^(2)))`
C. `(q)/(4piepsilon_(0))[(1)/(a)+(1)/(b)]`
D. `(q)/(4piepsilon_(0))[(a+b)/((a^(2)+b^(2)))]`

Explanation

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

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Let the charge distributed over the two spheres be \( q \).
- The radii of the two spheres are \( a \) (inner sphere) and \( b \) (outer sphere), where \( b > a \).
- The surface charge densities of both spheres are equal.

2. **Define Surface Charge Densities:**
- Let the surface charge density be \( \sigma \).
- For the inner sphere (radius \( a \)):
\[ \sigma = \frac{q_1}{4\pi a^2} \]
- For the outer sphere (radius \( b \)):
\[ \sigma = \frac{q_2}{4\pi b^2} \]

3. **Equate the Surface Charge Densities:**
- Since the surface charge densities are equal:
\[ \frac{q_1}{4\pi a^2} = \frac{q_2}{4\pi b^2} \]
- This simplifies to:
\[ q_1 \cdot b^2 = q_2 \cdot a^2 \]

4. **Express Total Charge:**
- The total charge on both spheres is:
\[ q = q_1 + q_2 \]

5. **Substituting for \( q_2 \):**
- From the earlier equation, we can express \( q_2 \) in terms of \( q_1 \):
\[ q_2 = \frac{q_1 \cdot b^2}{a^2} \]
- Substitute this into the total charge equation:
\[ q = q_1 + \frac{q_1 \cdot b^2}{a^2} \]
- Factor out \( q_1 \):
\[ q = q_1 \left(1 + \frac{b^2}{a^2}\right) \]
- Solving for \( q_1 \):
\[ q_1 = \frac{q}{1 + \frac{b^2}{a^2}} = \frac{q \cdot a^2}{a^2 + b^2} \]

6. **Finding \( q_2 \):**
- Substitute \( q_1 \) back to find \( q_2 \):
\[ q_2 = \frac{q_1 \cdot b^2}{a^2} = \frac{q \cdot b^2}{a^2 + b^2} \]

7. **Calculate the Potential at the Center:**
- The potential due to the inner sphere at the center:
\[ V_1 = \frac{k \cdot q_1}{a} \]
- The potential due to the outer sphere at the center:
\[ V_2 = \frac{k \cdot q_2}{b} \]
- The total potential at the center \( V \):
\[ V = V_1 + V_2 = \frac{k \cdot q_1}{a} + \frac{k \cdot q_2}{b} \]

8. **Substituting \( q_1 \) and \( q_2 \):**
- Substitute \( q_1 \) and \( q_2 \) into the potential equation:
\[ V = \frac{k \cdot \frac{q \cdot a^2}{a^2 + b^2}}{a} + \frac{k \cdot \frac{q \cdot b^2}{a^2 + b^2}}{b} \]
- Simplifying gives:
\[ V = k \cdot \frac{q}{a^2 + b^2} \left( \frac{a^2}{a} + \frac{b^2}{b} \right) = k \cdot \frac{q}{a^2 + b^2} (a + b) \]

9. **Final Expression for Potential:**
- Thus, the potential at the common center is:
\[ V = \frac{k \cdot q (a + b)}{a^2 + b^2} \]
- In terms of \( \epsilon_0 \):
\[ V = \frac{q (a + b)}{4 \pi \epsilon_0 (a^2 + b^2)} \]