A charge Q is distributed over two concentric hollow spheres of radii r and `R (gt r)` such that the surface charge densities are equal. Find the potential at the common centre.

Options

A. `1/(4piepsilon_(0)) ((2R +r))/((R^(2)+r^(2))) Q`
B. `1/(4piepsilon_(0)) ((R+r))/((R^(2)+r^(2)) )Q`
C. ` 1/(4piepsilon_(0)) ((R+r))/(2(R^(2)+r^(2)))Q`
D. `1/(4piepsilon_(0)) ((R+2r)Q)/(2(R^(2)+r^(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 1: Define the Variables
Let:
- \( r \) = radius of the inner sphere
- \( R \) = radius of the outer sphere (where \( R > r \))
- \( Q \) = total charge distributed over both spheres
- \( \sigma \) = surface charge density on both spheres

### Step 2: Express Surface Charge Densities
The surface charge density \( \sigma \) for the inner sphere (radius \( r \)) and outer sphere (radius \( R \)) can be expressed as:
\[ \sigma = \frac{Q_1}{4\pi r^2} = \frac{Q_2}{4\pi R^2} \]
where \( Q_1 \) is the charge on the inner sphere and \( Q_2 \) is the charge on the outer sphere.

### Step 3: Relate the Charges
From the equality of surface charge densities, we can derive:
\[ \frac{Q_1}{Q_2} = \frac{r^2}{R^2} \]
This implies:
\[ Q_1 = Q_2 \cdot \frac{r^2}{R^2} \]

### Step 4: Use the Total Charge Condition
Since the total charge \( Q \) is distributed over both spheres:
\[ Q = Q_1 + Q_2 \]
Substituting \( Q_1 \) from the previous step:
\[ Q = Q_2 \cdot \frac{r^2}{R^2} + Q_2 \]
Factoring out \( Q_2 \):
\[ Q = Q_2 \left( \frac{r^2}{R^2} + 1 \right) \]
Thus, we can solve for \( Q_2 \):
\[ Q_2 = \frac{Q}{1 + \frac{r^2}{R^2}} = \frac{Q R^2}{R^2 + r^2} \]

### Step 5: Find \( Q_1 \)
Now substituting \( Q_2 \) back to find \( Q_1 \):
\[ Q_1 = Q_2 \cdot \frac{r^2}{R^2} = \frac{Q R^2}{R^2 + r^2} \cdot \frac{r^2}{R^2} = \frac{Q r^2}{R^2 + r^2} \]

### Step 6: Calculate the Potential at the Center
The potential \( V \) at the common center due to both spheres is given by:
\[ V = k \left( \frac{Q_1}{r} + \frac{Q_2}{R} \right) \]
Substituting \( Q_1 \) and \( Q_2 \):
\[ V = k \left( \frac{\frac{Q r^2}{R^2 + r^2}}{r} + \frac{\frac{Q R^2}{R^2 + r^2}}{R} \right) \]
This simplifies to:
\[ V = k \left( \frac{Q r}{R^2 + r^2} + \frac{Q R}{R^2 + r^2} \right) = k \frac{Q (r + R)}{R^2 + r^2} \]

### Final Answer
Thus, the potential at the common center is:
\[ V = k \frac{Q (r + R)}{R^2 + r^2} \]