Charge `Q` is distributed on two concentric spheres of radius r and R resp. if charge density of both spheres is same then electric potential at centre will be

Options

A. `KQ(1/r + 1/R)`
B. `KQ(r+R)/(r^2 + R)`
C. KQrR/(R+r)
D. `KQ(R^2 + r^2)/(R+r)`

Answer

To find the electric potential at the center of two concentric spheres with the same charge density, we can follow these steps:

### Step 1: Define the Charge Density
Let the charge density of both spheres be denoted as \( \sigma \). The charge on the inner sphere of radius \( r \) can be expressed as:
\[ Q_1 = \sigma \cdot 4\pi r^2 \]
Similarly, the charge on the outer sphere of radius \( R \) is:
\[ Q_2 = \sigma \cdot 4\pi R^2 \]

### Step 2: Total Charge
The total charge \( Q \) distributed on both spheres can be written as:
\[ Q = Q_1 + Q_2 = \sigma \cdot 4\pi r^2 + \sigma \cdot 4\pi R^2 \]
This can be factored as:
\[ Q = \sigma \cdot 4\pi (r^2 + R^2) \]

### Step 3: Electric Potential at the Center
The electric potential \( V_c \) at the center due to a charged sphere is given by the formula:
\[ V = \frac{kQ}{r} \quad \text{(for a sphere of radius \( r \))} \]
For the inner sphere (radius \( r \)):
\[ V_1 = \frac{kQ_1}{r} = \frac{k(\sigma \cdot 4\pi r^2)}{r} = k\sigma \cdot 4\pi r \]
For the outer sphere (radius \( R \)):
\[ V_2 = \frac{kQ_2}{R} = \frac{k(\sigma \cdot 4\pi R^2)}{R} = k\sigma \cdot 4\pi R \]

### Step 4: Total Electric Potential at the Center
The total electric potential at the center \( V_c \) is the sum of the potentials from both spheres:
\[ V_c = V_1 + V_2 = k\sigma \cdot 4\pi r + k\sigma \cdot 4\pi R \]
Factoring out \( k\sigma \cdot 4\pi \):
\[ V_c = k\sigma \cdot 4\pi (r + R) \]

### Step 5: Substitute for Charge Density
From the earlier equation for total charge \( Q \):
\[ \sigma = \frac{Q}{4\pi (r^2 + R^2)} \]
Substituting this into the expression for \( V_c \):
\[ V_c = k \left(\frac{Q}{4\pi (r^2 + R^2)}\right) \cdot 4\pi (r + R) \]
This simplifies to:
\[ V_c = \frac{kQ (r + R)}{r^2 + R^2} \]

### Final Answer
Thus, the electric potential at the center of the two concentric spheres is:
\[ V_c = \frac{kQ (r + R)}{r^2 + R^2} \]

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