Potential difference beween centre and surface of the sphere of radius R and uniorm volume charge density `rho` within it will be

Options

A. `(rhoR^2)/(6epsilon_0)`
B. `(rhoR^2)/(4epsilon_0)`
C. `(rhoR^2)/(3epsilon_0)`
D. `(rhoR^2)/(2epsilon_0)`

Explanation

To find the potential difference between the center and the surface of a sphere of radius \( R \) with a uniform volume charge density \( \rho \), we can follow these steps:

### Step 1: Understand the formula for electric potential
The electric potential \( V \) at a distance \( r \) from the center of a uniformly charged sphere can be calculated using the formula:
\[ V = \frac{KQ}{r} \]
where \( K = \frac{1}{4\pi\epsilon_0} \) and \( Q \) is the total charge enclosed within the sphere.

### Step 2: Calculate the total charge \( Q \)
The total charge \( Q \) of the sphere can be calculated using the volume charge density \( \rho \):
\[ Q = \rho \cdot V_{\text{sphere}} = \rho \cdot \left(\frac{4}{3}\pi R^3\right) \]

### Step 3: Substitute \( Q \) into the potential formula
Now, substituting \( Q \) into the potential formula, we get:
\[ V = \frac{K \cdot \rho \cdot \left(\frac{4}{3}\pi R^3\right)}{r} \]

### Step 4: Calculate the potential at the center \( V_c \)
For the center of the sphere (where \( r = 0 \)), the potential is given by:
\[ V_c = \frac{3}{2} \cdot \frac{KQ}{R} \]
Substituting \( Q \):
\[ V_c = \frac{3}{2} \cdot \frac{K \cdot \rho \cdot \left(\frac{4}{3}\pi R^3\right)}{R} = 2K\rho\pi R^2 \]

### Step 5: Calculate the potential at the surface \( V_s \)
For the surface of the sphere (where \( r = R \)), the potential is:
\[ V_s = \frac{KQ}{R} = \frac{K \cdot \rho \cdot \left(\frac{4}{3}\pi R^3\right)}{R} = \frac{4}{3}K\rho\pi R^2 \]

### Step 6: Find the potential difference \( V_c - V_s \)
Now, we can find the potential difference between the center and the surface:
\[ V_c - V_s = \left(2K\rho\pi R^2\right) - \left(\frac{4}{3}K\rho\pi R^2\right) \]

### Step 7: Simplify the expression
To simplify:
\[ V_c - V_s = K\rho\pi R^2 \left(2 - \frac{4}{3}\right) = K\rho\pi R^2 \left(\frac{6}{3} - \frac{4}{3}\right) = K\rho\pi R^2 \cdot \frac{2}{3} \]

### Step 8: Final expression for potential difference
Substituting \( K = \frac{1}{4\pi\epsilon_0} \):
\[ V_c - V_s = \frac{1}{4\pi\epsilon_0} \cdot \rho \cdot R^2 \cdot \frac{2}{3} = \frac{\rho R^2}{6\epsilon_0} \]

Thus, the potential difference between the center and the surface of the sphere is:
\[ \Delta V = \frac{\rho R^2}{6\epsilon_0} \]