There is a hemisphere of radius R having a uniform volume charge density `rho`. Find the electric potential and field at the centre

Answer

To solve the problem of finding the electric potential and electric field at the center of a hemisphere with radius \( R \) and uniform volume charge density \( \rho \), we can follow these steps:

### Step 1: Calculate the Total Charge
The total charge \( Q \) contained in the hemisphere can be calculated using the volume charge density \( \rho \) and the volume of the hemisphere.

The volume \( V \) of a hemisphere is given by:
\[ V = \frac{2}{3} \pi R^3 \]
Thus, the total charge \( Q \) is:
\[ Q = \rho \cdot V = \rho \cdot \frac{2}{3} \pi R^3 \]

### Step 2: Calculate the Electric Potential at the Center
The electric potential \( V \) at a point due to a charge \( Q \) at a distance \( r \) is given by:
\[ V = \frac{Q}{4 \pi \epsilon_0 r} \]
For a hemisphere, we consider the potential at the center, where \( r = 0 \). However, we can use the formula for the potential due to a continuous charge distribution. The potential at the center of the hemisphere can be expressed as:
\[ V = \frac{Q}{4 \pi \epsilon_0 R} \]
Substituting for \( Q \):
\[ V = \frac{\rho \cdot \frac{2}{3} \pi R^3}{4 \pi \epsilon_0 R} = \frac{\rho R^2}{6 \epsilon_0} \]

### Step 3: Calculate the Electric Field at the Center
The electric field \( E \) at the center of the hemisphere can be derived from the total charge. The electric field due to a uniformly charged hemisphere at its center is given by:
\[ E = \frac{Q}{2 \pi \epsilon_0 R^2} \]
Substituting for \( Q \):
\[ E = \frac{\rho \cdot \frac{2}{3} \pi R^3}{2 \pi \epsilon_0 R^2} = \frac{\rho R}{3 \epsilon_0} \]

### Final Results
Thus, the electric potential \( V \) at the center of the hemisphere is:
\[ V = \frac{\rho R^2}{6 \epsilon_0} \]
And the electric field \( E \) at the center of the hemisphere is:
\[ E = \frac{\rho R}{3 \epsilon_0} \]