Electric potentiall at centre of non conducting sphere of radii .r. is ____ `["q is charge in it" k = (1)/(4 pi epsilon_(0))]`

Options

A. A) `(3)/(2)(kq)/(r )`
B. B) `(2)/(3)(kq)/(r )`
C. C) `(3kq)/(r )`
D. D) `(kq)/(r )`

Answer

To find the electric potential at the center of a non-conducting sphere of radius \( R \) with charge \( Q \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We need to find the electric potential \( V \) at the center of a non-conducting sphere with radius \( R \) and total charge \( Q \).

2. **Use the Formula for Electric Potential Inside a Sphere**: The electric potential \( V \) at a distance \( r \) from the center of a uniformly charged non-conducting sphere is given by:
\[ V = \frac{KQ}{2R} \left(3 - \frac{r^2}{R^2}\right) \]
where \( K = \frac{1}{4\pi \epsilon_0} \).

3. **Substitute the Value for \( r \)**: Since we are interested in the potential at the center of the sphere, we set \( r = 0 \):
\[ V = \frac{KQ}{2R} \left(3 - \frac{0^2}{R^2}\right) \]
This simplifies to:
\[ V = \frac{KQ}{2R} \cdot 3 \]

4. **Simplify the Expression**: Now we simplify the expression:
\[ V = \frac{3KQ}{2R} \]

5. **Final Result**: Thus, the electric potential at the center of the non-conducting sphere is:
\[ V = \frac{3KQ}{2R} \]

### Answer:
The electric potential at the center of the non-conducting sphere of radius \( R \) is \( \frac{3KQ}{2R} \).