Potential at a point x-distance from the centre inside the conducting sphere of radius `R` and charged with charge `Q` is

Options

A. `(1)/(4pi epsilon_(0))(Q)/(R)`
B. `(1)/(4pi epsilon_(0))(Q)/(x)`
C. `(1)/(4 pi epsilon_(0))xQ`
D. zero

Answer

To find the potential at a point located at a distance \( x \) from the center inside a conducting sphere of radius \( R \) and charged with charge \( Q \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Conducting Sphere**:
- A conducting sphere with charge \( Q \) distributes its charge uniformly on its surface. Inside the conducting material, the electric field (\( E \)) is zero due to electrostatic shielding.

2. **Applying Gauss's Law**:
- According to Gauss's law, the electric field inside a conductor in electrostatic equilibrium is zero. Therefore, for any point inside the conducting sphere (at a distance \( x < R \)), we have:
\[ E = 0 \]

3. **Relating Electric Field to Potential**:
- The electric potential \( V \) is related to the electric field by the equation:
\[ V = -\int E \, dr \]
- Since the electric field \( E \) is zero inside the conducting sphere, the potential \( V \) remains constant throughout the interior of the sphere.

4. **Determining the Potential**:
- The potential inside the sphere is equal to the potential on the surface of the sphere. The potential \( V \) at the surface of a charged sphere is given by:
\[ V = \frac{1}{4 \pi \epsilon_0} \frac{Q}{R} \]
- Therefore, the potential at any point inside the conducting sphere (including at a distance \( x \) from the center) is also:
\[ V = \frac{1}{4 \pi \epsilon_0} \frac{Q}{R} \]

### Final Answer:
Thus, the potential at a point \( x \) distance from the center inside the conducting sphere is:
\[ V = \frac{1}{4 \pi \epsilon_0} \frac{Q}{R} \]

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