A conducting sphere of radius R is charged to a potential of V volts. Then the electric field at a distance `r ( gt R)` from the centre of the sphere would be

Options

A. `(V)/(r)`
B. `(R^(2)V)/(r^(3))`
C. `(RV)/(r^(2))`
D. `(rV)/(R^(2))`

Answer

To solve the problem, we need to determine the electric field at a distance \( r \) from the center of a conducting sphere of radius \( R \) that is charged to a potential \( V \) volts, where \( r > R \).

### Step-by-Step Solution:

1. **Understand the Concept of Electric Field and Potential:**
The electric potential \( V \) at the surface of a charged conducting sphere is given by the formula:
\[ V = \frac{kQ}{R} \]
where \( k \) is Coulomb's constant, \( Q \) is the charge on the sphere, and \( R \) is the radius of the sphere.

2. **Relate Charge to Potential:**
From the above formula, we can express the charge \( Q \) in terms of the potential \( V \):
\[ Q = \frac{VR}{k} \]

3. **Electric Field Outside the Sphere:**
For a conducting sphere, the electric field \( E \) at a distance \( r \) from the center (where \( r > R \)) can be calculated using the formula for the electric field due to a point charge:
\[ E = \frac{kQ}{r^2} \]

4. **Substituting Charge into Electric Field Formula:**
Now substitute the expression for \( Q \) into the electric field formula:
\[ E = \frac{k \left(\frac{VR}{k}\right)}{r^2} \]
Simplifying this gives:
\[ E = \frac{VR}{r^2} \]

5. **Final Expression for Electric Field:**
Therefore, the electric field \( E \) at a distance \( r \) from the center of the sphere (where \( r > R \)) is:
\[ E = \frac{V}{r^2} \cdot R \]

### Conclusion:
The electric field at a distance \( r \) from the center of the conducting sphere is given by:
\[ E = \frac{VR}{r^2} \]