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 of finding the electric field at a distance \( r \) (where \( r > R \)) from the center of a charged conducting sphere of radius \( R \) that is charged to a potential of \( V \) volts, we can follow these steps:

### Step 1: Understand the properties of a conducting sphere
A conducting sphere has the property that any excess charge resides on its surface. Inside the conducting material, the electric field is zero, and the potential is constant throughout the conductor.

### Step 2: Relate the potential to the charge
The potential \( V \) at the surface of the conducting sphere is given by the formula:
\[ V = \frac{kQ}{R} \]
where \( k \) is Coulomb's constant, \( Q \) is the total charge on the sphere, and \( R \) is the radius of the sphere.

### Step 3: Solve for the charge \( Q \)
From the equation above, we can express the charge \( Q \) in terms of the potential \( V \):
\[ Q = \frac{VR}{k} \]

### Step 4: Determine the electric field outside the sphere
For points outside the conducting sphere (where \( r > R \)), the electric field \( E \) can be calculated using Gauss's law. The electric field at a distance \( r \) from the center of the sphere is given by:
\[ E = \frac{kQ}{r^2} \]

### Step 5: Substitute the expression for \( Q \)
Now, substitute the expression for \( Q \) into the electric field equation:
\[ E = \frac{k \left( \frac{VR}{k} \right)}{r^2} \]
This simplifies to:
\[ E = \frac{VR}{r^2} \]

### Conclusion
Thus, the electric field at a distance \( r \) (where \( r > R \)) from the center of the sphere is:
\[ E = \frac{VR}{r^2} \]