Question

A conducting sphere of radius R is charged to a potential of V volt. Then the electric field at a distance \(r(>R)\) from the centre of sphere would be :

• A. \( \frac{RV}{r^2} \)
• B. \( \frac{V}{r} \)
• C. \( \frac{rV}{R^2} \)
• D. \( \frac{R^2V}{r^3} \)

Hint:

Outside a charged sphere, it behaves like a point charge at centre.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A conducting sphere acts as a point charge concentrated at its center for all points outside the sphere. We can relate the surface potential \( V \) to the total charge \( Q \).
: Key Formula or Approach:
1. Potential on surface: \( V = \frac{1}{4\pi\epsilon_0} \frac{Q}{R} \).
2. Electric field outside (\( r>R \)): \( E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2} \).
Step 2: Detailed Explanation:
From the expression for potential \( V \):
\[ \frac{Q}{4\pi\epsilon_0} = V \cdot R \]
Now, substitute this into the expression for electric field \( E \) at distance \( r \):
\[ E = \left( \frac{Q}{4\pi\epsilon_0} \right) \cdot \frac{1}{r^2} \]
\[ E = (V \cdot R) \cdot \frac{1}{r^2} \]
\[ E = \frac{RV}{r^2} \]
Step 3: Final Answer:
The electric field at distance \( r \) is \( RV/r^2 \).