Question

A hollow charged metal sphere has a radius ' \(r\) '. If the potential difference between its surface and a point at a distance ' \(3r\) ' from the centre is ' \(v\) ', then the electric field intensity at a distance ' \(3r\) ' is

• A. \(\frac{V}{2r}\)
• B. \(\frac{v}{3r}\)
• C. \(\frac{v}{6r}\)
• D. \(\frac{v}{4r}\)

Hint:

Outside sphere: Acts like point charge

Answer 1

The Correct Option is C

Concept: Potential outside a sphere: \[ V = \frac{kQ}{r} \]

Step 1: Potential difference. \[ \Delta V = \frac{kQ}{r} - \frac{kQ}{3r} = \frac{2kQ}{3r} \] Given: \[ \frac{2kQ}{3r} = v \Rightarrow kQ = \frac{3vr}{2} \]

Step 2: Electric field at $3r$. \[ E = \frac{kQ}{(3r)^2} = \frac{kQ}{9r^2} \] \[ E = \frac{3vr/2}{9r^2} = \frac{v}{6r} \]

Step 3: Conclusion.
$E = \frac{v}{6r}$ Final Answer: Option (C)