Question

A hollow metal sphere has a radius $r$. The potential difference between a point on its surface and at a point at a distance $3r$ from its centre is $V$. The electric intensity at the distance $3r$ from the centre of the sphere will be

• A. $\frac{V}{3r}$
• B. $\frac{3V}{r}$
• C. $\frac{V}{r}$
• D. $\frac{V}{6r}$

Hint:

To remember this connection easily, look at the ratio between potential and field at any point outside a sphere: $E_x = \frac{V_x}{x}$. Here, the potential at $3r$ is $V_{3r} = \frac{kQ}{3r}$. Since $V = \frac{2kQ}{3r}$, we know $V_{3r} = \frac{V}{2}$. Plugging this into the ratio gives $E = \frac{V/2}{3r} = \frac{V}{6r}$ instantly!

Answer 1

The Correct Option is D

Step 1: The set up.
A hollow metal sphere of radius $r$ carries some charge $Q$. Outside the sphere it behaves as if all the charge sat at the centre. The potential difference between the surface (distance $r$) and a point at distance $3r$ is $V$. We want the electric field at distance $3r$.
Step 2: Potential and field formulas.
Outside the sphere, at distance $x$, \[ V_x = \frac{kQ}{x}, \qquad E_x = \frac{kQ}{x^2} \]
Step 3: Write the two potentials.
\[ V_{\text{surface}} = \frac{kQ}{r}, \qquad V_{\text{outer}} = \frac{kQ}{3r} \]
Step 4: Use the given difference.
\[ V = \frac{kQ}{r} - \frac{kQ}{3r} = \frac{kQ}{r}\left(1 - \frac{1}{3}\right) = \frac{2kQ}{3r} \] Solve for the charge term: \[ kQ = \frac{3rV}{2} \]
Step 5: Field at distance 3r.
\[ E = \frac{kQ}{(3r)^2} = \frac{kQ}{9r^2} \]
Step 6: Substitute and simplify.
\[ E = \frac{3rV/2}{9r^2} = \frac{3rV}{18r^2} = \frac{V}{6r} \] This is option (4). \[ \boxed{E = \frac{V}{6r}} \]