question_answer

A hollow charged metal sphere has a radius r. If the potential difference between its surface and a point at a distance \[3\text{ }r\]from the centre is V, then electrical intensity at distance \[3\text{ }r\]from the centre is

A)  \[\frac{V}{2r}\]

B)  \[\frac{V}{3r}\]

C)  \[\frac{V}{4r}\]

D)  \[\frac{V}{6r}\]

Correct Answer: D

Solution :

The potential at a distance r, due to charge q is \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q}{r}\] Potential at a distance \[(3r)\] is \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q}{3r}\] Difference in potential \[=\frac{q}{4\pi {{\varepsilon }_{0}}}\left[ \frac{1}{r}-\frac{1}{3r} \right]\] \[\Rightarrow \] \[V=\frac{2q}{4\pi {{\varepsilon }_{0}}\times 3r}\] Intensity of electric field \[E=\frac{q}{4\pi {{\varepsilon }_{0}}}\frac{1}{{{(3r)}^{2}}}\] \[\therefore \] \[\frac{E}{V}=\frac{q}{4\pi {{\varepsilon }_{0}}q{{r}^{2}}}\times \frac{4\pi {{\varepsilon }_{0}}3r}{2q}\] \[\Rightarrow \] \[\frac{E}{V}=\frac{1}{6r}\] \[\Rightarrow \] \[E=\frac{V}{6r}\]