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 r from the centre is V, then electrical intensity at distance 3 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                 \[E=\frac{q}{4\pi {{\varepsilon }_{0}}}\frac{1}{{{(3r)}^{2}}}\] \[\therefore \]  \[\frac{E}{V}=\frac{q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\times \frac{4\pi {{\varepsilon }_{0}}3r}{2q}\] \[\Rightarrow \]               \[\frac{E}{V}=\frac{1}{6r}\] \[\Rightarrow \]               \[E=\frac{V}{6r}\]