A hollow charged metal sphere has radius ' $R$ '. If the potential difference between its surface and a…

A hollow charged metal sphere has radius ' $R$ '. If the potential difference between its surface and a point at a distance ' $5 \mathrm{R}$ ' from the centre is V, then magnitude of electric field Intensity at a distance ' $5 \mathrm{R}$ ' from the centre of sphere is
  1. $\frac{\mathrm{V}}{2 \mathrm{R}}$
  2. $\frac{\mathrm{V}}{20 \mathrm{R}}$
  3. $10 \mathrm{VR}$
  4. $20 \mathrm{VR}$

Solution

$\begin{aligned} & \mathrm{V}=\frac{1}{4 \pi \in_0} \frac{\mathrm{q}}{\mathrm{R}}-\frac{1}{4 \pi \in_0} \frac{\mathrm{q}}{5 \mathrm{R}} \\ & =\frac{1}{4 \pi \in_0} \frac{\mathrm{q}}{\mathrm{R}}\left(1-\frac{1}{5}\right) \\ & =\frac{1}{4 \pi \in_0} \frac{\mathrm{q}}{\mathrm{R}}\left(\frac{4}{5}\right) \\ & \mathrm{E}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{(5 \mathrm{R})^2} \\ & \therefore \frac{\mathrm{E}}{\mathrm{V}}=\frac{1}{20 \mathrm{R}} \text { or } \mathrm{E}=\frac{\mathrm{V}}{20 \mathrm{R}}\end{aligned}$

Asked in: MHT CET 2021 (20 Sep Shift 1)

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