A charged spherical conductor has radius ' $r$ '. The potential difference between its surface and a point…

A charged spherical conductor has radius ' $r$ '. The potential difference between its surface and a point at a distance ' $3 r$ ' from the center is ' $\mathrm{V}$ ' The electric intensity at a distance ' $3 \mathrm{r}$ ' from the centre of the conductor is
  1. $\frac{\mathrm{V}}{8 \mathrm{r}}$
  2. $\frac{\mathrm{V}}{2 \mathrm{r}}$
  3. $\frac{\mathrm{V}}{4 \mathrm{r}}$
  4. $\frac{\mathrm{V}}{6 \mathrm{r}}$

Solution

$\begin{aligned} & \mathrm{V}=\frac{\mathrm{q}}{4 \pi \varepsilon_0 \mathrm{r}}\left[1-\frac{1}{3}\right]=\frac{2}{3} \cdot \frac{\mathrm{q}}{4 \pi \varepsilon_0 \mathrm{r}} \\ & \mathrm{E}=\frac{\mathrm{q}}{4 \pi \varepsilon_0} \frac{1}{(3 \mathrm{r})^2}=\frac{\mathrm{q}}{4 \pi \varepsilon_0} \cdot \frac{1}{9 \mathrm{r}^2} \\ & \frac{\mathrm{E}}{\mathrm{V}}=\frac{1}{6 \mathrm{r}} \\ & \therefore \mathrm{E}=\frac{\mathrm{V}}{6 \mathrm{r}}\end{aligned}$ .

Asked in: MHT CET 2021 (23 Sep Shift 2)

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