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A hollow metal sphere has a radius ' $r$ '. The potential difference between a point on its surface and at a point at a distance ' $3 r$ ' from its center is ' $\mathrm{V}$ '. The electric intensity at the distance ' $3 \mathrm{r}$ ' from the center of the sphere will be
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Verified Answer
The correct answer is:
$\frac{\mathrm{V}}{6 \mathrm{r}}$
Let the charge on the sphere is Q.
$$
\begin{aligned}
& \mathrm{V}=\frac{\mathrm{kQ}}{\mathrm{r}}-\frac{\mathrm{kQ}}{3 \mathrm{r}}=\frac{2 \mathrm{kQ}}{3 \mathrm{r}} \\
& \Rightarrow \mathrm{kQ}=\frac{3 \mathrm{Vr}}{2} \\
& \mathrm{E}=\frac{\mathrm{kQ}}{(3 \mathrm{r})^2}=\frac{1}{(3 \mathrm{r})^2} \cdot \frac{3 \mathrm{Vr}}{2}=\frac{\mathrm{V}}{6 \mathrm{r}}
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{V}=\frac{\mathrm{kQ}}{\mathrm{r}}-\frac{\mathrm{kQ}}{3 \mathrm{r}}=\frac{2 \mathrm{kQ}}{3 \mathrm{r}} \\
& \Rightarrow \mathrm{kQ}=\frac{3 \mathrm{Vr}}{2} \\
& \mathrm{E}=\frac{\mathrm{kQ}}{(3 \mathrm{r})^2}=\frac{1}{(3 \mathrm{r})^2} \cdot \frac{3 \mathrm{Vr}}{2}=\frac{\mathrm{V}}{6 \mathrm{r}}
\end{aligned}
$$
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