MHT CET · Physics · Electrostatics
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
- A \(\frac{\mathrm{V}}{3 \mathrm{r}}\)
- B \(3 \mathrm{Vr}\)
- C \(\frac{\mathrm{V}}{\mathrm{r}}\)
- D \(\frac{\mathrm{V}}{6 \mathrm{r}}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{V}}{6 \mathrm{r}}\)
Step-by-step Solution
Detailed explanation
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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