MHT CET · Physics · Electrostatics
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
- A \(\frac{\mathrm{V}}{8 \mathrm{r}}\)
- B \(\frac{\mathrm{V}}{2 \mathrm{r}}\)
- C \(\frac{\mathrm{V}}{4 \mathrm{r}}\)
- D \(\frac{\mathrm{V}}{6 \mathrm{r}}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{V}}{6 \mathrm{r}}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{V}=\frac{\mathrm{q}}{4 \pi \varepsilon_0 \mathrm{r}}\left[1-\frac{1}{3}\right]=\frac{2}{3} \cdot \frac{\mathrm{q}}{4 \pi \varepsilon_0 \mathrm{r}} \\ & \mathrm{E}=\frac{\mathrm{q}}{4 \pi \varepsilon_0} \frac{1}{(3 \mathrm{r})^2}=\frac{\mathrm{q}}{4 \pi \varepsilon_0} \cdot \frac{1}{9 \mathrm{r}^2} \\ & \frac{\mathrm{E}}{\mathrm{V}}=\frac{1}{6 \mathrm{r}} \\ & \therefore \mathrm{E}=\frac{\mathrm{V}}{6 \mathrm{r}}\end{aligned}\)
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