Search any question & find its solution
Question:
Answered & Verified by Expert
A solid sphere of radius $R_{1}$ and volume charge density $\rho=\frac{\rho_{0}}{\mathrm{r}}$ is enclosed by a hollow sphere of radius $R_{2}$ with negative surface charge density $\sigma$, such that the total charge in the system is zero. $\rho_{0}$ is a positive constant and $r$ is the distance from the centre of the sphere. The ratio $\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}$ is
Options:
Solution:
1038 Upvotes
Verified Answer
The correct answer is:
$\sqrt{\rho_{0} /(2 \sigma)}$
Let us first calculate the total charge on the soild sphere.
Let us consider a concentric sphere of radius $r$ and thickness dr.
Then volume of thesphere, $\mathrm{d} \mathrm{V}=4 \pi \mathrm{r}^{2} \mathrm{dr}$
Given, the volume charge density of the sphere $=\rho=\frac{\rho_{0}}{\mathrm{r}}$
$\therefore$ Charge on this sphere,
$\mathrm{dQ}=\rho \cdot \mathrm{dV}=\frac{\rho_{0}}{\mathrm{r}} \cdot 4 \pi \mathrm{r}^{2} \mathrm{dr}=4 \pi \rho \cdot \mathrm{rdr}$
$\therefore$ Total charge on the whole solid sphere,
$\begin{array}{l}
\begin{aligned}
\mathrm{Q}_{\mathrm{S}} &=\int_{0}^{\mathrm{R}_{1}} \mathrm{~d} \mathrm{Q}=4 \pi \rho_{0} \int_{0}^{\mathrm{R}_{1}} \mathrm{rdr} \\
&=4 \pi \rho_{0} \frac{\mathrm{R}_{1}^{2}}{2}=2 \pi \rho_{0} \mathrm{R}_{1}^{2}
\end{aligned} \\
\therefore \mathrm{Q}_{\mathrm{S}}=2 \pi \rho_{0} \mathrm{R}_{1}^{2} ...(1)
\end{array}$
Now, the total charge on the hollow sphere, $\mathrm{Q}_{\mathrm{h}}=-\left(4 \pi \mathrm{R}_{2}^{2}\right) \sigma$ ...(2)
By question, $Q_{s}+Q_{h}=0$
$\begin{array}{l}
\Rightarrow 2 \pi \rho_{0} \mathrm{R}_{1}^{2}=4 \pi \mathrm{R}_{2}^{2}{ }^{\mathrm{h}} \\
\Rightarrow\left(\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}\right)^{2}=\frac{\rho_{0}}{2 \sigma} \Rightarrow \frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\sqrt{\frac{\rho_{0}}{2 \sigma}}
\end{array}$
Let us consider a concentric sphere of radius $r$ and thickness dr.
Then volume of thesphere, $\mathrm{d} \mathrm{V}=4 \pi \mathrm{r}^{2} \mathrm{dr}$
Given, the volume charge density of the sphere $=\rho=\frac{\rho_{0}}{\mathrm{r}}$
$\therefore$ Charge on this sphere,
$\mathrm{dQ}=\rho \cdot \mathrm{dV}=\frac{\rho_{0}}{\mathrm{r}} \cdot 4 \pi \mathrm{r}^{2} \mathrm{dr}=4 \pi \rho \cdot \mathrm{rdr}$
$\therefore$ Total charge on the whole solid sphere,
$\begin{array}{l}
\begin{aligned}
\mathrm{Q}_{\mathrm{S}} &=\int_{0}^{\mathrm{R}_{1}} \mathrm{~d} \mathrm{Q}=4 \pi \rho_{0} \int_{0}^{\mathrm{R}_{1}} \mathrm{rdr} \\
&=4 \pi \rho_{0} \frac{\mathrm{R}_{1}^{2}}{2}=2 \pi \rho_{0} \mathrm{R}_{1}^{2}
\end{aligned} \\
\therefore \mathrm{Q}_{\mathrm{S}}=2 \pi \rho_{0} \mathrm{R}_{1}^{2} ...(1)
\end{array}$
Now, the total charge on the hollow sphere, $\mathrm{Q}_{\mathrm{h}}=-\left(4 \pi \mathrm{R}_{2}^{2}\right) \sigma$ ...(2)
By question, $Q_{s}+Q_{h}=0$
$\begin{array}{l}
\Rightarrow 2 \pi \rho_{0} \mathrm{R}_{1}^{2}=4 \pi \mathrm{R}_{2}^{2}{ }^{\mathrm{h}} \\
\Rightarrow\left(\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}\right)^{2}=\frac{\rho_{0}}{2 \sigma} \Rightarrow \frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\sqrt{\frac{\rho_{0}}{2 \sigma}}
\end{array}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.