Question

A solid sphere of radius ${{R}_{1}}$ and volume charge density $\rho =\dfrac{{{\rho }_{\circ }}}{r}$ 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, ${{\rho }_{\circ }}$ is a positive constant and r is the distance from the center of the sphere. The ratio $\dfrac{{{R}_{1}}}{{{R}_{2}}}$ is
a) $\dfrac{\sigma }{{{\rho }_{\circ }}}$
b)$\sqrt{\dfrac{2\sigma }{{{\rho }_{\circ }}}}$
c)$\sqrt{\dfrac{{{\rho }_{\circ }}}{2\sigma }}$
d)$\dfrac{{{\rho }_{\circ }}}{\sigma }$

Answer 1

The net charge in the above system is the sum of the total charge in the solid sphere and the total charge on the surface of the solid sphere. The charge distribution of the solid sphere varies with r and hence the total charge has to be integrated. Once the equation for the total charge in the system is obtained equating this to zero will give the ratio of the radius of the two spheres.

Formula used:
$\sigma =\dfrac{{{q}_{2}}}{4\pi R_{2}^{2}}$
${{q}_{1}}=\int\limits_{0}^{{{R}_{1}}}{\rho dV}$

Complete step-by-step answer:
In the question it is given that the solid hollow sphere of radius ${{R}_{1}}$ has a volume charge distribution of $\rho =\dfrac{{{\rho }_{\circ }}}{r}$ . The charge distribution varies with the distance from the center of the sphere. The volume charge density is defined as the ratio of charge enclosed to that of the volume of the of the particular object.
Hence integrating the total charge ${{q}_{1}}$ of the solid sphere using volume integral from r=0 to r=${{R}_{1}}$ we get,
$\begin{aligned} & {{q}_{1}}=\int\limits_{0}^{{{R}_{1}}}{\rho dV} \\\ & \because dV=4\pi {{r}^{2}}dr \\\ & \Rightarrow {{q}_{1}}=\int\limits_{0}^{{{R}_{1}}}{\dfrac{{{\rho }_{\circ }}}{r}}(4\pi {{r}^{2}}dr)=4\pi {{\rho }_{\circ }}\int\limits_{0}^{{{R}_{1}}}{rdr} \\\ & \Rightarrow {{q}_{1}}=4\pi {{\rho }_{\circ }}\left[ \dfrac{{{r}^{2}}}{2} \right]_{0}^{{{R}_{1}}}=4\pi {{\rho }_{\circ }}\left[ \dfrac{{{R}_{1}}^{2}}{2}-\dfrac{{{(0)}^{2}}}{2} \right] \\\ & \therefore {{q}_{1}}=2\pi {{\rho }_{\circ }}{{R}_{1}}^{2} \\\ \end{aligned}$
The surface charge density is defined as the ratio of the charge on the surface to that the surface area of the object. Let us say the sphere of radius ${{R}_{2}}$ has a surface charge density of $\sigma$ . Then the total charge ${{q}_{2}}$ on the surface of the sphere is equal to,
$\begin{aligned} & \sigma =\dfrac{{{q}_{2}}}{4\pi R_{2}^{2}} \\\ & \therefore {{q}_{2}}=\sigma 4\pi R_{2}^{2} \\\ \end{aligned}$
It is given that the net charge in the system is zero implying ${{q}_{1}}={{q}_{2}}$ . Hence by equating the charges on the surface and the charge in the solid sphere we get,
$\begin{aligned} & {{q}_{1}}={{q}_{2}} \\\ & \sigma 4\pi R_{2}^{2}=2\pi {{\rho }_{\circ }}{{R}_{1}}^{2} \\\ & \Rightarrow \sigma 2R_{2}^{2}={{\rho }_{\circ }}{{R}_{1}}^{2} \\\ & \Rightarrow \dfrac{R_{2}^{2}}{{{R}_{1}}^{2}}=\dfrac{{{\rho }_{\circ }}}{2\sigma } \\\ & \therefore \dfrac{{{R}_{2}}}{{{R}_{1}}}=\sqrt{\dfrac{{{\rho }_{\circ }}}{2\sigma }} \\\ \end{aligned}$
Hence the correct answer of the above question is option c.

So, the correct answer is “Option C”.

Note: It is to be noted that the charge in the system is zero i.e. the difference in the charges on the either bodies is zero and hence they are equal. For a variable charge distribution, the total charge is always to be integrated. For volume distribution volume integral and for surface distribution surface integral is needed to be used.