Question

A charge $Q$ is distributed over two concentric hollow spheres of radii $r$ and $R(R > r)$ such that their surface densities are equal. The charge on smaller and bigger shells is:

& \text{A}\text{. }\dfrac{Q{{r}^{2}}}{{{r}^{2}}+{{R}^{2}}}\text{and}\dfrac{Q{{R}^{2}}}{{{r}^{2}}+{{R}^{2}}}\text{,respectively} \\\ & \text{B}\text{. Q}\left( 1+\dfrac{{{r}^{2}}}{{{R}^{2}}} \right)\text{ and Q}\left( 1+\dfrac{{{R}^{2}}}{{{r}^{2}}} \right)\text{,respectively} \\\ & \text{C}\text{. Q}\left( 1-\dfrac{{{r}^{2}}}{{{R}^{2}}} \right)\text{ and Q}\left( 1-\dfrac{{{R}^{2}}}{{{r}^{2}}} \right)\text{,respectively} \\\ & \text{D}\text{. }\dfrac{Q{{R}^{2}}}{{{r}^{2}}+{{R}^{2}}}\text{and}\dfrac{Q{{r}^{2}}}{{{r}^{2}}+{{R}^{2}}}\text{,respectively} \\\ \end{aligned}$$

Answer 1

To find the charge, assume the charges on the sphere be continuously distributed. Given that the, surface densities are equal, then $\dfrac{Q_{A}}{4\pi R^{2}} =\dfrac{Q_{B}}{4\pi r^{2}}$ also $_{A}+Q_{B}=Q$. Then using these equations we can find the charge on the surfaces of the hollow sphere.

Formula used:
$\dfrac{Q_{A}}{4\pi R^{2}} =\dfrac{Q_{B}}{4\pi r^{2}}$ And $Q_{A}+Q_{B}=Q$

Complete step by step answer:
We know that the charge density is the measure of continuous charge accumulated over a particular area. It can be measured in terms of linear charge density, which is the measure of total charge per unit length, or surface-area charge density is the charge density, which is the charge per unit area, as used here, or volume charge density which is the charge per unit volume of the object.
Let the charge on the surface of the sphere with radius $R$ be $Q_{A}$ and the charge on the surface of the sphere with radius $r$ be $Q_{B}$, assume the charge is continuously distributed on the hollow sphere surfaces.

Then given that, surface charge density i.e. the charge density, which is the charge per unit area, is equal, i.e. $\dfrac{Q_{A}}{4\pi R^{2}} =\dfrac{Q_{B}}{4\pi r^{2}}$ or $\dfrac{ Q_{A}}{ R^{2}} =\dfrac{Q_{B}}{r^{2}}$. Where $A=4\pi R^{2}$ is due to the spherical shape.
Also given that $Q_{A}+Q_{B}=Q$
Then we can say that $Q_{A}=\dfrac{QR^{2}}{R^{2}+r^{2}}$ and $Q_{B}=\dfrac{Qr^{2}}{R^{2}+r^{2}}$
Thus the charge on smaller and bigger shells is, $Q_{B}=\dfrac{Qr^{2}}{R^{2}+r^{2}}$and $Q_{A}=\dfrac{QR^{2}}{R^{2}+r^{2}}$ respectively.

Hence the answer is A. $Q_{B}=\dfrac{Qr^{2}}{R^{2}+r^{2}}$and $Q_{A}=\dfrac{QR^{2}}{R^{2}+r^{2}}$ respectively.

Note:
This question is a little tricky. Kindly read the question properly. Be careful with the options A. and D. since the question asks: the charge on smaller and bigger shells, here $Q_{B}$and $Q_{A}$. Hence A. is the answer. If the question asks the charge on bigger and smaller shells, then since $Q_{A}$ and $Q_{B}$, D. will be the answer.