Question

A conducting sphere of radius 10 cm is charged with 10 $\mu$C. Another uncharged sphere of radius 20 cm is allowed to touch it for some time. After that if the spheres are separated, then surface density of charges on the spheres will be in the ratio of

• A. 1 : 1
• B. 2 : 1
• C. 1 : 3
• D. 4 : 1

Answer 1

Answer: (B)

Let the common potential after the touch is V. So, applying conservation of charge
$10 \times10^{-6} = V \times C_{1} + V \times C_{2} $
$V = \frac{10 \times10^{-6}}{\left(C_{1} + C_{2}\right)} $
Charge on first sphere
$= C_{1}V = \frac{10 \times10^{-6}}{\left(C_{1} + C_{2}\right)} \times C_{1}$
Charge on second sphere
$ = C_{2}V = \frac{10 \times10^{-6}}{\left(C_{1} + C_{2}\right)} \times C_{2}$
Charge densities are,
$ = \frac{10 \times10^{-6} \times C_{1}}{\left(C_{1} + C_{2}\right)e \pi r_{1^{2}}} \& \frac{10 \times10^{-6} \times C_{2}}{\left(C_{1} + C_{2}\right)4 \pi r_{2^{2}}} $
and their ratio $= \frac{C_{1}}{C_{2}} \times \frac{r_{2^{2}}}{r_{1^{2}}}$
$ = \frac{4\pi\in_{0} r_{1}}{4 \pi\in_{0} r_{2}} \times\frac{r_{2^{2}}}{r_{1^{2}}} $
$= \frac{r_{2}}{r_{1}} = \frac{20}{10} = 2:1$ [Capacity of spherical capacitor = $4 \pi \in_0 R$]