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
$10\times 10^{-6}=V\times C_1+V\times C_2$
$V=\frac{10\times 10^{-6}}{\left(C_1+C_2\right)}$
Charge on first sphere
$=C_{1}V=\frac{10\times 10^{-6}}{\left(C_1+C_2\right)}\times C_1$
Charge on second sphere
$=C_{2}V=\frac{10\times 10^{-6}}{\left(C_1+C_2\right)}\times C_2$
Charge densities are,
$=\frac{10\times 10^{-6}\times C_1}{\left(C_1+C_2\right)e\pi r_{1^2}}\&\frac{10\times 10^{-6}\times C_2}{\left(C_1+C_2\right)4\pi r_{2^2}}$
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 = $4\pi {\in }_{0}R$]