Question

A conducting sphere of radius $10cm$ is charged with $10\mu C$. Another uncharged sphere of radius 20 cm is allowed to touch it for sometime, 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)

After touch if the common potential is V, according to charge conservation

$Q=C_{1}V+C_{2}V$ here $Q=10\mu C$ and $C_1,C_2$ are the capacitances of two sphere.

or $V=\frac{Q}{C_1+C_2}$

After touch the charge on first sphere is $Q_1=C_{1}V=\frac{C_{1}Q}{C_1+C_2}$ and charge on second sphere is $Q_2=C_{2}V=\frac{C_{2}Q}{C_1+C_2}$

Charge densities are ${\sigma }_1=\frac{Q_1}{S_1}=\frac{C_{1}V}{4\pi r_1^2}$ and ${\sigma }_2=\frac{Q_2}{S_1}=\frac{C_{2}V}{4\pi r_2^2}$

Thus, $\frac{{\sigma }_1}{{\sigma }_2}=\frac{C_1}{C_2}\times \frac{r_2^2}{r_1^2}=\frac{4\pi {ϵ}_0{r}_1}{4\pi {ϵ}_0{r}_2}\times \frac{r_2^2}{r_1^2}=\frac{r_2}{r_1}=20/10=2/1$ as the capacitance of isolated sphere $=4\pi {ϵ}_{0}R$