Question

Two spherical conductors $A_1$ and $A_2$ of radii $r_1$ and $r_2$ and carrying charges $q_1$ and $q_2$ are connected in air by a copper wire as shown in the figure. Then the equivalent capacitance of the system is:

• A. $\frac{4\pi {ϵ}_0{r}_1{r}_2}{r_2-r_1}$
• B. $4\pi {ϵ}_0(r_1+r_2)$
• C. $4\pi {ϵ}_0{r}_2$
• D. $4\pi {ϵ}_0{r}_1$

Answer 1

Answer: (B)

When we calculate capacitance of single sphere shell , we assume outer shell is earth and it has infinite radius.
By Gauss's law the electric field at a distance r in between the shell and earth is
$E.4\pi r^2=\frac{q_1}{{ϵ}_0}\Rightarrow E=\frac{q_1}{4\pi {ϵ}_0{r}^2}$
The potential difference between shell and earth is ${\int }_0^{V}dV=-{\int }_{\infty }^{r_1}Edr$
$V=-\frac{q_1}{4\pi {ϵ}_0}{\int }_{\infty }^{r_1}\frac{dr}{r^2}=\frac{q_1}{4\pi {ϵ}_0{r}_1}$
Thus the capacitance of $A_1$ is $C_1=\frac{q_1}{V}=4\pi {ϵ}_0{r}_1$
Similarly for sphere $A_2$ the capacitance $C_2=4\pi {ϵ}_0{r}_2$
Here both are connected in parallel so the equivalent capacitance is
$C_{eq}=C_1+C_2=4\pi {ϵ}_0(r_1+r_2)$