Question

Two spheres of radii 2 cm and 3 cm have equal surface charge density. Calculate the ratio of their charges.

• A. 2 : 3
• B. 3 : 2
• C. 8 : 27
• D. 4 : 9

Answer 1

The correct option is B 3 : 2

Both the spherical conductors charged to the same potential.

$\therefore$ Potential of sphere, at surface=Potential of sphere $2$ at surface.

$\begin{array}{c}\Rightarrow \frac{k{q}_1}{r_1}=\frac{k{q}_2}{r_2}\\ \Rightarrow \frac{q_1}{r_1}=\frac{q_2}{r_2}.........\left(i\right)\end{array}$

Now surface charge density $\sigma =\frac{q}{4\pi r^2}$

$\begin{array}{c}\frac{{\sigma }_1}{{\sigma }_2}=\frac{\frac{q}{4\pi {r_1}^2}}{\frac{q}{4\pi {r_2}^2}}\\ =\frac{r_2}{r_1}\left[from\left(i\right)\right]\\ =\frac{3}{2}\\ \Rightarrow {\sigma }_1:{\sigma }_2=3:2\end{array}$

Hence, Option $B$ is correct.