Question

Two metal spheres, one of radius R and the other of radius 2R, both have same surface charge density $\sigma$. If they are brought in contact and separated, then the new surface charge densities on each of the sphere are respectively

• A. $\frac{5}{2}\sigma ,\frac{5}{4}\sigma$
• B. $\frac{5}{3}\sigma ,\frac{5}{6}\sigma$
• C. $\frac{3}{5}\sigma ,\frac{6}{5}\sigma$
• D. $\frac{2}{3}\sigma ,\frac{1}{2}\sigma$

Answer 1

Answer: (B)

$\frac{5}{3}\sigma ,\frac{5}{6}\sigma$

Before contact, charges of each spheres,

$q_1=\sigma 4\pi R^2;q_2=\sigma 4\pi {\left(2R\right)}^2=4q_1$ When the two sphere are brought in contact, their charges are shared till their potentials become equal i.e., $V_1=V_2$

$\therefore \frac{q_1^{\prime }}{4\pi {\epsilon }_{0}R}=\frac{q_2^{\prime }}{4\pi {\epsilon }_0\left(2R\right)}$

$\therefore q_2^{\prime }=2q_1^{\prime }....\left(i\right)$

As there is no loss of charge in the process

$\therefore q_1^{\prime }+q_2^{\prime }=q_1+q_2=q_1+4q_1=5q_1=5\left(\sigma 4\pi R^2\right)$

or $q_1^{\prime }+2q_1^{\prime }=5\sigma 4\pi R^2$ (using (i))

$q_1^{\prime }=\frac{5}{3}\sigma 4\pi R^2;q_2^{\prime }=2q_1^{\prime }=\frac{10}{3}\left(\sigma 4\pi R^2\right)$

${\sigma }_1=\frac{q_1^{\prime }}{4\pi R^2}=\frac{5}{3}\sigma ,{\sigma }_2=\frac{q_2^{\prime }}{4\pi {\left(2R\right)}^2}=\frac{5}{6}\sigma$