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 one 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)

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

Answer 2

: Before contact, charges of each spheres,

$q_{1} = \sigma 4\pi R^{2}$ and $q_{2} = \sigma 4\pi(2R)^{2} = 4q_{1}$

When the two sphere sphere aer brought in contact, their charges are shared till their potentials become equal i.e.

$V_{1} = V_{2}.$

$\therefore\frac{q'_{l}}{4\pi\varepsilon_{0}R} = \frac{q'_{2}}{4\pi\varepsilon_{0}(2R)}$

$\therefore q'_{2} = 2q'_{l}$ … (i)

As there is no loss of charge in the process

$\therefore q _ { 1 } ^ { \prime } + q _ { 2 } ^ { \prime } = q _ { 1 } + q _ { 2 } = q _ { 1 } + 4 q _ { 1 } = 5 q _ { 1 } = 5 \left( \sigma 4 \pi R ^ { 2 } \right)$

or $q'_{1} + 2q'_{1} = 5\sigma 4\pi R^{2}$ (using (i))

$q'_{1} = \frac{5}{3}\sigma 4\pi R^{2}$ and $q'_{2} = 2q'_{l} = \frac{10}{3}(\sigma 4\pi R^{2})$

Hence, $\sigma_{1} = \frac{q'_{l}}{4\pi R^{2}} = \frac{5}{3}\sigma,\sigma_{2} = \frac{q'_{2}}{4\pi(2R)^{2}} = \frac{5}{6}\sigma$