Question

Two metal spheres, one of radius R and the other of radius 2R respectively have the same surface charge density s. They are brought in contact and separated.
What will be the new surface charge densities on them?​

Answer 1

Suppose $\sf q_1$ is the charge on sphere with radius R and $\sf q_2$ is the charge of the sphere with radius 2R.

Since, they have equal surface charge density.

$\sf \: \sigma = \dfrac{q_1}{4\pi {r}^{2} } = \dfrac{q_2}{4\pi(2 {r}^{2}) } \\ \\ \implies \sf \: q_2 = 4q_1 - - - - - - - (1)$

When two charged surfaces are brought in contact, the same configuration is maintained until both of them reach common potential. In other words, after separation the spheres are at same potential and have lost or gained some charge. Therefore (let $\sf q_3$ and $\sf q_4$ be the new charges) -

$\sf V = \dfrac{kq_3}{r} = \dfrac{kq_4}{2r} \\ \\ \implies \sf \: q_4 = 2q_3 - - - - - - - (2)$

Since, charge is always conserved.

$\sf \: q_1 + q_2 = q_3 + q_4$

Using relations (1) and (2),

$\implies \sf \: 5q_1 = 3q_3 \\ \\ \implies \sf \: q_3 = \dfrac{5}{3} q_1$

Dividing by 4πr² on both sides,

$\implies \sf \: \dfrac{q_3}{4\pi {r}^{2} } = \dfrac{5}{3(4\pi {r}^{2}) } q_1 \\ \\ \implies \boxed{ \boxed{ \sf \: \sigma_{r} = \dfrac{5}{3} \sigma}}$

Note : Sigma subscript r denotes surface charge density of sphere with radius r.

Likewise, using relations (1) and (2),

$\implies \sf \: \dfrac{5}{4} q_2 = \dfrac{3}{2} q_4 \\ \\ \implies \sf \: q_4 = \dfrac{5}{6} q_2$

Dividing by 4π(2r)² on both sides,

$\implies \sf \: \dfrac{q_4}{4\pi {(2r)}^{2} } = \dfrac{5}{6(4\pi {(2r)}^{2}) } q_2 \\ \\ \implies \boxed{ \boxed{\sf \: \sigma_{2r} = \dfrac{5}{6} \sigma}}$

Note : Sigma subscript 2r denotes surface charge density of sphere with radius 2r.

Answer 2

Here is the answer...hope it helps ❄️