Question

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

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

Answer 1

Answer: (D)

$\sigma_{1} =\frac{5}{3}\sigma,\quad\sigma_{2} =\frac{5}{6}\sigma$

Answer 2

Before contact, $Q_{1}=\sigma\cdot4\pi R^{2}$ and $Q_{2}=\sigma\cdot4\pi\left(2R\right)^{2}$
As, surface charge density, $\sigma=\frac{\text{Net charge} \left(Q\right)}{\text{Surface area} \left(A\right)}$
Now, after contact, $Q'_{1}+Q'_{2}=Q_{1}+Q_{2}=5Q_{1}$
$=5\left(\sigma\cdot4\pi R^{2}\right)$
They will be at equal potentials, so,
$\frac{Q'_{1}}{R}=\frac{Q'_{2}}{2R}$
$\Rightarrow Q'_{2}=2Q'_{1}$
$\therefore 3Q'_{1}=5\left(\sigma\cdot4\pi R^{2}\right)$ (From equation (i))
and $Q'_{2}=\frac{10}{3}\left(\sigma\cdot4\pi R^{2}\right)$
$\therefore \sigma_{1}=\frac{5}{3}\sigma$ and $\sigma_{2}=\frac{5}{6}\sigma$