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. ${\sigma }_1=\frac{5}{6}\sigma ,{\sigma }_2=\frac{5}{2}\sigma$
• B. ${\sigma }_1=\frac{5}{2}\sigma ,{\sigma }_2=\frac{5}{6}\sigma$
• C. ${\sigma }_1=\frac{5}{2}\sigma ,{\sigma }_2=\frac{5}{3}\sigma$
• D. ${\sigma }_1=\frac{5}{3}\sigma ,{\sigma }_2=\frac{5}{6}\sigma$

Answer 1

Answer: (D)

Before contact, $Q_1=\sigma \cdot 4\pi R^2$ and $Q_2=\sigma \cdot 4\pi {\left(2R\right)}^2$
As, surface charge density, $\sigma =\frac{\text{Net charge}\left(Q\right)}{\text{Surface area}\left(A\right)}$
Now, after contact, $Q_1^{\prime }+Q_2^{\prime }=Q_1+Q_2=5Q_1$
$=5\left(\sigma \cdot 4\pi R^2\right)$
They will be at equal potentials, so,
$\frac{Q_1^{\prime }}{R}=\frac{Q_2^{\prime }}{2R}$
$\Rightarrow Q_2^{\prime }=2Q_1^{\prime }$
$\therefore 3Q_1^{\prime }=5\left(\sigma \cdot 4\pi R^2\right)$ (From equation (i))
and $Q_2^{\prime }=\frac{10}{3}\left(\sigma \cdot 4\pi R^2\right)$
$\therefore {\sigma }_1=\frac{5}{3}\sigma$ and ${\sigma }_2=\frac{5}{6}\sigma$