Question

A large conducting sphere of radius $r$, having a charge $Q$ is placed in contact with a small neutral conducting sphere of radius $r^{\prime }$ and is then separated. The charge on smaller sphere will now be:

• A. $\frac{Q(r+r^{\prime })}{r}$
• B. $\frac{Q(r-r^{\prime })}{r^{\prime }}$
• C. $\frac{Qr}{r^{\prime }+r}$
• D. $\frac{Q{r}^{\prime }}{r^{\prime }+r}$

Answer 1

The correct option is D $\frac{Q{r}^{\prime }}{r^{\prime }+r}$

Charges will be redistributed when both spheres are placed in contact with each other, such that common potential is achieved.

Charges are distributed in ratio of capacitance. Let $Q_1^{\prime }$ and $Q_2^{\prime }$ be the charges of bigger and smaller spheres, respectively.

$\because V_1=V_2$

$\therefore \frac{Q_1^{\prime }}{C_1}=\frac{Q_2^{\prime }}{C_2}$

$\Rightarrow \frac{Q_1^{\prime }}{Q_2^{\prime }}=\frac{C_1}{C_2}=\frac{r}{r^{\prime }}(\because C=4\pi {ϵ}_{0}r)$

$\Rightarrow \frac{Q_1^{\prime }}{Q_2^{\prime }}+1=\frac{r}{r^{\prime }}+1$

$\therefore \frac{Q_1^{\prime }+Q_2^{\prime }}{Q_2^{\prime }}=\frac{r+r^{\prime }}{r^{\prime }}......\left(1\right)$

From conservation of charge on system of two spheres,

$Q_1^{\prime }+Q_2^{\prime }=Q+0=Q........\left(2\right)$

From eqs.$\left(1\right)$ and $\left(2\right)$

$\Rightarrow \frac{Q}{Q_2^{\prime }}=\frac{r+r^{\prime }}{r^{\prime }}$

$\therefore Q_2^{\prime }=\frac{Q{r}^{\prime }}{r+r^{\prime }}$

Hence, option $\left(d\right)$ is correct.