Question

Two metallic spheres of radii $1\text{cm}$ and $3\text{cm}$ are given charges of $-1\times {10}^{-2}\text{C}$ and $5\times {10}^{-2}\text{C}$ respectively. If these are connected by a conducting wire (potential on them becomes equal), the final charge on bigger sphere is

• A. $1\times {10}^{-2}\text{C}$
• B. $3\times {10}^{-2}\text{C}$
• C. $2\times {10}^{-2}\text{C}$
• D. $4\times {10}^{-2}\text{C}$

Answer 1

The correct option is B $3\times {10}^{-2}\text{C}$

Given:
$Q_1=5\times {10}^{-2}\text{C};Q_2=-1\times {10}^{-2}\text{C}$

$R=3\text{cm};r=1\text{cm}$

The Capacitance of the spherical body of radius $r$ is

$C=4\pi {ϵ}_{0}r.........\left(1\right)$

From the definition of capacitance of a conductor :

$C=\frac{Q}{V}$

Here $Q$ is the charge on conductor, $V$ be the potential

from eq.$\left(1\right)$

$\Rightarrow V=\frac{Q}{C}=\frac{Q}{4\pi {ϵ}_{0}r}.......\left(2\right)$


After connecting with conducting wire, let $q$ be the net charge transfer from $Q_1$ to $Q_2$ such that the potential of both spheres become equal.

Thus, $V_1=V_2$

from eq. $\left(2\right)$

$\frac{Q_1-q}{4\pi {ϵ}_{0}R}=\frac{Q_2+q}{4\pi {ϵ}_{0}r}$

$\Rightarrow \frac{Q_1-q}{R}=\frac{Q_2+q}{r}\Rightarrow \frac{Q_1-q}{3}=\frac{Q_2+q}{1}$

$\Rightarrow q=\frac{Q_1-3Q_2}{4}=\frac{5\times {10}^{-2}-(-3\times {10}^{-2})}{4}$

$\Rightarrow q=2\times {10}^{-2}\text{C}$

Therefore, the final charge on bigger sphere is,
$Q_1-q=(5-2)\times {10}^{-2}=3\times {10}^{-2}\text{C}$

$\therefore$ Option $($b) is correct choice.

$\begin{array}{c}\text{Why this question?}\\ \text{Tip: In such problems charge will flow}\\ \text{from one conductor to another such that}\\ \text{the final potential of both conductors is same.}\\ \text{Concept : It can be visualized in same manner}\\ \text{as if two different level tanks are connected}\\ \text{through a pipe. Thus water will flow}\\ \text{from higher level to lower level, right!!}\\ \text{similarly, charge will also flow from high}\\ \text{potential conductor towards lower potential conductor}\end{array}$