Question

Two identical charges each having $Q$ are fixed at a distance $2d$ apart. A third particle having a mass $m$ and charge$-q$ is projected from the midpoint of the line joining the two charges along a direction perpendicular to the line joining the two charges. Find the minimum velocity to be imparted, such that $-q$ charge reaches infinity.

• A. $\sqrt{\frac{KQq}{md}}$
• B. $2\sqrt{\frac{KQq}{md}}$
• C. $\sqrt{\frac{KQq}{2md}}$
• D. $\frac{1}{2}\sqrt{\frac{KQq}{md}}$

Answer 1

The correct option is B

$2\sqrt{\frac{KQq}{md}}$

Applying the energy conservation principle,

${K.E}_i+{P.E}_i={K.E}_f+{P.E}_f...\left(1\right)$

where,
$K.E_i=\frac{1}{2}m{V}_{min}^2=$ Initial kinetic energy of the system.

here, $V_{min}$ is the velocity to be imparted to $-q$, such that it reaches infinity.

Initial potential energy of the system,
$P.E_i=\frac{-KQq}{d}+\frac{-KQq}{d}+\frac{K{Q}^2}{2d}$

When the particle just reaches infinity, its potential and kinetic energy becomes zero.

$\therefore$ Final potential energy of the system $P.E_f=0+0+\frac{K{Q}^2}{2d}$

For minimum velocity, $K.E_f=0$, Final potential energy, $P.E_f=\frac{K{Q}^2}{2d}$

Substituting the values in $\left(1\right)$, we have

$\frac{1}{2}m{V}_{min}^2-\frac{KQq}{d}-\frac{KQq}{d}+\frac{K{Q}^2}{2d}=0+\frac{K{Q}^2}{2d}$

$\Rightarrow \frac{1}{2}m{V}_{min}^2=\frac{2KQq}{d}$

$\Rightarrow V_{min}=\sqrt{\frac{4KQq}{md}}$

$\therefore V_{min}=2\sqrt{\frac{KQq}{md}}$

Hence, option (b) is the correct answer.