Question

A charge $3\mu \text{C}$ is released from rest from a point $P$ in an electric field where electric potential is $20\text{V}$. When the charge reaches slowly at infinity its kinetic energy becomes:

• A. $60\mu \text{J}$
• B. $45\mu \text{J}$
• C. $30\mu \text{J}$
• D. $120\mu \text{J}$

Answer 1

The correct option is A $60\mu \text{J}$

Given:
charge $q=3\times {10}^{-6}\text{C};V_p=20\text{V}$

When charge reaches infinity its potential becomes zero.

$\Rightarrow V_f=V_{\infty }=0$ and $V_i=V_p=20\text{V}$

Work done by external force in taking a charge from point $\text{P}$ to $\infty$ very slowly is given by

$\Rightarrow \frac{W_{ext(P\to \infty )}}{q}=V_{\infty }-V_P$

$\Rightarrow W_{ext}=q(V_{\infty }-V_P)=-q{V}_p$

$\Rightarrow W_{ext}=-60\mu \text{J}$

Thus, $W_{elec}=-W_{ext}=60\mu \text{J}$

Now let us come to the situation when charge is released from rest and there is no external force to resist the electrostatic force.

Now, applying work- energy theorem,

$W_{allforce}=\Delta K.E$

$\Rightarrow W_{elec}=\Delta K.E=K.E_f-K.E_i$

$\Rightarrow 60\mu \text{J}=K.E_f-0(\because K.E_i=0)$

$\therefore K.E_f=60\mu \text{J}$

So, option $\left(a\right)$ is correct.

$\begin{array}{c}\text{Why this question}\\ \text{It challenges you to apply the understanding of}{W}_{ext}\\ \text{and}{W}_{elec}\text{at finest level.}\\ \\ \text{Tip: In the problem, charge is taken from region of}\\ \text{High potential}(V_p=20\text{V})\text{to low potential}(V_{\infty }=0)\\ \text{This is a natural tendency of +ve charge movement to}\\ \text{gain more stability}\\ \\ \text{Trick: First Find}{W}_{elec}\text{by taking charge slowly}(\Delta K.E=0),\\ \text{then use it to find}K.E_f\text{when it is set free to move}\end{array}$