Question

A particle of mass $0.01\text{kg}$ and charge ${10}^{-6}\text{C}$ is placed at rest in a uniform electric field $2\times {10}^6\text{N/C}$ and then released. The kinetic energy attained by the particle after moving a distance of $2\text{m}$ is
[Assume gravity free space]

• A. $1\text{J}$
• B. $2\text{J}$
• C. $3\text{J}$
• D. $4\text{J}$

Answer 1

The correct option is D $4\text{J}$

Given:
$m=0.01\text{kg};q={10}^{-6}\text{C}$
$E=2\times {10}^6\text{N/C};d=2\text{m}$

So, the force $F$ experienced by charge in the electric field

$F=qE={10}^{-6}\times 2\times {10}^6=2\text{N}$

On applying work - energy theorem,

Work done by all forces = change in kinetic energy

$\Rightarrow W_{electric}=K.E_f-K.E_i$

$\Rightarrow W_{electric}=\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2$

$\because$ charge released from rest, so $v_i=0$

$\Rightarrow F.d=\frac{1}{2}m{v}_f^2-0=K.E_f$

$\Rightarrow K.E_f=2\times 2$

$\Rightarrow K.E_f=4\text{J}$

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