Question

A particle having charge $q$ and mass $m$ is released from rest at origin, in a region with uniform electric field of magnitude $E$ which is directed along the positive $x-$axis. The speed of the particle as it crosses point $x=x_0$ will be -

​​​​​​​$[$0.77 Mark$]$

• A. $\sqrt{\frac{qE{x}_0}{m}}$
• B. $\sqrt{\frac{2qE{x}_0}{m}}$
• C. $\sqrt{\frac{qE}{m{x}_0}}$
• D. $\sqrt{\frac{4qE{x}_0}{m}}$

Answer 1

The correct option is B $\sqrt{\frac{2qE{x}_0}{m}}$

From work-energy theorem,

Work done by electric field $=$ Change in kinetic energy

$\Rightarrow \text{Force}\times \text{displacement}=K.E_f-K.E_i$

$\Rightarrow qE{x}_0=\frac{1}{2}m{v}^2-0$

$\Rightarrow v=\sqrt{\frac{2qE{x}_0}{m}}$