A particle of mass m and charge q is released from rest in a...

The particle is at rest initially.

So, the initial velocity $u = 0m{s}^{- 1}$.

The force on the charged particle in the uniform electric field,

$F = qE$

$⇒ ma = qE$

$⇒ a = \dfrac{qE}{m}$

Applying the velocity-displacement equation under constant acceleration,

${v}^{2} = {u}^{2} + 2ax$

$⇒ {v}^{2} = {0}^{2} + 2\left(\dfrac{qE}{m}\right)x$

$⇒ {v}^{2} = \dfrac{2qE}{m}x$