A particle of mass m and charge q is released from rest in a uniform electric field. If there is no other…

The particle is at rest initially.

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

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

$F=qE$

$\Rightarrow ma=qE$

$\Rightarrow a=\frac{qE}{m}$

Applying the velocity-displacement equation under constant acceleration,

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

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

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