Question

A uniform electric field is directed along the x-axis between the parallel plated of a charged capacitor, separated by a distance. A positive point charge of mass is released from rest at point A next to the positive plate and accelerates to point B next to the negative plat.

Find the speed of the particle at point B if the motion is under constant acceleration.

Answer 1

Work done = Change in kinetic energy

We know that work done in electric charge, $W=qEd=\frac{1}{2}mv^2$

The speed of the particle at point B can be found by making v the subject of the formulae

$qEd=\frac{1}{2}mv^2$

$\frac{2qEd}{m}=v^2$

$v^2=\frac{2qEd}{m}$

$v=\sqrt\frac{2qEd}{m}$