Question

A particle of mass $m$ and charge $q$ at rest is released in a uniform electric field between parallel planes of charge $+q$ and $-q$ respectively. The particle accelerates towards the other place a distance ${}^{\prime }{d}^{\prime }$ away. The speed at which it strikes the opposite plane is:

• A. $\sqrt{qEd/m}$ and $j$
• B. $\sqrt{2qEd/m}$ along $-j$
• C. $\sqrt{qm/Ed}$ along $j$
• D. $\sqrt{qm/Ed}$ along $-j$

Answer 1

The correct option is B $\sqrt{2qEd/m}$ along $-j$

We have

$\frac{1}{2}m{v}^2=qEd$

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

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

Thus the particle strikes the opposite plane with speed $\sqrt{\frac{2qEd}{m}}$ along $-j$.