Question

A particle of mass 'm' and charge 'q' is placed at rest in uniform electric field E and then released. The momentum gained by the particle after moving a distance 'y' is \[\sqrt{2x}.\] Then x is equal to ______.

Options

• Eqy

• \[\frac{2\text{Eqy}}{m}\]

• \[\sqrt{2\text{mEqy}}\]

• mEqy

Answer 1

A particle of mass 'm' and charge 'q' is placed at rest in uniform electric field E and then released. The momentum gained by the particle after moving a distance 'y' is \[\sqrt{2x}.\] Then x is equal to mEqy.

Explanation:

Force on the particle, F = qE

Acceleration \[\mathrm{a=\frac{F}{m}=\frac{qE}{m}}\]

\[\mathrm{v^2=2as=2\frac{qE}{m}y}\]     (given, distance (s) = y)

\[\therefore\quad\mathrm{v}=\sqrt{\frac{2\mathrm{qEy}}{\mathrm{m}}}\]

Momentum, \[\mathrm{mv=\sqrt{\frac{2Eqy}{m}}\times m=\sqrt{2mEqy}}\]

\[\therefore\] x = mEqy