Question

A charged particle of mass m and charge q is released from rest in a uniform electric field. The momentum attained by particle after moving a distance x is​

Answer 1

The momentum attained by particle after moving a distance x is √mqEx

Explanation:

The force on the particle of mass m and charge q in a uniform electric field E is

$F=qE$

Therefore, the acceleration of the particle

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

The velocity of particle after moving a distance x, if it starts from rest

Using the third equation of motion

$v^2=u^2+2as$

$\implies v^2=0^2+2\times\frac{qE}{m}\times x$

$\implies v=2\sqrt{\frac{qEx}{m}}$

Thus, the momentum of the particle

$p=mv$

$\implies p=m\times2\sqrt{\frac{qEx}{m}}$

$\implies p=2\sqrt{mqEx}$

Hope this answer is helpful.

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Q: A particle of mass m and charge q is placed at rest in a uniform electric field e and then released. The kinetic energy attained by the particle after moving a distance y is:

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