# A charged particle of charge q and mass m is released from rest in an uniform electric feild E.The kinetic energy of the charged particle after time"t" seconds is

Vikas TU
14149 Points
11 years ago

1) qE + mg = ma
a = qE/m + g

2) v = 0 + at

v = (qE/m + g)t

3) K.E =1/2mv^2

= 1/2m [(qE/m + g)t]^2 Joule

Chetan Mandayam Nayakar
312 Points
11 years ago

force=qE,impulse=qEt=momentum=p, k.e.=p2/2m

dhaval Rohilla
13 Points
6 years ago
If the effect of gravity is neglected then, ma =qE. Then from kinematic eqn v=u+at as the motion starts from rest then u=0,putting the value of v^2 in the kinetic energy formula we get E^2q^2t^2 \ 2m
Udit Thacker
15 Points
5 years ago
The particle here will perform uniformly accelerated motion as it is acted upon by a constant electrostatic force qE. The acceleration of the particle is thus qE/m. Now just apply the equation of motion v = u + at .
initial velocity u=0
acceleration a=qE/m
by v=u+at
v=0 + (qE/m)t

v= (qE/m)t
KE= ½(mv2)
KE= 1/2m(qE/m)2t2
Yash Chourasiya
4 years ago
Dear Student

A force is generated because of the elective field
F = qE
⇒ a = qE/m​.

As the particle starts from rest and acceleration is constant applying equation of motion.
V = u + at
V = 0 + qEt/m​
V = qET/m​
Then,KE = 1/2(​mv2)
= (1/2​)m(qEt/m​)2
​= (q2E2t2​)/2m