Question

A point particle of charge 2.5 nC and mass 3.25 x10-^3 is in uniform electric field directed to the right .It is released from rest and moves to the right .After it is travel ed 12.0 cm,it's speed us 25m/s.Find the

a) qork done on the particle

b)change un the electric potential energy of the particle

c) magnitude of the electric field.

Answer 1

Solution

a) The work done on the particle can be obtained 

Work done =change in kinetic energy

$W=0.5mv^2\\=0.5*3.25*10^{-3}*625\\=1.02J$

b) change in potential energy

U=-K=-1.02J

c) electric field is

$E=\frac{dU}{q}\\=\frac{0.12*1.05}{2.5*10^{-9}}\\=0.05*10^{-9}V/m$