Question #294870

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

a)work done on the particle

b) change in the electric potential energy of the particle and

c) magnitude of the electric field.







1
Expert's answer
2022-02-08T08:16:21-0500

a)

W=mv22=1.02 J,W=\frac{mv^2}2=1.02~J,

b)

U=Wq=4.06 MV,U=\frac Wq=4.06~MV,

c)

E=Ud=33.85 MNC.E=\frac Ud=33.85~\frac{MN}C.


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