Answer to Question #295949 in Electricity and Magnetism for Lhen

Question

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

a) work done on the partcle

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

c) magnitude of the electric field.

Answer 1

We know that

$V^2=U^2+2aS$

$25^2=0+2\times a\times0.12$

$$a=\frac{25^2}{2\times0.12}=\frac{625}{2\times0.12}=2604m/sec^2$$

$F=ma$

$$F=3.25\times10^{-3}\times2604=8.463N$$

Work

W=F.dr

$W=8.4263\times0.12=1.0155J$

$U=-\frac{kq_1q_2}{r}$

$$U=-\frac{9\times10^9\times2.5\times2.5\times10^{-18}}{0.12}=-4.68\times10^{-7}J$$

Electric field

$$E=\frac{kq}{r^2}=\frac{9\times10^9\times2.5\times10^{-9}}{0.12^2}=1562.5N/C$$