Question

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

Answer 1

To be given in question

Chage $(q)= 2.5\times 10^{-9} =2.5nc$

mass$(m)=3.25\times10^{-3} kg$

distance$(d)=12cm =12\times10^{-2}meter$

Speed (v)=25 $meter/sec$

To be asked in question

Work done (w)=?

Potential ∆U=?

Electric field (E)=?

We know that

(a)

Work energy relation r

$∆W=\frac{1}{2}mv^2$

Put value

$∆W=\frac{1}{2}\times{3.25}\times10^{-3}\times{25}^2$

$∆W=1.016 jule$

(b)change in electric potential of the particle is work done on the particle to

$∆U=1.016 jule$

(C)

Magnitude of electric field

$F=qE$

$F.d=1.016$

$F=\frac{1.016}{12\times 10^{-2}} =qE$

$E=\frac{1.016}{12\times 10^{-2}\times2.5\times10^{-19}}$

$E=3.39\times 10^ {19} N/C$