Answer to Question #299755 in Physics for kelly

Question

A point charge of 3.0 nC with a mass of 4.0 g is moved from x=1.0m to x= 1.5m in an electric field of 5,0 N/C with the same direction as the motion of the charge. (a) How much work is done on the charge by the electric force? (b) What is the change in the potential energy of the charge? (c) Assuming that the charge started from rest, what is its speed at x=1.5 m?

 

Answer 1

Given:

$q=3.0\:{\rm nC},\quad m=4.0\:\rm g$

$x_1=1.0\:{\rm m},\quad x_2=1.5\:{\rm m}$

$E=5.0\:\rm N/C$

(a)

$$W=qEd\\ =3.0*10^{-9}*5.0*(1.5-1.0)=7.5*10^{-9}\:\rm J$$

(b)

$$V=\frac{W}{q}=\frac{7.5*10^{-9}}{3.0*10^{-9}}=2.5\:\rm V$$

(c)

$$v=\sqrt{2W/m}\\ =\sqrt{2*7.5*10^{-9}/4.0*10^{-3}}=1.9*10^{-3}\:\rm m/s$$