Question

A +2.00 nC point charge is at the origin, and a second -5.00 nC point charge is on the xaxis at x = 0.800 m.

a) Find the electric field (magnitude and direction) at each of the following points on the xaxis: i) x = 0.200 m; ii) x = 1.20 m; iii) x = -0.200 m.

b) Find the net electric force that the two charges would exert on the electron placed at

each point in part (a).

Answer 1

(a)

i.

$$E=E_1+E_2=k\frac{|q_1|}{r_1^2}+k\frac{|q_2|}{r_2^2}$$

$$E_x=9\times 10^9\left(\frac{2\times 10^{-9}}{0.200^2}+\frac{5\times 10^{-9}}{0.600^2}\right)=575\:\rm N/C$$

ii.

$$E=E_1-E_2=k\frac{|q_1|}{r_1^2}-k\frac{|q_2|}{r_2^2}$$

$$E_x=9\times 10^9\left(\frac{2\times 10^{-9}}{1.200^2}-\frac{5\times 10^{-9}}{0.400^2}\right)=-269\:\rm N/C$$

iii.

$$E=-E_1+E_2=-k\frac{|q_1|}{r_1^2}+k\frac{|q_2|}{r_2^2}$$

$$E_x=9\times 10^9\left(-\frac{2\times 10^{-9}}{0.200^2}+\frac{5\times 10^{-9}}{1.00^2}\right)=-405\:\rm N/C$$

(b)

$$F_x=qE_x\\ =-1.6\times 10^{-19}\times575=-9.2\times 10^{-17}\:\rm N$$