Answer to Question #313040 in Physics for Archi

Question #313040

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).



1
Expert's answer
2022-03-18T09:16:11-0400

(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\\\\\n=-1.6\\times 10^{-19}\\times575=-9.2\\times 10^{-17}\\:\\rm N"


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