Asked by gopi
23.1 .. A point charge q 1 = +2.40 mC is held stationary at the origin. A second point charge q 2 = -4.30 mC moves from the point x = 0.150 m, y = 0 to the point x = 0.250 m, y = 0.250 m. How much work is done by the electric force on q 2 ?
Answers
Answered by
bobpursley
Find the electric potential at both points.
v1=kq1/x=k(2.4e-3)/.150
v2=kq1/(sqrt(.250^2+.250^2)
now the work is (v2-V1)(-4.3e-3), where q1 is 2.40e-3 C, and k is 9.0 x 109
v1=kq1/x=k(2.4e-3)/.150
v2=kq1/(sqrt(.250^2+.250^2)
now the work is (v2-V1)(-4.3e-3), where q1 is 2.40e-3 C, and k is 9.0 x 109
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.