Question
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
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