Answer to Question #135629 in General Chemistry for Yaaseen Andor

Question

A point charge q1 = + 2.4 µC is held stationary at the origin. A second point
charge q2 = - 4.3 µC 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 q2?
Note: Show your free body diagram (FBD) and solutions accurately,
systematically and orderly with the standard unit, and express it into
prefix word if possible.

Answer 1

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