Apoint charge q1 = +2.90 µc is held stationary at the origin. a second point charge q2 = -5.30 µc moves from the point x = 0.110 m, y = 0 to the point x = 0.280 m, y = 0.280 m. how much work is done by the electric force on q2?

A point charge with a charge q1 = + 2.90 uC is held stationary at the origin hence coordinates are (0, 0).

A second point charge with a charge q2= - 5.40 uC moves from the point x = 0.110 m, y = 0 to the point x = 0.280 m, y = 0.280 m  [(0.110, 0) to (0.28, 0.28)]

The initial distance between the two charges is:

Initial distance = r1 = 0.110 m

The final distance between the two charges is calculated using hypotenuse formula:

Final distance = r2 = sqrt [0.28^2 + 0.28^2]

r2 = sqrt [2 * 0.0784]

r2 = sqrt [0.1568]

r2 = 0.396 m

Δr = r1 - r2 = 0.110 - 0.396 = - 0.286 m

r1r2 = 0.04356 m^2

Work is done by the electric force on q2= W. The formula to use with change in location is:

W =k q1 q2 [1/r2 - 1/r1]

W = 9*10^9 * 2.9*10^-6 * -5.3*10^-6 [(1/0.396) – (1/0.110)]

W = 0.908 J = 9.08*10^-1 J            (ANSWER)

answer: option c.

explanation: