A second point charge q2 = 20 μC is moved from the point (x, y) = (3 m, 4 m) to the point (x, y) = (1 m, 0 m) where it is held stationary. How much work was done by the electric force on q2?

Work done by the electric force on q2 = -216 X 10³ X q1 J.

Explanation:

Charge on q2 = 20μC

x1 = 3m , y1 = 4m

x2 = 1m , y2 = 0m

Work done on q2 = ?

We know,

Work done = -(Ub - Ua)

Ua = kq1 X q2 / √x² + y²

Ua = k q1 X 20 μC / √3² + 4²

Ua = k q1 X 20 μC / 5

Ub = k q1 X q2 / 1

Ub = k q1 X 20 μC

work done = - k . q1 ( 4 μC + 20 μC)

                  = - k . q1 X 24 μC

Value of k = 9 X 10⁹

So, W = - 9 X 10⁹ X q1 X 24 X 10⁻⁶ J

W = -216 X 10³ X q1 J

Therefore, work done by the electric force on q2 = -216 X 10³ X q1 J.

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.