Question

A negative charge of 0.198 C exerts an attractive force of 83,506 N on a second charge that is 3,004 m away. What is the magnitude of the second charge? ( solve for one of the charges)

Answer 1

-606.73C

Step-by-step explanation:

According to coulombs law

F = kq1q2/r^2

Substitute the given value

83,506 = 9*10^9(-0.198)q2/(3004)^2

83,506 = -1.242*10^9q2/9,024,016

-1.242*10^9q2 = 7.54*10^11

q2 = -7.54*10^11/1.242*10^9

q2 = -606.73C

hence the magnitude of the other charge is -606.73C