Question

An object carries a +15.5 µC charge. It is 0.525 m from a -7.25 µC charge. What is the magnitude of the electric force on the object?

Answer 1

the force of attraction between the two charges is 3.55 N.

Step-by-step explanation:

Given;

first charge carried by the object, q₁ = 15.5 µC

second charge carried by the q₂ = -7.25 µC

distance between the two charges, r = 0.525 m

The force of attraction between the two charges is calculated as;

`F = (kq_1q_2)/(r^2) \\\\F = ((9* 10^9)(15* 10^(-6))(7.25* 10^(-6)))/((0.525)^2) \\\\F = 3.55 \ N`

Therefore, the force of attraction between the two charges is 3.55 N.