Answer:
the force of attraction between the two charges is 3.55 N.
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;
[tex]F = \frac{kq_1q_2}{r^2} \\\\F = \frac{(9\times 10^9)(15\times 10^{-6})(7.25\times 10^{-6})}{(0.525)^2} \\\\F = 3.55 \ N[/tex]
Therefore, the force of attraction between the two charges is 3.55 N.
