*Charges Q1 and Q2 exert repulsive forces of 17N on eachother. What is the Repulsive force in N when separation is decreased so that final separation is 72% of their initial separation?

Question

Please explain in steps

Answer 1

Coulomb's law states that the Force exerted by the electric field on each of two charged particles is:

F = (k)(q1)(q2) / r^2, where q1 and q2 are the charges of the particles, and k is Coulomb's constant.

We are given

17 N = (k)(q1)(q2) / r^2

If we decrease r to .72 r and hold the rest constant, then how does that affect the rest? Well, we see that we have an inverse relation to the square of distance (inverse square law), so by moving them closer together, we are going to have a (1/(0.72))^2 stronger interaction.

Final Force = (17 N) (1/(0.72))^2

Final Force = 32.8 N

Let me know if that makes sense, and feel free to send me a DM if you have any other questions or need tutoring!

-Ethan