Find the electric force if charge a is 25 c and charge b is 40 c and they are placed .005m apart. is this force attracting or repelling?

question 9 options:

3.6 x 10 17 n; repelling or repulsion

3.6 x 10 17 n; attracting or attraction

5.2 x 10 17 n; attracting or attraction

5.2 x 10 17 n; repelling or repulsion

3.6 X 10^17 N; repelling or repulsion

Explanation:

According to coulombs law of electrostatic attraction which states that the force of attraction between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Mathematically,

F = kqAqB/d² where;

qA and qB are the charges

d is the distance between the charges.

k is the coulombs constant = 9×10^9Nm²/C²

Given qA = 25C, qB = 40C, d = 0.005m

Substituting the values in the formula we will have;

F = 9×10^9×25×40/0.005²

F = 9×10^12/0.005²

F = 3.6×10^17N

Since the charges are both positive (like charges), they will repel.

According to the law of electrostatic attraction, like charges repel, unlike charges attracts.

the answer is c.)

explanation: