Two charges 5 c and 15 c are seperated by 10 cm. what is the electric force between them?

6.75*10^13N

Explanation:

The electric force between the charges can be determined using coulombs law 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 their distance between them.

Mathematically, F = kq1q2/r² where;

q1 and q2 are the charges

r is the distance between the charges

F is the force of attraction

k is the coulombs constant

Given q1 = 5C q2 = 15C r = 10cm = 0.1m k = 9*10^9Nm²/C²

Substituting the given values in the formula we have;

F = 9*10^9*5*15/0.1²

F = 6.75*10^11/0.01

F = 6.75*10^13N

Therefore the electric force between them is 6.75*10^13N