Question

Two one-coulomb point charges are at a distance of 1.25 m from each other. What is the size of the electric force between the two charges?

Answer 1

Answer:

5.76N

Explanation:

We are given that

$q_1=1C$

$q_2=1 C$

Distance between two charges ,$r=1.25m$

Electrostatic force=$F=k\frac{q_1q_2}{r^2}$

Where k=$9\times 10^9Nm^2/C^2$

Substitute the values then we get

$F=\frac{9\times 10^9\times 1\times 1}{(1.25)^2}$

$F=5.76N$

Hence, the electric force between two charges=5.76N