Question

suppose that two point charges each with a charge of +1.00 coulomb are separated by a distance of 1.00 metre. determine the magnitude of the electric force of repulsion between them

Answer 1

Apply Coulomb's law,

F = ( k q₁ q₂ ) ÷ r²

F         = electric force

k         = Coulomb constant

q₁, q₂ = charges

r         = distance of separation

Answer 2

Given: Two point charges of 1 coulomb is separated by 1 metre.

To find: Electric force of repulsion between them

Explanation: The two point charge are taken to be positive or negative in nature. There is a force of repulsion between two point charges because in electrostatics, same charges repel each other.

Here, q1=q2 = 1 coulomb, r= 1 m

The formula for calculating electric force is:

F= kq1q2/ r^2 where k is a constant whose value is 9*10^9

F= (9*10^9 * 1 *1) / (1^2)

= 9* 10^9 N

Therefore, the force of electronic repulsion between two point charges is 9*10^9 N.