Question

Calculate the electric force between two point charges of -54.0 μC and +36.0 μC when they are 2.00 cm apart.

Answer 1

Answer:

43740 N and attractive force

Explanation:

Given that,

Charge 1, q₁ = -54.0 μC

Charge 2, q₂ = +36.0 μC

The distance between charges, r = 2 cm = 0.02 m

We need to find the force between charges. The formula for the force between charges is given by :

$F=k\dfrac{q_1q_2}{r^2}$

Put all the values,

$F=9\times 10^9\times \dfrac{54\times 10^{-6}\times 36\times 10^{-6}}{(0.02)^2}\\\\F=43740\ N$

As the charges are opposite, the force between them is attractive. Hence, the required force is 43740 N.