Two point charges 40 nL and 100 nL are in vacuum at a distance of r = 2 cm

Two point charges 40 nL and 100 nL are in vacuum at a distance of r = 2 cm from each other. What is the force F of the interaction between charges?

Data: vacuum; qz1 – the first charge (qz1 = 40 nC = 4 * 10 ^ -8 C); qz2 – the second charge (qz2 = 100 nC = 10 ^ -7 C); r is the distance between charges (r = 2 cm = 2 * 10 ^ -2 m).

Const: k – coeff. proportionality (k = 9 * 10 ^ 9 N * m2 / Cl2).

To find out the strength of the interaction between the presented point charges, consider the formula: Fx = k * qz1 * qz2 / r ^ 2.

Let’s perform the calculation: Fx = 9 * 10 ^ 9 * 4 * 10 ^ -8 * 10 ^ -7 / (2 * 10 ^ -2) ^ 2 = 0.18 N.

Answer: The force of interaction between the presented point charges, according to the calculation, is 0.18 N.