Question

Two charged particles are a distance of 1.82 m from each other. One of the particles has a charge of 7.52 nC, and the other has a charge of 4.46 nC. (a) What is the magnitude (in N) of the electric force that one particle exerts on the other?(b) Is the force attractive or repulsive?

Answer 1

Answer:

(a) $F=9.10*10^{-8}N$

(b) The force is repulsive

Explanation:

a) According to Coulomb's law, the magnitude of the electrice force that one particle exerts on the other is defined as:

$F=\frac{kq_1q_2}{d^2}$

Here k is the coulomb constant, $q_1$ and $q_2$ are the signed magnitudes of the charges and d is the distance between them.

$F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(7.52*10^{-9}C)(4.46*10^{-9}C)}{(1.82m)^2}\\F=9.10*10^{-8}N$

b) According to Coulomb's law, if the two charges have the same sign, the electrostatic force between them is repulsive.