Question

Two particles with charges of 2nC and 5nC are separated by a distance of 3m. The charge 2nC is placed on the left. Find the force on charge 2nC

Answer 1

Answer:

$1\cdot 10^{-8} N$ to the left

Explanation:

The magnitude of the electrostatic force between two charges is given by the following equation:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=9\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the magnitude of the two charges

r is the distance between the two charges

Moreover, the force is:

- Attractive if the charges have opposite sign

- Repulsive if the charges have same sign

In this problem, we have:

$q_1=2nC=2\cdot 10^{-9}C$ is the magnitude of charge 1

$q_2=5nC =5\cdot 10^{-9}C$ is the magnitude of charge 2

r = 3 m is the distance between the two charges

Substituting, we find the force on both charges:

$F=\frac{(9\cdot 10^9)(5\cdot 10^{-9})(2\cdot 10^{-9}}{3^2}=1\cdot 10^{-8} N$

Here, the two charges are both positive, so the force is repulsive; since the 2 nC charge is on the left, this means that the force on this charge is to the left (away from the 5 nC charge).