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

`1\cdot 10^(-8) N` to the left

Step-by-step explanation:

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

`F=k(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=((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).