A charge Q₁ of 3.0 μC is placed at point (36 cm, 0) and another charge Q₂ of 9.0 μC is placed at a point (0, -27 cm).
Let i and j be the unit vectors in x- and y- directions, respectively. The force exerted by Q₂ on Q₁ is:

Question

Question

This question was previously asked in

AAI ATC Junior Executive 27 July 2022 Shift 1 Official Paper

Answer 1

Concept:

Coulomb's Law:

According to Coulomb’s law, the force of attraction or repulsion between two charged bodies is directly proportional to the product of their charges
And inversely proportional to the square of the distance between them.
It acts along the line joining the two charges considered to be point charges.

Coulomb's Formula

F =

In vector form, ,

where ϵ₀ = permittivity in the air,

r = magnitude of the distance between two charge

Explanation:

Given,

The charge at point (36cm, 0), q₁ = 3.0 μC

The charge at point (0,-27 cm), q₂ = 9.0 μC

The position vector,

The magnitude of the position vector,

From Coulomb's law,

F = (0.96 N) i + (0.72 N) j