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:

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 = \(\rm\frac{1}{4\pi\varepsilon_0\varepsilon_r}.\frac{q_1q_2}{d^2}=\rm\frac{1}{4\pi\varepsilon_0K}.\frac{q_1q_2}{d^2}\)

F = \(\rm\frac{1}{4\pi\varepsilon}.\frac{q_1q_2}{d^2}\)

In vector form, \(F=\frac{1}{4\pi ϵ_0} \frac{q_1q_2}{r^3} ⃗ r\),

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, \(⃗ r= (0.36\hat i+0.27\hat j)m\)

The magnitude of the position vector, \(|⃗ r| =\sqrt{(0.36)^2+(0.27)^2}= 0.45m \)

From Coulomb's law, \(F=\frac{1}{4\pi ϵ_0} \frac{q_1q_2}{r^3} ⃗ r\)

\(F=9\times 10^{9} \times \frac{3\times 10^{-6}\times 9\times 10^{-6}}{(0.45)^3} (0.36\hat i+0.27\hat j)\)

F = (0.96 N) i + (0.72 N) j