Two point charges 3 μC and 4 μC repel each other with a force of 10 N. If - 6 μC of charge is added to each of them, the new repulsive force is

The correct answer is option 1) i.e. 5 N less

CONCEPT:

• Coulomb's law: It states that the magnitude of the electrostatic force F between two point charges q1 and q2 is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance r between them. 

It is represented mathematically by the equation:

\(F = \frac{1}{4\pi ϵ_0}\frac{q_1 q_2}{r^2}\)

Where ϵ0 is the permittivity of free space (8.854 × 10-12 C2 N m-2).

CALCULATION:

Given that: q₁ = 3 μC, q₂ = 4 μC, and F = 10 N

Let the distance of separation be 'r'.

Using, \(F = \frac{1}{4\pi ϵ_0}\frac{q_1 q_2}{r^2}\)

\(⇒ F = \frac{1}{4\pi ϵ_0}\frac{(3μ)(4μ)}{r^2}\)      ----(1)

When -6 μC is added to each of them,

⇒ q₁' = 3 + (-6) = -3 μC

⇒ q₂' = 4 + (-6) = -2 μC

\(⇒ F' = \frac{1}{4\pi ϵ_0}\frac{(-3μ)(-2μ)}{r^2}\)      ----(2)

Dividing (2) by (1),

\(\frac{F'}{F} = \frac{(-3μ)(-2μ)}{(3μ)(4μ)} = \frac{6}{12} = \frac{1}{2}\)

\(⇒ F' = F \times \frac{1}{2} = 10 \times \frac{1}{2} = 5\: N\)

• Therefore, the new force of attraction F' is 5 N less than the initial force F.