In fig., two equal positive point charges q₁ = q₂ = 2.0 µC interact with a third point charge Q = 4.0 µC. The magnitude, as well as direction, of the net force on Q is

Concept :

The electrostatic force between two point charges is given by Coulomb's law:

F = k × |q₁ × q₂| / r²

Where,

k is Coulomb's constant (8.99 × 10⁹ N·m²/C²)

r is the distance between the charges.

Calculation:

Let the distance between each q₁ (or q₂) and Q be d. Assume Q is at the origin (0,0) and q₁ and q₂ are symmetrically placed along the x-axis at positions (-d,0) and (d,0) respectively.

Force due to q₁ on Q, F₁:

⇒ F₁ = k × |q₁ × Q| / d²

Force due to q₂ on Q, F₂:

⇒ F₂ = k × |q₂ × Q| / d²

Since q₁ and q₂ are equal and symmetrically placed, the net force on Q will be the sum of the forces due to q₁ and q₂, which will add up in the x-direction:

Net force, F_net = F₁ + F₂

⇒ F_net = 2 × k × |q × Q| / d²

Substitute the values:

⇒ F_net = 2 × 8.99 × 10⁹ N·m²/C² × |2.0 × 10^-6 C × 4.0 × 10^-6 C| / d²

⇒ F_net = 2 × 8.99 × 10⁹ × 8.0 × 10^-12 / d²

⇒ F_net = 2 × 71.92 × 10^-3 / d²

⇒ F_net = 0.14384 / d²

Given that the distance d is such that the force is 0.46 N:

Hence, the magnitude of the net force on Q is 0.46 N in the + x-direction.

∴ The correct answer is option 2.