Question

Three charges +Q, q, +Q are placed respectively at distance 0, d/2 and d from the origin on the X-axis

If the net force experienced by +Q placed at x = 0 is zero, then value of q is [Main 2019, 9 Jan Shift I]

Answer 1

q = –Q/4

Answer 2

Let the charge at x = 0 be +Q. The forces on this charge are due to:

1. The charge q at x = d/2, and
2. The charge +Q at x = d.

Since q must exert an attractive force (to the right) to cancel the repulsive force (to the left) from the other +Q, q must be negative.

• Force due to q:

  Magnitude, $F₁ = k \cdot Q \cdot |q|/(d/2)² = k \cdot Q \cdot |q|/(d²/4) = (4k \cdot Q \cdot |q|)/d²$ directed to the right.

• Force due to +Q at x = d:

  Magnitude, $F₂ = k \cdot Q²/d²$ directed to the left (repulsive).

Setting F₁ = F₂ for equilibrium: $(4k \cdot Q \cdot |q|)/d² = k \cdot Q²/d²$ $⇒ 4|q| = Q$ $⇒ |q| = Q/4$

Since q is negative, $q = –Q/4$