Two point charges Q each are placed at a distance d apart. A third point charge q is placed at a distance x from mid-point on the perpendicular bisector. The value of x at which charge q will experience the maximum Coulomb's force is :

Solution

Physics sir · 4 steps

Only the force along the bisector survives, so write it as a function of x and find its peak.

1. Net force. Each charge is at distance √(x² + d²/4) from q. The components parallel to the line joining the charges cancel. The components along the bisector add, each with factor cos θ = x/√(x² + d²/4).

   F(x) = (2kQq x)/(x² + d²/4)^(3/2)

2. Set dF/dx to zero. Write a = d/2. Differentiating gives dF/dx ∝ (a² - 2x²)/(x² + a²)^(5/2). This is zero when x² = a²/2, so x = a/√2.
3. Check it is maximum. F = 0 at x = 0, because the two pushes are opposite. F also falls to zero as x → ∞. So the single turning point in between must be the maximum.
4. So. With a = d/2, the force is greatest at x = d/2√2, which is option (d).

The idea to keep. For two equal charges a distance 2a apart, the field on the perpendicular bisector peaks at x = a/√2. Always check whether a is the half-separation or the full separation.

Answer from the NTA final answer key. Solution by PrepWiser.