Find the electric field at point P (as shown in figure) on the perpendicular bisector of a uniformly charged thin wire of length L carrying a charge Q. The distance of the point P from the centre of the rod is a = √3/2L.

Solution

Physics sir · 4 steps

Each half of the rod makes 30° at P, so the finite-wire formula finishes it in one line.

1. Use symmetry. Charge elements at equal distances above and below O give fields whose components along the wire cancel. Only the components along OP add up. Adding these gives the standard result for a point on the perpendicular bisector. Here λ = Q/L and θ is the angle each half of the rod makes at P.

   E = (2λ sin θ)/(4πε₀ a), tan θ = (L/2)/a

2. Find the angle. tan θ = (L/2)/(√3L/2) = 1/√3, so θ = 30° and sin θ = 1/2. Each end of the rod is at distance √(a² + L²/4) = √(3L²/4 + L²/4) = L from P.
3. Put in numbers. E = (2 × Q/L × 1/2)/(4πε₀ × √3/2L) = Q/(4πε₀ L) × 2/√3L = Q/(2√3πε₀ L²).
4. So. The field at P points away from the rod along OP and has magnitude Q/(2√3πε₀ L²). That is option (c).

The idea to keep. For a finite charged rod, on the bisector E = (2kλ sin θ)/a. Always find the half-angle θ first. JEE picks a so that θ is 30°, 45° or 60°.

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