A point charge q=1μC is located at a distance 2 cm from one end of a thin insulating wire of length 10 cm having a charge Q=24μC, distributed uniformly along its length, as shown in figure. Force between q and wire is ____ N.
(Use : 1/4πε₀=9× 10⁹ N· m²/C² )

Solution

Physics sir · 3 steps

The wire is not a point charge, so cut it into tiny bits and add the Coulomb forces from end to end.

1. Slice the wire. The charge per unit length is λ = Q/L. A small piece dx at distance x from q carries λ dx and pushes on q along the wire's axis. All the pieces push in the same direction, so their forces add with no cancellation.

   F = ∫ₐ^(a+L) kqλ/x² dx = kqλ(1/a-1/(a+L)) = kqQ/a(a+L)

2. Put in numbers. Here a = 0.02 m and a + L = 0.12 m, so a(a+L) = 2.4 × 10⁻³ m². The numerator is kqQ = 9 × 10⁹ × 10⁻⁶ × 24 × 10⁻⁶ = 0.216 N m².
3. So. F = 0.216/(2.4 × 10⁻³) = 90 N. The answer is 90.

The common slip: 44. Treating the whole charge Q as if it sat at the centre of the wire, 7 cm from q, gives about 44 N, not 90.

The idea to keep. For a point charge on the axis of a uniform rod, F = kqQ/a(a+L), where a and a+L are the distances to the near and far ends. Remember it as the product of the two end distances.

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