At the centre of a half ring of radius R=10 cm and linear charge density 4 nC m⁻¹, the potential is x π V. The value of x is _________.

Solution

Physics sir · 4 steps

Every bit of the half ring sits at the same distance R from the centre, so just add up kq/R.

1. Same distance. Potential is a scalar, so direction does not matter. Each small charge dq on the half ring is exactly R away from the centre, so each contributes (k dq)/R.
2. Total charge. The half ring has length π R, so its charge is Q = λ π R. Adding all the pieces gives V = kQ/R.

   V = (k λ π R)/R = kλπ

3. Put in numbers. The radius cancels. V = 9 × 10⁹ × 4 × 10⁻⁹ × π = 36π V.
4. So. Comparing with xπ V gives x = 36.

The common slip: 72. Taking the full circumference 2π R instead of π R doubles the charge and gives 72, not 36.

The idea to keep. For potential at the centre of any arc, the shape does not matter. Every charge is at distance R, so V = kQ/R with Q = λ × arc length.

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