At the centre of a half ring of radius R = 10 cm and linear charge density 4 nC m⁻¹, the potential is xπ V. The value of x is _________. Numerical answer type. Work it out, then open the answer.

Answer: 36. Half ring: V = (k(λπ R))/(R) = kλπ V = 9×10⁹×4×10⁻⁹×π = 36π V So x = 36.

At the centre of a half ring of radius R = 10 cm and linear charge… | JEE Main 2024 (9 Apr, Shift 1)

Q52JEE Main 2024 · 9 APR · SHIFT 1Medium

Electric Potential & Equipotentials

At the centre of a half ring of radius $R = 10\text{ cm}$ and linear charge density $4\text{ nC m}^{-1}$ , the potential is $x\pi\text{ V}$ . The value of $x$ is _________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 36

Half ring: $V = \dfrac{k(\lambda\pi R)}{R} = k\lambda\pi$
$V = 9\times10^9\times4\times10^{-9}\times\pi = 36\pi$ V
So $x = 36$ .

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