Q 12-01-024JEE MainJEE Main 2026 (6 Apr, Shift 1)Medium
A thin half ring of radius $35$ cm is uniformly charged with a total charge of $Q$ coulomb. If the magnitude of the electric field at centre of the half ring is $100\ \text{V/m}$, then the value of $Q$ is ______ nC. ($\epsilon_0 = 8.85 \times 10^{-12}\ \text{C}^2/\text{Nm}^2$ and $\pi = 3.14$)
Answer: (A) $2.14$
For a half ring with linear charge density $\lambda = \dfrac{Q}{\pi R}$, the field at the centre is
$$E = \frac{2k\lambda}{R} = \frac{\lambda}{2\pi\epsilon_0R} = \frac{Q}{2\pi^2\epsilon_0R^2}$$
$$Q = 2\pi^2\epsilon_0R^2E = 2 \times 9.86 \times 8.85 \times 10^{-12} \times 0.1225 \times 100 \approx 2.14 \times 10^{-9}\ \text{C}$$
Solution by Sreeraj P, M.Sc Physics