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) $2.14$
Concept. For a uniformly charged arc of angle $2\alpha$ at its centre: $E=\dfrac{2k\lambda\sin\alpha}{R}$. A half ring has $\alpha=90^\circ$, so $E=\dfrac{2k\lambda}{R}$ with $\lambda=\dfrac{Q}{\pi R}$.
Step 1. Field. $E=\dfrac{2kQ}{\pi R^2}=\dfrac{Q}{2\pi^2\varepsilon_0R^2}$.
Step 2. Solve. $Q=2\pi^2\varepsilon_0R^2E=2\times9.86\times8.85\times10^{-12}\times0.1225\times100=2.14\times10^{-9}$ C.
Answer. $2.14$ nC (Option A).
Solution by Sreeraj P, M.Sc Physics
