JEE Mains · Physics · STD 12 - 1. Electric charges and fields
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 \(100\) V/m, then the value of \(Q\) is _______ nC. (\(\epsilon_o = 8.85 \times 10^{-12}\) C\(^2\)/Nm\(^2\) and \(\pi = 3.14\))
- A \(2.14\)
- B \(2.44\)
- C \(3.25\)
- D \(0.7\)
Answer & Solution
Correct Answer
(A) \(2.14\)
Step-by-step Solution
Detailed explanation
The electric field at the center of a uniformly charged half ring is given by: \(E = \dfrac{2k\lambda}{R}\) where \(\lambda = \dfrac{Q}{\pi R}\) is the linear charge density. Substituting \(\lambda\), we get:…
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