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A thin half ring of radius 35 cm is uniformly charged with a total charge of \(Q\) coulomb. If the magnitude of the elec…

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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 \(100V/m\), then the value of \(Q\) is ______ nC.

\(\left({ϵ}_{0}=8.85\times {10}^{-12}{C}^{2}/N{m}^{2}\text{ and }\pi =3.14\right)\)

[JEE Main 2026, 6 Apr (Shift 1)]

a

2.14

b

2.44

c

3.25

d

0.7

✓ Correct answer: a)

2.14

Explanation

(a) At the center of charged semicircular ring,

\(E=\frac{2k\lambda }{R}\text{ Where }:k=\frac{1}{4\pi {ϵ}_{0}}\\\)
\(\lambda =\text{ linear charge density }=\frac{Q}{\pi R}\)

\(\text{ Substituting }\lambda \text{ into the field formula: }\\\)

\(E=\frac{2kQ}{\pi {R}^{2}}=\frac{2Q}{4{\pi }^{2}{\in }_{0}{R}^{2}}=\frac{Q}{2{\pi }^{2}{\in }_{0}{R}^{2}}\\\)
\(\Rightarrow Q=E\times 2{\pi }^{2}{\in }_{0}{R}^{2}\\\)
\(\Rightarrow Q=100\times 2\times (3.14{)}^{2}\times \left(8.85\times {10}^{-12}\right)\times (0.35{)}^{2}\\\)
\(Q\approx 2.1378\times {10}^{-9}C=2.14nC\)

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