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JEE Main 2026 (6 Apr, Shift 1) · Electric Field & Field Lines

Electric Field & Field Lines question from JEE Main 2026 (6 Apr, Shift 1)

Q4JEE Main 2026 · 6 APR · SHIFT 1Medium

Electric Field & Field Lines

A thin half ring of radius 35 cm is uniformly charged with a total charge of QQ coulomb. If the magnitude of the electric field at centre of the half ring is 100 V/m100\text{ V/m}, then the value of QQ is ______ nC. (ϵ0=8.85×10−12 C2/N m2(\epsilon_0 = 8.85 \times 10^{-12}\text{ C}^{2}/\text{N m}^{2} and π=3.14)\pi = 3.14)

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Answer: (a)

  1. For a half ring: E=2kλRE = \dfrac{2k\lambda}{R} with λ=QπR\lambda = \dfrac{Q}{\pi R}, so E=Q2π2ε0R2E = \dfrac{Q}{2\pi^2\varepsilon_0 R^2}.
  2. Q=2π2ε0R2E=2(3.14)2(8.85×10−12)(0.35)2(100)Q = 2\pi^2\varepsilon_0 R^2 E = 2(3.14)^2(8.85\times10^{-12})(0.35)^2(100)
  3. Q=1971.9×8.85×10−12×0.1225≈2.14×10−9 CQ = 1971.9 \times 8.85\times10^{-12} \times 0.1225 \approx 2.14\times10^{-9}\text{ C}
  4. So the answer is (a) 2.14 nC.
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