JEE Main 2026 Physics question
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. (ε₀=8.85×10⁻¹² C²/Nm² and π=3.14)
Answer: (A)
Solution
For a half ring, E=2kQ/(πR²)=Q/(2π²ε₀R²). So Q=2π²ε₀R²E=2(3.14)²(8.85×10⁻¹²)(0.35)²(100)≈2.14×10⁻⁹ C=2.14 nC.