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At t = 0, a capacitor carrying its full initial charge q0 is joined to an inductor of self-inductance L. The instant when the energy is shared equally by the electric field and the magnetic field is

1. (π/4)√(LC)
2. 2π√(LC)
3. √(LC)
4. π√(LC)

Correct answer: (π/4)√(LC)

Solution

Charge q = q0*cos(wt) with w = 1/sqrt(LC); capacitor (electric) energy is proportional to q^2. Equal sharing needs q^2 = q0^2/2, i.e. cos^2(wt) = 1/2, so wt = pi/4 and t = (pi/4)*sqrt(LC).

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