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Question

A capacitor made of two parallel circular plates of area A holds a charge Q0 initially.  Suppose that it discharges as Q(t)=Q0eλt. During the discharge, the ratio ϵBϵE of the magnetic field energy ϵB to the electric field energy ϵE between the two plates is given by
  1. 116ϵ0μ0π(Aλ2)
  2. 18ϵ0μ0π(Aλ2)
  3. 1161ϵ0μ0(Aλ2)
  4. 181ϵ0μ0(Aλ2)


Solution

The correct option is B 18ϵ0μ0π(Aλ2)
B.ds=μ0ϵ0dϕdtB2πr=μ0ϵ0(Eπr2)B2πr=μ0ϵ0πr2dEdtB=μ0ϵ02ddt[QAϵ0]B=μ0ϵ0r21Aϵ0dQdtB=μ0r2AQ0(λ)eλtB=λμ0Q02AeλtrUB=R0B22μ02πrdrl=πlμ0R0λ2μ20Q204A2e2λtr2rdr=πlμ0λ2μ20Q204A2e2λtR44=πtλ2μ20Q2016A2R4e2λt=lλ2μ20Q2016πe2λtUE=Q22C=Q20e2λt2ϵ0πR2l=Q2e2λt2ϵ0πR2e2λtUBUE=lλ2Q20μ016πe2λtQ20l2ϵ0πR2e2λt=λ2μ0ϵ08R2

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