A leaky parallel plate capacitor is filled completely with a material having dielectric constant `K = 5` and electrical conductivity `sigma= 7.4 xx 10^-12 Omega^-1 m^-1`. If the charge on the capacitor at the instant `t = 0` is `q_0 = 8.55 muC`, then calculate the leakage current at the instant `t = 12 s`.

Answer

0.198

Explanation

The problem is basically os discharging of CR circuit,because the plates of the capacitor,there is a capacitor as well as resistance
`R=d/(sigmaA) (R=1/(sigmaA)) and C=(K epsilon_0A)/d`
Time constant,`tau_c=CR=(Kepsilon_0)/(sigma)`
`tau_c=(5xx8.86xx10^(-12))/(7.4xx10^(-12))=5.98s`
Charge at any time decreass exponentially as
`q=q_0e^(-t//tau_e),` Here ,`q_0=8.85xx10^(-6)C` (Charge at time t=0)
Therefore,discharging (leoakage) current at time t will be given by
`i=(-(dq)/(dt))=(q_0)/(tau_c)e^(-t//tau_e)`
Current t=12 s is `i=((8.85xx10^(-6))/(5.98))e^(-12//5.98)=0.198xx10^(-6)A=0.198 mu A ,i=0.198 mu A`