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InterviewSolution
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The space between the plates orf a parallel plate capacitor is completely filled with a meterail of resistaivity `2xx10^(11) Omegam` and dielectric constant 6. Capacity of the capacitor with the given dielectric medium between the paltes is 20uF. Find the leakage current if a potential difference 2500 V is applied across the capacitor. |
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Answer» Charge on the capacitor, `Q = CV = 20xx10^(-19) xx2500 = 5xx10^(-5)C` Surface charge density is `sigma_s = (A)/(Q)` [A is the plate area.] Electric field strength between the plates is `E = (sigma_(s))/(K epsilon_0) = (Q)/(Kaepsilon_0)` Using `J = simgaE = (E )/(rho)` [Here `simga` is the conducitivity and `rho` is the resistivity.] Current density is `J = Q//KA epsion_0rho` and current is `JA = (Q)/(Kepsilonrho) =(5xx10^(-5))/(6xx(8.85xx10^(-22))xx(2xx10^(11))` `=4.7xx10^(-6) A = 4.7muA` |
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