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The gap between the plates of a parallel-plate capacitor is filled with glass of resistivity `rho = 100 G Omega m`, The capacitance of the capacitors equals `C = 4.0 nF`. Find the leakeage current of the capacitor when a voltage `V = 2.0 kV` is applied to it. |
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Answer» Let us mentally impart the charges `+q` and `-q` to the plates of the capacitor. Then capacitance of the network , `C = (q)/(varphi) = (epsilon epsilon_(0) int E_(n) dS)/(varphi)` ....(1) Now, electric current, `i = int vec(j) d vec(S) = int sigma E_(n) as vec(j) uarr uarr vec(E)`. ....(2) Hence, using (1) IN (2), we get, `i = (C varphi)/(epsilon epsilon_(0)) = 1.5 mu A` |
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