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.

Explanation

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`