Air isn\'t a perfect electric insulator, but it has a very high resistivity. Dry

Question

Air isn't a perfect electric insulator, but it has a very high resistivity. Dry air has a resistivity of approximately 3× 1013 m. A capacitor has square plates 12 cm on a side separated by 1.4 mm of dry air. Assuming that the air between the capacitor's plates can be treated like any other resistor, find its resistance value. If the capacitor is charged to 290 V , what current will flow between the plates of the capacitor? Make the approximation that the potential difference doesn't change as the charge flows and that air obeys Ohm's law. What fraction of its initial charge will the capacitor lose after one minute?

Explanation / Answer

First of all, determine the resistance between plates.

This is, R = L/A

=> R = (3x10^13)(1.4 x 10^-3m) / (12^2x10^-4m²)

=> R = 2.92 x 10^12

So, current between the plates, I = potential difference / R = 290 / (2.92 x 10^12) = 9.93 x 10^-11 A

So, charge lost = i(C/s) x t(s) = (9.93 x 10^-11)(60s)

=> q = 5.96 x 10^-9 C

Again, original charge Q -
Cap C = A/L = Q/V
=> Q = AV / L = (8.85x10^-12)(0.0144)(290) / (1.40 x 10^-3)

=> Q = 2.64 x 10^-8 C

Therefore, q / Q = 5.96x10^-9 C / 2.64x10^-8 C = 0.226

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