Question

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8. A capacitor is made from a 80 μm thick teflon filminserted between two parallel copper plates. a) what is the effective area of the plates of this capacitor if it has a capacitance of 2.5 μF? b) What is the maximum working voltage of the capacitor, that is, howmuch voltage can it stand before breakdown?

Answer 1

Q8.

Thickness of the teflon film, d = 80 X10^-6 m

dielectric strength of teflon, E_max = 600 kV/cm = 60000 KV/m

dielectric constant of teflon, k = 2.1.

capacitance of the parallel plate combination, C = 2.5 X10^-6 F

For parallel plate capacitor, C = (k* epsilon_0*A)/ d

effective area of the plates, A = (C * d) / (k * epsilon_0) = ( 2.5 X10^-6 * 80 X10^-6 ) / (2.1 * 8.854X10^-12) = 10.76 m² (Ans (a))

The maximum working voltage of the capacitor should be such that the dielectric inside the capacitor does not breakdown.

E_max = (V_max / d )  

V_max = (E_max * d ) = (60000 * 80 X10^-6 ) = 4.8 V (Ans (b))