1.

Find the capacitance of a capacitor whose area of cross section of the plates is 4m^2 and distance of separation between the plates is 2m. The capacitor is placed in vacuum.(a) 1.77*10^-11F(b) 1.34*10^-11F(c) 2.33*10^-11F(d) 5.65*10^-11FThis question was posed to me in my homework.My question is based upon Relative Permittivity in chapter Capacitance and Capacitors of Basic Electrical Engineering

Answer»

The correct answer is (a) 1.77*10^-11F

To ELABORATE: The expression for finding the VALUE of capacitance is:

C=epsilon*A/d.

The MEDIUM is free SPACE hence, epsilon = 8.85*10^-12.

Therefore, C=8.85*10^-12*4/2 = 1.77*10^-12F = 1.77*10^-11F.



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