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an air-filled capacitor is made from two flat parallel plates 1.0 mm apart. the inside area of each plate is 8.0 cm2. (a) what is the capacitance of this set of plates? (b) if the region between the plates is filled with a material whose dielectric constant is 6.0, what is the new capacitance?



Answer :

bharathparasad577

The new capacitance is C' = 4.25 * 10-11 F.

The ratio of a system's change in electric charge to the equivalent change in its electric potential is known as capacitance.

Depending on how it is used, any capacitor can have either a fixed or variable capacitance. It may appear from the equation that "C" relies on charge and voltage. Actually, it depends on the size, shape, and insulator employed between the conducting plates of the capacitor.

Capacitance (C) = EKA/ d

where, k = air-filled

A = Area of plates

d = separation distance

Given,

A = 8.0 * 10-4 m2

d = 1 * 10-3m

a) C = (8.85 * 10-12 c2/Nm2) (1) (8.0* 10-4 m2) / 1 * 10-3 m

C = 7.08 * 10-12 F

b) Now dielectric (k = 6) inserted, new capacitance

C' = kC

   = 6 (7.08 *10-12 F)

C' = 4.25 * 10-11 F

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