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Question -

A parallel plate capacitor with air betweenthe plates has a capacitance of 8 pF (1pF = 10−12 F). What willbe the capacitance if the distance between the plates is reduced by half, andthe space between them is filled with a substance of dielectric constant 6?



Answer -

Capacitance between the parallel plates of the capacitor,C = 8 pF

Initially, distance between the parallelplates was and it was filled with air. Dielectric constantof air, k = 1

Capacitance, C, is given by theformula,

Where,

A =Area of each plate

= Permittivity of free space

If distance between the plates is reduced to half,then new distance, d = 
Dielectric constant of thesubstance filled in between the plates,  = 6
Hence, capacitance of the capacitor becomes

Taking ratios of equations (i) and (ii), we obtain

Therefore, the capacitance between theplates is 96 pF.

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