Question

-6 (Leader_Phase - 2, 2A & 2B] - 2020
The capacity of a parallel plate condenser is Colf a
dielectric of relative permitivity &, and thickness equal
to one fourth the plate separation is placed between the
plates, then its capacity becomes C. The value of C.
will be:-​

Answer 1

Explanation:

★ CAPACITANCE

★ GIVEN ;

» Thickness is 1 / 4 th of the Plate seperation

» t = d / 4

★ New Capacitance = ???

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We know that ;

★ C = Ë × A / [ (d - t ) + t / K ]

★ C = Ë × A / [ d - d / 4 ] + d / 4K ]

★ C = Ë × A / [ 3d / 4 ] + d / 4K ]

★ C = Ë × A / [ 3d / 4 + d / 4 ( 1 / K ) ]

★ C = Ë × A [ d × 1 / K ]

★ C = Ë × A × K / d ★