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Standard VII

Geography

Ocean Currents

A parallel pl...

Question

A parallel plate capacitor with air between the plates has a capacitance of $8 p E (1 p E = 10^{- 12} F)$ . What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant $6$ ?

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Solution

Given condition:
Capacitance $(C) = \frac{ϵ_{0} A}{d} = 8 p F$

Now, $C^{'} = \frac{K ϵ_{0} A}{d^{'}}$

Given, $d^{'} = \frac{d}{2}$

$K = 6$

$C^{'} = \frac{6 ϵ_{0} A}{\frac{d}{2}}$

$C^{'} = \frac{12 ϵ_{0} A}{d}$

Putting values of $C, C^{'} = 12 C$

$= 12 \times 8 \times 1 p F$

$= 96 p F$

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

Given condition: Capacitance (C)=ϵ0Ad=8pF Now, C′=Kϵ0Ad′ Given, d′=d2 K=6 C′=6ϵ0Ad2 C′=12ϵ0Ad Putting values of C,C′=12C =12×8×1pF =96pF

A parallel plate capacitor with air between the plates has a capacitance of $8 p E (1 p E = 10^{- 12} F)$ . What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant $6$ ?

Given condition:
Capacitance $(C) = \frac{ϵ_{0} A}{d} = 8 p F$

Now, $C^{'} = \frac{K ϵ_{0} A}{d^{'}}$

Given, $d^{'} = \frac{d}{2}$

$K = 6$

$C^{'} = \frac{6 ϵ_{0} A}{\frac{d}{2}}$

$C^{'} = \frac{12 ϵ_{0} A}{d}$

Putting values of $C, C^{'} = 12 C$

$= 12 \times 8 \times 1 p F$

$= 96 p F$