Q

A parallel plate capacitor with air between the plates has a capacitance of 8 p 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

2.5) A parallel plate capacitor with air between the plates has a capacitance of 8 pF $\left ( 1pF = 10^{-12}F \right )$. 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?

Views

As we know,

$C=\frac{\epsilon_r\epsilon_0 A}{d}$

where A= area of the plate

$\epsilon_0$ = permittivity of the free space

d = distance between the plates.

Now, Given

The capacitance between plates initially

$C_{initial}=8pF=\frac{\epsilon A}{d}$

Now, Capacitance when the distance is reduced  half  and filled with the substance of dielectric 6

$C_{final}=\frac{6\epsilon_0 A}{d/2}=12\frac{\epsilon _0A}{d}=12*8pF=96pF$

Hence new capacitance is 96pF.

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