Question

The capacity of a parallel plate capacitor formed by the plates of same are A is 0.02μF with air as dielectric. Now one plate is replaced by a plate of area 2A and dielectric (K = 2) is introduced between the plates, the capacity is :

Answer 1

Af**k off

nswer:

Explanation:

Answer 2

Answer:

The capacity of the new introduced plate is equal to 0.08μF whose area 2A and dielectric constant K=2.

Explanation:

Given, the capacitor is formed with plates of Area 'A' and the capacitance is equal to 0.02μF. The dielectric constant of air is K=1.

The capacitance of the capacitor is given by:

$C=\frac{K\epsilon_{o}A}{d}$                                                                             ............(1)

where ε₀ is the absolute permittivity of the space.

A is the area of the plate

and, d is the separation between the plates of the capacitor.

So equation (1) becomes :

$0.02\mu F=\frac{\epsilon _{o}A}{d}$                                                                       ............(2)

When plates of capacitor replaced by plates with area 2A then capacitance will be C'. Area of plate A'=2A and dielectric constant K=2.

$C'=\frac{K'\epsilon_{o}A'}{d}$

$C'=\frac{(2)\epsilon_{o}(2A)}{d}$

$C'=4(\frac{\epsilon_{o}A}{d})$

Put the value from equation no. (2);

$C'=4(0.02\mu F)$

$C'=0.08\mu F$

Therefore, the capacity of the introduced plate will be 0.08μF.