Question

The capacitance of a parallel plate capacitor having air

between its plates is 10 µf if the distance between the plates is

reduced to half and the medium between plates is replaced by

a medium of dielectric constant k=5, then find the value how

much increase or decrease in the capacitance of the capacitor.​

Answer 1

Given : The capacitance of a parallel plate capacitor having air  between its plates is 10 µf  . the distance between the plates is  reduced to half and the medium between plates is replaced by  a medium of dielectric constant k=5,  

To Find : the value how  much increase or decrease in the capacitance of the capacitor.​

Solution:

C = Kε₀ A / d

The capacitance of a parallel plate capacitor having air

between its plates is 10 µf

C =  10 µf

K = 1  for dielectric constant of air

=> 1. ε₀ A / d  = 10  µf

=> ε₀ A / d  = 10  µf

distance between the plates is  reduced to half  

=> New d  = d/2

medium between plates is replaced by  a medium of dielectric constant k=5

=> k = 5

new C  = 5 ε₀ A / (d/2)

=> new C  =  10 ε₀ A / d

=> new C  =  10 * 10 µf

=> new C  =  100  µf

increase in capacitance = 100 - 10 = 90 µf

Learn More:

A parallel plate capacitor is made of two rectangular plates of area ...

https://brainly.in/question/8806301

In a circular parallel plate capacitor radius of each plate is 5cm and ...

https://brainly.in/question/8489492

A parallel plate capacitor having plate area 'A' , and separation ...

https://brainly.in/question/18896298

Answer 2

Answer:

90

Explanation:

The capacitance of a parallel plate capacitor is given as

C = Kε₀ A / d

Here K is the dielectric constant value of medium between plates, A is the cross sectional area and d is the distance between the plates.

Case 1

Here K = 1 ( for air)

C = 1 x ε₀ A / d

10 = ε₀ A / d

Case 2

Here D' =  d/2

A' = A

K' = 5

C' = K'ε₀ A' / d'

Putting the values

C' = 5 ε₀ A / d/2

C' = 10C

C' = 10 x 10 = 100uf

Difference between C' and C = 100 - 10 = 90