Question

A parallel plate capacitor with its b/w the plates has a capacitor of 12pf what will be the Capacitor of distance bie the plates reduced by half and the space blu them is filled wi o material of dielectric constant 6 (2016) 6 ( /

​

Answer 1

Answer: 96 pf

Explanation:Step 1: Initial capacitance

Let the initial distance between the plates be d.

Let A be the area of each plate.

Capacitance of the parallel plate capacitor is given by:

                C  

i

​

=  

d

A∈  

0

​

 

​

                  ....(1)

Step 2: Capacitance after inserting dielectric

Let k be the dielectric constant.

               C  

f

​

=  

d  

′

 

kA∈  

0

​

 

​

 

As the distance between plates is reduced to half, new distance is d  

′

=  

2

d

​

 

               ∴C  

f

​

=  

d

2kA∈  

0

​

 

​

=2kC  

i

​

=2×6×8pF=96pF