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Asked by Vishwa Nath Last Modified
Answer 
Apoorva Purohit
Capacitance between the parallel plates of the capacitor, C = 8 pF
Initially, distance between the parallel plates was d and it was filled with air. Dielectric constant of air, k = 1
Capacitance, C, is given by the formula,
Where,
A = Area of each plate
= Permittivity of free space
If distance between the plates is reduced to half, then new distance, d’ = 
Dielectric constant of the substance filled in between the plates, 
= 6
Hence, capacitance of the capacitor becomes
Taking ratios of equations (i) and (ii), we obtain
Therefore, the capacitance between the plates is 96 pF.
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Gourab Saha
Clear Your Concepts and Score Higher
Arnab Saha
ELECTRICAL ENGINEERING GRADUATE WITH MORE THAN 6 YRS OF TEACHING EXPERIENCE FROM CLASS 9 TO B.TECH
initially d,
therefore, C= EA/d=8 -(1)
now d1=d/2, and dielectric is put,
therefore C1=KEA/d1. -(2) here k=6 which is dielectric constant
we know d1=d/2 substituting it in (1)
C1=6EA/(d/2)=12EA/d. -(3)
(3)÷(1)
C1÷8=12EA/d÷EA/d
C1=96 uF
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