Question

A parallel plate capacitor with air between the plates has a
capacitance of 8 pF (1pF = 10-12 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 constant 6?
​

Answer 1

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