Question

A dielectric of thickness 5cm and dielectric
constant 10 is introduced between the
plates of a parallel plate capacitor having
plate area 500 sq. cm and separation
between the plates 10cm. The capacitance
of the capacitor with dielectric slab is​

Answer 1

Answer:

if the parallel plate capacitor without any dielectric is charged , its potential V is given by , V = Q / C   = Q d / (A εo )  ................(1)

 

where A is area of plates, d is the distance between plates, Q is charge on plates and εo is permitivity of free space.

 

when dielectric of 4 mm thickness increased and plates are separated by another 3.2 mm to maintain the same potential difference,

 

we have electric field  in free space E1 = σ / εo  and  electric field in dielectric E2 = σ / ( εo εr )  , where σ is the charge density and

 

εr is relative permitivity of dielectric.

 

hence potential difference V = V1 + V2 = E1 ( d - 0.8) + E24  = ( σ / εo ) [ (d - 0.8) + ( 4/εr ) ] ......................(2)

 

let us substitute σ = Q/A in (2), then we have V = ( Q / (A εo ) ) [ (d - 0.8) + ( 4/εr ) ]  ............................ ( 3 )

 

since potential differenceis maintained as 200 V with and without dieclectric, we can equate eqn.(1)  and eqn.(3)

 

Qd / (A εo )  = ( Q / (A εo ) ) [ (d - 0.8) + ( 4/εr ) ]     or   d  = (d - 0.8) + ( 4/εr ) or  4/εr  = 0.8    or  εr = 5