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A parallel plate capacitor has a certain capacitance. When 2/3rd of the distance between the plates is filled with dielectric the capacitance is found to be 2.25 times the initial capacitance. The dielectric constant of the dielectric is
Asked by sirib942254 | 21 Jul, 2021, 10:00: PM
Potential difference V between plates is given as

V =   [ E1 × (2/3)d ]+  [ E× (1/3)d ]  .......................(1)

where E1 is electric field in dielectric material,  E2 is electric field in air and d is distance between plates .

If σ is surface charge density on plates , then we have

E1 = σ / ( εr εo )     and     E2 = σ / ( εo )

where εr is dielectric constant and εo is permittivity of free space .

Hence , we rewrite eqn.(1) as

.....................(2)

If Q is magnitude of total charge on each plate and A is area of plate , then surface charge density
is given as  σ = Q / A .

Hence  eqn.(2) is rewritten as

............................ (3)

Capacitance C is defined as   C = Q / V ,  Hence we get capacitance C from eqn.(3) as

But we are given that capacitance with dielectric  is increased by 2.25 times of initial capacitance with only air gap.

Capacitance with only air gap , C' =  [ ( εo A ) / d ]

Hence  we get ,

Afiter simplifying above expression , we get dielectric constant  εr = 6
Answered by Thiyagarajan K | 21 Jul, 2021, 11:37: PM
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