# NEET Class neet Answered

**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

Expert Answer

Potential difference V between plates is given as

V = [ E

_{1}× (2/3)d ]+ [ E_{2 }× (1/3)d ] .......................(1)where E

_{1}is electric field in dielectric material, E_{2}is electric field in air and d is distance between plates .If σ is surface charge density on plates , then we have

E

_{1}= σ / ( ε_{r}ε_{o}) and E_{2}= σ / ( ε_{o})where ε

_{r}is dielectric constant and ε_{o}^{ }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' = [ ( ε

_{o}A ) / d ]Hence we get ,

Afiter simplifying above expression , we get dielectric constant ε

_{r}= 6
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