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 = [ E₁ × (2/3)d ]+ [ E₂× (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 / ( ε_r ε_o ) and E₂ = σ / ( ε_o )

where ε_r is dielectric constant and ε_ois permittivity of free space .

Hence , we rewrite eqn.(1) as

begin mathsize 14px style V space equals space open square brackets fraction numerator sigma over denominator epsilon subscript r epsilon subscript o end fraction cross times fraction numerator 2 d over denominator 3 end fraction space plus space sigma over epsilon subscript o cross times d over 3 space close square brackets space equals space fraction numerator sigma space d over denominator 3 epsilon subscript o end fraction open square brackets 2 over epsilon subscript r plus 1 close square brackets end style

.....................(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

begin mathsize 14px style V space equals space fraction numerator Q space d over denominator 3 space A space epsilon subscript o space end fraction open square brackets 2 over epsilon subscript r plus 1 close square brackets end style

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

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

begin mathsize 14px style C space equals space 3 space cross times fraction numerator epsilon subscript o A over denominator d end fraction cross times fraction numerator 1 over denominator open square brackets begin display style 2 over epsilon subscript r end style plus 1 close square brackets end fraction end style

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 ,

begin mathsize 14px style 3 space cross times fraction numerator epsilon subscript o A over denominator d end fraction cross times fraction numerator 1 over denominator open square brackets begin display style 2 over epsilon subscript r end style plus 1 close square brackets end fraction space space equals space space 2.25 space cross times fraction numerator epsilon subscript o A over denominator d end fraction end style

Afiter simplifying above expression , we get dielectric constant ε_r = 6

Answered by Thiyagarajan K | 21 Jul, 2021, 11:37: PM

Please Login or Sign up to continue!

NEET Class neet Answered

Choose Filters

Open In Ask A Doubt App ?

Get in Touch

Call / Whats App