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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
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Asked by sirib942254 | 21 Jul, 2021, 10:00: PM
answered-by-expert Expert Answer
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
 
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' =  [ ( εo 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  εr = 6 
Answered by Thiyagarajan K | 21 Jul, 2021, 11:37: PM
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