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
![question image](https://images.topperlearning.com/topper/new-ate/topr_395233748629222324e4199aaf44affb469829fc06099f2.jpeg)
Asked by sirib942254 | 21 Jul, 2021, 22:00: PM
![](https://images.topperlearning.com/topper/tinymce/imagemanager/files/35bf465c8faee266c7ab4b7d3794c14260f85a4fd90b65.97704299f7.png)
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 / ( 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](https://images.topperlearning.com/topper/tinymce/cache/680e8445061aaa23ebf99fa3911f63b7.png)
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](https://images.topperlearning.com/topper/tinymce/cache/99b31fed134027a497d488451b64b8bf.png)
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](https://images.topperlearning.com/topper/tinymce/cache/2c1ddd0fd36ea479c2ee1a93f844663a.png)
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](https://images.topperlearning.com/topper/tinymce/cache/a49a674a92a9465dc0aecae9e897f14b.png)
Afiter simplifying above expression , we get dielectric constant εr = 6
Answered by Thiyagarajan K | 21 Jul, 2021, 23:37: PM
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