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Asked by Prashant DIGHE | 07 Mar, 2020, 22:12: PM
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Capacitance C' of partially filled capacitor is given by,  begin mathsize 14px style C apostrophe space equals space fraction numerator A space epsilon subscript o over denominator open square brackets d space plus space t left parenthesis 1 over epsilon subscript r minus 1 right parenthesis close square brackets end fraction space end style  ......................(1)
where A is area of parallel plates, εo is permittivity of free space,  εr is dielectric constant of dielectric material,
d is the distance between parallel plates and t is thickness of dielectric material
 
if t = 0.5 d, then we have, begin mathsize 14px style C apostrophe space equals space fraction numerator C over denominator open square brackets 1 space plus space 0.5 left parenthesis 1 over epsilon subscript r minus 1 right parenthesis close square brackets end fraction space end style ..........................(2)
 
If Capacitance is changed to 4/3 of its original value after inserting dielectric material, then we have   ( C' / C ) = 4/3
 
Hence from eqn.(2) , we get       begin mathsize 14px style fraction numerator 1 over denominator 1 plus 0.5 open parentheses begin display style 1 over epsilon subscript r end style space minus space 1 close parentheses end fraction space equals space 4 over 3 end style
By solving above equation, we get εr = 2
Answered by Thiyagarajan K | 08 Mar, 2020, 08:33: AM

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