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NEET Class neet Answered

The gap between the plates
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Asked by padamkumar52010 | 11 Jul, 2022, 12:11: AM
answered-by-expert Expert Answer
Potential difference between plates that are separated by a distance d is given as
 
begin mathsize 14px style V space equals space integral subscript 0 superscript d E space d y end style  ............................ (1)
where E is elctric field at a distance y
 
begin mathsize 14px style E space equals space fraction numerator sigma over denominator space epsilon left parenthesis y right parenthesis space epsilon subscript o end fraction end style
 
where σ is charge density on plates and ε(y) is variable dielectric constant and εo is permittivity of free space
 
ε(y) = ε1 + m y  , where m = ( ε2 - ε1 ) /d
 
Hence potential difference is obtained from eqn.(1) as
 
 
begin mathsize 14px style V space equals space sigma over epsilon subscript o integral subscript 0 superscript d fraction numerator d y over denominator epsilon subscript 1 plus m y end fraction space equals space fraction numerator sigma over denominator m space epsilon subscript o end fraction open square brackets ln left parenthesis epsilon subscript 1 plus m y right parenthesis close square brackets subscript 0 superscript d space end style

begin mathsize 14px style V space equals space fraction numerator sigma over denominator m space epsilon subscript o end fraction space ln space open parentheses epsilon subscript 2 over epsilon subscript 1 close parentheses space space equals space fraction numerator sigma space d over denominator left parenthesis epsilon subscript 2 minus epsilon subscript 1 right parenthesis space epsilon subscript o end fraction space ln space stretchy left parenthesis epsilon subscript 2 over epsilon subscript 1 stretchy right parenthesis space end style
charge density σ = Q / A  , where Q is charge on plates and A is area of plate .
Hence above expression for potential difference becomes
 
 
begin mathsize 14px style V space equals Q over A space cross times fraction numerator space d over denominator space left parenthesis epsilon subscript 2 minus epsilon subscript 1 right parenthesis space epsilon subscript o end fraction space ln space stretchy left parenthesis epsilon subscript 2 over epsilon subscript 1 stretchy right parenthesis space end style ........................... (2)
Capacitance C = Q / V
 
begin mathsize 14px style C space equals space fraction numerator open parentheses epsilon subscript 2 minus epsilon subscript 1 close parentheses over denominator ln open parentheses begin display style epsilon subscript 2 over epsilon subscript 1 end style close parentheses end fraction space cross times fraction numerator epsilon subscript o A over denominator d end fraction end style
 

 

 
Answered by Thiyagarajan K | 11 Jul, 2022, 09:08: PM
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