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Home / Doubts and Solutions/JEE/Physics

Question

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Asked by rbhatt16 26th June 2018, 6:10 PM
Answered by Expert
Answer:
 
Capacitance C of a capacitor formed between two coaxial cylinder is given by
 
begin mathsize 12px style C space equals space fraction numerator 2 pi space epsilon subscript 0 space K space L over denominator log space open parentheses begin display style b over a end style close parentheses end fraction end style
 
where K is dielectric constant of the dielectric material placed between cylinders, L is length of cylinder,
a and b are inner and outer radius respectively.
 
when a potential difference V is applied across the cyllinders and dipped in water, due to fringe effect of electric field,
water is attracted inside upto a height h as shown in figure.
 
When water is inside the cylinderical capacitor upto a height h, then the capacitor is a parallel combination of two capacitors,
one with dielectric material(water) of length h and other one without dielectric of length (H-h) as shown in figure.
H is full length of cylinders.
 
Hence equivalent capacitance begin mathsize 12px style C space equals space fraction numerator 2 space pi space epsilon subscript 0 over denominator log open parentheses begin display style b over a end style close parentheses end fraction open curly brackets K space h space plus space left parenthesis H minus h right parenthesis space close curly brackets end style
Potential Energy E begin mathsize 12px style equals space 1 half C space V squared space equals space fraction numerator pi space epsilon subscript 0 over denominator log open parentheses begin display style b over a end style close parentheses end fraction open curly brackets K space h space plus space left parenthesis H minus h right parenthesis space close curly brackets space V squared space end style
attractive force F experienced by water is given by, begin mathsize 12px style F space equals space fraction numerator partial differential E over denominator partial differential h end fraction space equals space fraction numerator pi space epsilon subscript 0 over denominator log open parentheses begin display style b over a end style close parentheses end fraction open curly brackets K space minus space 1 close curly brackets V squared end style...................(1)
Force given in eqn.(1) is balanced by force due to water's weight
 
hence   we have    begin mathsize 12px style space fraction numerator pi space epsilon subscript 0 over denominator log open parentheses begin display style b over a end style close parentheses end fraction open curly brackets K space minus space 1 close curly brackets V squared space space equals space pi open parentheses b squared space minus space a squared close parentheses cross times h cross times rho cross times g end style  ...........................(2)
 
From Eqn.(2), we write for h as,  begin mathsize 12px style h space equals space fraction numerator open parentheses K minus 1 close parentheses epsilon subscript 0 space V squared over denominator log open parentheses begin display style b over a end style close parentheses space left parenthesis b squared minus a squared right parenthesis rho g end fraction end style
 
 
Answered by Expert 28th June 2018, 3:57 PM
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