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Electrostatic Potential And Capacitance

Question
CBSEENPH12038969

A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate capacitor but has the thickness d/2, where d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor. 

Solution

Capacitance having a dielectric of thickness ‘t’ is given by, 
                                straight C space equals space fraction numerator straight epsilon subscript straight o space straight A over denominator straight d space minus space straight t space plus space begin display style straight t over straight K end style end fraction 

When the thickness of the plates is reduced to half, t = d/2 then,

 Capacitance becomes, 
straight C space equals space fraction numerator straight epsilon subscript straight o straight A over denominator straight d space minus space begin display style straight d over 2 end style plus begin display style fraction numerator straight d over denominator 2 straight K end fraction end style end fraction space

space space space equals space fraction numerator straight epsilon subscript straight o straight A over denominator begin display style straight d over 2 end style plus space begin display style fraction numerator straight d over denominator 2 straight K end fraction end style end fraction space

space space space space equals space fraction numerator straight epsilon subscript straight o straight A over denominator begin display style straight d over 2 end style open parentheses 1 plus begin display style 1 over straight K end style close parentheses end fraction space

space space space space equals space fraction numerator 2 straight epsilon subscript straight o AK over denominator straight d space left parenthesis straight K space plus space 1 right parenthesis thin space end fraction

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