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CBSE Class 12-science Answered

Pls answer 
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Asked by maheshnanjappa120 | 26 Feb, 2019, 11:25: PM
answered-by-expert Expert Answer
as shown in figure, the charge 1μC placed at centre of square is having equal separation distance d from the all charges situated at corner of square.
 
This equal distance d is obtained from  2d2 = (0.02√2)2   or  d = 0.02 m
 
Hence potential energy V of  1 μC charge = begin mathsize 12px style fraction numerator 4 cross times 1 cross times 10 to the power of negative 15 end exponent over denominator 4 πε subscript straight o space 0.02 end fraction plus fraction numerator begin display style 6 cross times 1 cross times 10 to the power of negative 15 end exponent end style over denominator begin display style 4 πε subscript straight o space 0.02 end style end fraction plus fraction numerator begin display style open parentheses negative 8 close parentheses cross times 1 cross times 10 to the power of negative 15 end exponent end style over denominator begin display style 4 πε subscript straight o space 0.02 end style end fraction plus fraction numerator begin display style 10 cross times 1 cross times 10 to the power of negative 15 end exponent end style over denominator begin display style 4 πε subscript straight o space 0.02 end style end fraction space space space V o l t end style
let us consider 1/(4πεo ) = 9×109 N m2C-2 , then begin mathsize 12px style V space equals space fraction numerator 9 cross times 10 to the power of 9 cross times 12 cross times 10 to the power of negative 15 end exponent over denominator 0.02 end fraction space equals space 54 cross times 10 to the power of negative 4 space end exponent V o l t end style
Answered by Thiyagarajan K | 27 Feb, 2019, 12:58: PM

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