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Three concentric spherical metallic shells A , B and C of radii a , b and c $\;(a < b < c)\;$ have charge densities $\;\sigma , - \sigma \; and \; \sigma\;$ respectively . If the shells A and C are at the same potential then the relation between a , b and c is

$(a)\;a+b+c=0\qquad(b)\;a+c=b\qquad(c)\;a+b=c\qquad(d)\;2a+b=c$

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Answer : (c) a+b=c
Explanation :
$V_{A}=\large\frac{k\;\sigma\;4\;\pi\;a^2}{a} =\large\frac{k\;(-\sigma)\;4\;\pi\;b^2}{b}+\large\frac{k\;\sigma\;4\;\pi\;c^2}{c}$
$=k\; \sigma 4\;\pi\;(a-b+c)$
$V_{C}=\large\frac{k\;\sigma\;4\;\pi\;a^2}{c} -\large\frac{k\;\sigma\;4\;\pi\;b^2}{b}+\large\frac{k\;\sigma\;4\;\pi\;c^2}{c}$
$=k\;\sigma\;4\;\pi\;(\large\frac{a^2}{c}-\large\frac{b^2}{c}+c)$
A and C are at same potential
$V_{A}=V_{C}$
$(a-b+c)=\large\frac{a^2-b^2}{c}+c$
$(a-b)=\large\frac{(a-b)\;(a+b)}{c}$
$a+b=c\;.$
answered Feb 14, 2014 by yamini.v
 

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