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Potential difference between centre and surface of the sphere of radius R and $\rho$ uniform volume charge density within it will be

$(a)\;\large\frac{\rho R^2}{6 \in_{0}}\qquad(b)\;\large\frac{\rho R^2}{4\in_{0}} \qquad(c)\;\large\frac{\rho R^2}{3 \in_{0}}\qquad(d)\;\large\frac{\rho R^2}{2 \in_{0}}$

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Answer : (a) $\large\frac{\rho R^2}{6 \in_{0}}$
Explanation :
Potential at centre = $\; V_{C}|_{r=0}=\large\frac{kQ}{2R^3}\;(3R^2-r^2)|_{r=0}$
$V_{C}=\large\frac{3kQ}{2R}$
$V_{surface}=\large\frac{kQ}{R}$
$\bigtriangleup V=\large\frac{kQ}{2R}=\large\frac{k}{2R}\;\rho\large\frac{4}{3} \pi R^3$
$\bigtriangleup V= \large\frac{\rho R^2}{6 \in_{0}}\;.$
answered Feb 7, 2014 by yamini.v
 

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