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If the potential at the centre of a uniformly charged hollow sphere of radius R is V then electric field at a distance r from the centre of sphere will be $\;(r > R)$

$(a)\;\large\frac{VR}{r^2}\qquad(b)\;\large\frac{Vr}{R^2}\qquad(c)\;\large\frac{VR}{R^2+r^2}\qquad(d)\;\large\frac{VR}{r}$

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Answer : (a) $\;\large\frac{VR}{r^2}$
Explanation :
$E (r) = \large\frac{k\;q}{r^2}\quad\;for\;r > R$
Where q is the charge on hollow sphere and $\;V=\large\frac{k\;q}{R}$
$E (r) = \large\frac{VR}{r^2}\;.$
answered Feb 6, 2014 by yamini.v
 

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