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# Two concentric conducting spheres of radii R and 2R are carrying charges Q and -2Q respectively . If the charge on inner sphere is doubled the potential difference between the two spheres will becomes

$(a)\;two\;times\qquad(b)\;four\;times\qquad(c)\;be\;halved\qquad(d)\;remain\;same$

Explanation :
Initial potential difference between A & B
$V_{AB}=(\large\frac{kQ}{R}-\large\frac{k2Q}{2R})-(\large\frac{kQ}{2R}-\large\frac{k2Q}{2R})$
$V_{AB}=\large\frac{kQ}{2R}$
Final potential difference between A & B
$V^{|}_{AB}=(\large\frac{2kQ}{R}-\large\frac{2kQ}{2R})-(\large\frac{2kQ}{2R}-\large\frac{2kQ}{2R})$
$V^{|}_{AB}=\large\frac{kQ}{R}$
$V^{|}_{AB} = 2 V_{AB}\;.$
edited Feb 7, 2014 by yamini.v