A) V
B) 2V
C) 4V
D) ? 2V
Correct Answer: A
Solution :
In case of a charged conducting sphere \[{{V}_{\text{inside}}}={{V}_{\text{centre }}}={{V}_{\text{surface}}}=\frac{1}{4\pi {{\varepsilon }_{o}}}.\frac{q}{R}\], \[{{V}_{\text{outside}}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q}{r}\] If a and b are the radii of sphere and spherical shell respectively, then potential at their surface will be \[{{V}_{\text{sphere }}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{Q}{a}\] and \[{{V}_{\text{shell}}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{Q}{b}\] \[\therefore \]\[V={{V}_{\text{sphere }}}-{{V}_{\text{shell}}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\left[ \frac{Q}{a}-\frac{Q}{b} \right]\] Now when the shell is given charge (?3Q), then the potential will be \[V{{'}_{\text{sphere}}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{Q}{a}+\frac{(-3Q)}{b} \right],\]\[V{{'}_{\text{shell}}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{Q}{b}+\frac{(-3Q)}{b} \right]\] \[\therefore \]\[V{{'}_{\text{sphere }}}-V{{'}_{\text{shell}}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{Q}{a}-\frac{Q}{b} \right]=V\]
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