# please answer quickly

The charge Q is placed at a distance R from the centre of the shell. Due to this charge there will be induced charge on both inner and outer surface of the shell so that the inner surface has charge -Q and outer surface has charge +Q so that the net charge on the shell is zero.

Now the potential due to these charges at the centre is given by

$V=\frac{1}{4\pi {\epsilon}_{o}}\frac{Q}{R}+\frac{1}{4\pi {\epsilon}_{o}}\frac{-Q}{3R}+\frac{1}{4\pi {\epsilon}_{o}}\frac{+Q}{5R}\phantom{\rule{0ex}{0ex}}V=\frac{1}{4\pi {\epsilon}_{o}}\left(\frac{Q}{R}-\frac{Q}{3R}+\frac{Q}{5R}\right)\phantom{\rule{0ex}{0ex}}V=\frac{1}{4\pi {\epsilon}_{o}}\frac{13Q}{15R}\phantom{\rule{0ex}{0ex}}hencethecorrectoptionis\left(3\right)$

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