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Ten charges are placed on the circumference of a circle of radius $R$ with constant angular separation between successive charges. Alternate charges $1,3,5,7,9$ have charge $(+q)$ each, while $2,4,6,8,10$ have charge $(-q)$ each. The potential $V$ and the electric field $E$ at the centre of the circle are respectively
(Take $V =0$ at infinity $)$
$V =\frac{10 q }{4 \pi \epsilon_{0} R } ; E =\frac{10 q }{4 \pi \epsilon_{0} R ^{2}}$
$V =0, E =\frac{10 q }{4 \pi \epsilon_{0} R ^{2}}$
$V =0, E =0$
$V =\frac{10 q }{4 \pi \varepsilon_{0} R } ; E =0$
Solution
Sol. Potential of centre $= V =\quad \Sigma\left(\frac{ kq }{ R }\right)$
$V _{ C }=\frac{ K (\Sigma q )}{ R }$
$V _{ C }=\frac{ K (0)}{ R }=0$
Electric field at centre $\overrightarrow{ E }_{ B }=\Sigma \overrightarrow{ E }$
Let $E$ be electric field produced by each charge at the centre, then resultant electric field will be
$E _{ C }=0,$ since equal electric field vectors are acting at equal angle so their resultant is equal to zero.
