JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
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 \()\)
- A \(V =\frac{10 q }{4 \pi \epsilon_{0} R } ; E =\frac{10 q }{4 \pi \epsilon_{0} R ^{2}}\)
- B \(V =0, E =\frac{10 q }{4 \pi \epsilon_{0} R ^{2}}\)
- C \(V =0, E =0\)
- D \(V =\frac{10 q }{4 \pi \varepsilon_{0} R } ; E =0\)
Answer & Solution
Correct Answer
(C) \(V =0, E =0\)
Step-by-step Solution
Detailed explanation
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…
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