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Charge densities of the four quadrants of the ring are $λ$ , –2 $λ$ , – $λ$ and + 2 $λ$ respectively. The electric potential at the centre of ring is

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zero

$\frac{\sqrt{2} λ}{4 {πε}_{0}}$

$\frac{\sqrt{5} λ}{4 {πε}_{0}}$

$\frac{\sqrt{2} λ}{4 {πε}_{0} R}$

answer is B.

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Detailed Solution

$V = \frac{1}{4 {πε}_{0}} \sum \frac{q}{r} = \frac{1}{4 {πε}_{0}} \frac{(- 2 λ - λ + 2 λ + λ) \frac{πr}{2}}{r}$
= zero

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