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Q.

A uniformly charged ring of charge Q and radius R is folded across its diameter such that two halves make an angle 60° with each other then net electric field at centre 'O' of ring is equal to :

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a

$\frac{4 kQ}{{πR}^{2}}$

b

$\frac{2 kQ}{{πR}^{2}}$

c

$\frac{\sqrt{3} kQ}{{πR}^{2}}$

d

$\frac{2 \sqrt{3} kQ}{{πR}^{2}}$

answer is A.

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Detailed Solution

detailed_solution_thumbnail

Both would produce $\frac{2 kλ}{R}$ electric field and angle between field would be 60°.

$∴ E_{resultant} = 2 x (\frac{2 kλ}{R}) x \cos 30^{o} = \frac{\sqrt{3} kQ}{{πR}^{2}}$

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