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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

4kQπR2

b

2kQπR2

c

3kQπR2

d

23kQπR2

answer is A.

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Detailed Solution

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Both would produce 2kλR electric field and angle between field would be 60°.

∴Eresultant=2 x 2kλR x cos 30o=3kQπR2

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