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
3kQπR2
b
23kQπR2
c
2kQπR2
d
4kQπR2
answer is A.
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Detailed Solution
Both would produce 2kλR electric field and angle between field would be 60°.∴Eresultant=2 x 2kλR x cos 30o=3kQπR2
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 :