Question

# 5 equal charges ‘q′ each are placed at the corners of a regular hexagon ABCDEF of side ‘a′ as shown. Find the net electric field at the centre O of the hexagon?

A
5kqa2 towards F
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B
kqa2 towards F
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C
32kqa2 towards F
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D
52kqa2 towards C
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Solution

## The correct option is B kqa2 towards FInorder to solve, find electric field due to each charge at O and then add using principle of superposition. Distance between O and B is ‘a′ as side length of regular hexagon is given to be ‘a′. Now →EOB=kqa2 , directed radially away from B. →EOE=kqa2; same magnitude but direction exactly opposite to →EOB. →EOE and →EOB are equal and opposite vectors and hence will cancel each other out. Similarly, →EOA and →EOD are equal and opposite, so they will also have a zero net effect. Hence, net field will be only due to point charge at C. →EOC=kqa2 direction away from C i.e. towards F

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