Question

# 5 equal charges ‘q’ are placed at 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 A kqa2 towards F Find E due to each charge then add by principle of superposition. Distance between O and B is ‘a’ as side length of regular hexagon is given to be ‘a’ Now →EOB=kqa2 , directedradially 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. 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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