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Four identical point charges, each of charge q, are placed on the vertices of a regular hexagon of side a such that the net electric field at centre of hexagon is zero. If one of the charge is now removed, then the electric field at centre of hexagon is |
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Answer» Answer: ANSWER (A) In given situation field at center is E towards A which due to charge q D
Now if charge B is REMOVED then then field at center will be E towards A and E towards B Now resulting field will be 2Ecosθ/2=2Ecos30= 3
E A->4 (B)If charge at C is removed then field at center will be E towards A and E towards C RESULTANT field 2Ecos2θ/2=E B->2 (C)filed at center initially was due to charge at D now if it is removed then field will become zero. C->3 (D)Since when charge at C is removed field at center is E towards B and field at center due charge at B is balancing the field due to charge at E therefore, when charge at B is also removed then field due E will also have INFLUENCE and which is towards B therefore, both the fields will add up and resultant field at center will be 2E D->1 |
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