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Six charges of equal magnitude, 3 positive and 3 negative are to be placed on PQRSTU corners of a regular hexagon, such that field at the centre is double that of what it would have been if only one +ve charge is placed at R

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+,+,+,-,-,-

-,+,+,+,-,-

+,-,+,-,+,-

-,+,+,-,+,-

answer is C.

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Detailed Solution

The electric field due to charges at P, S, Q and T is zero . So, electric field due to charges due to U and R will only be added up. i.e. -,+,+,-,+,-

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