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Question

Five point charges, each of value +q are placed on five vertices of a regular hexagon of side L. The magnitude of the force on a point charge of value q placed at the centre of the hexagon is nq24πϵ0L2. Then the value of n?

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Solution

Among the five forces due to five charges, 2 pair (four) of forces are cancelled with each other since they are equal in magnitude and opposite in direction.



The resultant force is due to a single force. Also since it is a hexagon, distance between the centre and corner is equal to side length of the hexagon.
F=14πϵ0×q×qL2

F=14πϵ0×q2L2 (attarcative in nature)

On comparing with nq24πϵ0L2:
We get, n=1

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