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Bhavya Bhadauria Grade: 11
Five point charges each of value +q are placed on five vertices of a regular hexagon of side L m. What is the magnitude of force on a point charge of value -q coulomb placed at the centre of hexagon?
2 years ago

Answers : (3)

ankush singh
30 Points
The answer is kq2/L2 towards the direction of empty vertics.  Because force due to other 4 charges cancel out each other.
2 years ago
Heet Mehta
26 Points
										There are five vertices so it must be a Pentagon,  four charges cancel each other out and there is one charge left on the remaining vertex so apply the regular formula F = kq2/r2
10 months ago
12 Points
										If there were 6 charges at each corner then the net electric field at the centre would be zero since it is a regular hexagon . the electric field due to 5 charges will be equal and opposite to the net electric field due to sixth charge so as to make the net electric field at the centre to be zero . Hence electric field at the centre due to 5 charges is will be equal to electric field at the centre due to sixth charge. F=(kq^2)/L^2
2 months ago
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