Five point charges each of charge q are placed on the five vertices of a regular hexagon of side 'l' .Find the magnitude of the resultant force on a charge -q placed at the centre of the hexagon
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Answer:
The resultant force is zero
Explanation:
The force between any two particles placed at a distance 'd'apart and have charge of magnitudes q1 and q2 is given by (k×q1×q2)÷d^2.
Since the charge is placed symmetrically at the center the force of attraction due to one charge is cancelled out by an other charge placed at the opposite corner, as a result each and every force acting on the charge of -q is cancelled out by the force acted on it by the charge q placed at its corner, therefore the net charge acting on the charge of -q placed at the center is zero.
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