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Five point charges, each of charge +q are placed on 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.

 

 
 
 
 
 

Answers (1)

The force due to the charges placed diagonally opposite at the vertices of a hexagon, on the charge -q cancel in pairs.

Hence, the net force is due to one charge only.

Net force \left | \vec{f} \right |=\frac{1}{4\pi \epsilon _{0}}\; \frac{q^{2}}{l^{2}} is on charge -q placed at the centre of the hexagon.

Posted by

Safeer PP

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