Four point charges $q_{A}=2\; \mu C, q_{B}=-5\; \mu C,$ $q_{C}=2\; \mu C,$ and $q_{D}=-5\;\mu C$ are located at the corners of a square $ABCD$ of side $10\; cm .$ What is the force on a charge of $1 \;\mu C$ placed at the centre of the square?
The given figure shows a square of side $10\, cm$ with four charges placed at its corners. $O$ is the centre of the square. Where, (Sides) $AB = BC = CD = AD =10\, cm$
(Diagonals) $AC = BD =10 \sqrt{2}\, cm$
$AO = OC = DO = OB =5 \sqrt{2} \,cm$
A charge of amount $1 \,\mu\, C$ is placed at point $O$
Force of repulsion between charges placed at corner $A$ and centre $O$ is equal in magnitude but opposite in direction relative to the force of repulsion between the charges placed at corner $C$ and centre $O$. Hence, they will cancel each other. Similarly, force of attraction between charges placed at comer $B$ and centre $O$ is equal in magnitude but opposite in direction relative to the force of attraction between the charges placed at corner $D$ and centre $O$. Hence, they will also cancel each other. Therefore, net force caused by the four charges placed at the corner of the square on $1 \,\mu\, C$ charge at centre $O$ is zero.
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