Four charges are arranged at the corners of a square ABCD, as shown in the figure. The force on the positive charge kept at the centre O is

The correct option is C along the diagonal BD

$F_{netduetochargeplacedatAandC,}{F}_{net}={\overrightarrow{F}}_{OA}-\overrightarrow{F_{OC}}=0$

$\because \left|F_{OA}\right|=\left|F_{OC}\right|=\left|\frac{KqQ}{{\left(\frac{a}{\sqrt{2}}\right)}^2}\right|$

$F_{netduetochargeplacedatBandD,}{F}_{net}=\left|\overrightarrow{F_{OD}}\right|+\left|\overrightarrow{F_{OB}}\right|$

$=F_1+F_2⟶towardsD$

If we put a positive charge at O.

The resultant force due to the charge placed at A and C is zero and resultant force due to B and D is towards D, along the diagonal BD.