Four point charges Q.q Q and q are placed at the corners of a square side 'a' as shown in the figures.

Find the resultant electric force on a charge Q,

Explanation

Let us find the force on the charges Q at the point C Force due to the other charge Q
` F_1 =( 1)/( 4pi in _0) ( Q^(3))/( (asqrt2)^(2)) =(1)/( 4in _0) ((Q^(2))/( 2a^(2))) ` (along AC)
Force due to the charges q (at B) `,F_2`
` = ( 1)/( 4pi in _0) (qQ)/( a^(2)) ` along BC
Force due to the charge q (at D ) , ` F_3`
`= ( 1)/( 4pi in _0)( qQ)/( a^(2)) ` along DC
Resultant of these two equal forces
` = F_( 23) =(1)/( 4pi in _0) (qQ( sqrt2))/( a^(2))` (along AC)
Net force on charge Q ( at point C )
` F= F_1 + F_(23) =(1)/( 4pi in _0) (Q)/( a^(2)) [(Q)/( 2) + sqrt(2q)]`
This force is directed along AC
(For the charges Q . at the point A , the force will have the same magnitude will but will be directed along (CA)
[Note : Don.t deduct marks if the student does not write the direction of the net force F]