Four charges of +q, +q +q and +q are placed at the corners A, B, C and D of s square. The resultant force on the charge at D

Explanation

Let side = a, `BD=sqrt(2)a`
force `F_(1)=(1)/(4pi in_(0)) . (q^(2))/(a^(2)) ` (A to D)
force `F_(2)=(1)/(4pi in_(0)).(q^(2))/(a^(2))` (C to D) `:. F_(1) = F_(2) =F`

Resultant of `F_(1) and F_(2)=sqrt(F_(1)^(2) +F_(2)^(2)) =sqrt(2).F`
force due to charge at B is
`F_(3) =(1)/(4pi in_(0)) .(q^(2))/((sqrt(2)a)^(2)) =(1)/(2) (1)/(4pi in_(0)) (q^(2))/(a^(2))` (B to D)
Resultant force on change at `D=sqrt(2)F+F_(3)`
`=sqrt(2) (1)/(4pi in_(0)).(q^(2))/(a^(2))+(1)/(2) (1)/(4pi in_(0)) .(q^(2))/(a^(2))`
`=(1)/(4pi in_(0)) (q^(2))/(a^(2)) [sqrt(2) +(1)/(2)]=(q^(2))/(8pi in_(0) a^(2)).[1+2sqrt(2)]`