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 = sqrt2a`
Along AD force `F_(1) = (1)/(4pi in_(0)) .(q^(2))/(a^(2))`
Along CD force `F_(2) = (1)/(4pi in_(0)).(q^(2))/(a^(2)) therefore F_(1) = F_(2)`
Resultant of `F_(1) and F_(2) = sqrt(F_(1)^(2) + F_(2)^(2))= sqrt2.F_(1)`
Force along BD
`F_(3) = (1)/(4pi in_(0)) .(q^(2))/((sqrt2a)^(2)) = (1)/(2) (1)/(4pi in_(0)) (q^(2))/(a^(2))`
Resultant force at `D= sqrt2F_(1) + F_(3)`
`= sqrt2 (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)) [sqrt2 + (1)/(2)] = (q^(2))/(8pi in_(0)a^(2)).[1 + 2 sqrt2]`