Four point charges Q , q Q and q are placed at the corners of a square of side 'a' as shown in the Fig. Find the
resultant electric force on a charge Q

Explanation

Let us calculate electric force on a charge Q situated at the corner A in figure , then
`|vecF_(AB)| = (1)/(4 pi in_(0)) (Qq)/(q^(2)) , |vecF_(AC)| = (1)/(4pi in_(0)) * (Q^(2))/(2a^(2)) ` and `|vecF_(AD)| = (1)/(4pi in_(0)) * (Qq)/(a^(2))`
Directions of the force are as shown in Fig.

`therefore` Force along BA , `F_(1) = F_(AB) + F_(AC) sin 45^(@)`
`= (1)/(4pi in_(0)) (Qq)/(a^(2)) + (1)/(4 pi in_(0)) * (Q^(2))/(2a^(2)) (1)/(sqrt2)`
`= (Q)/(4pi in_(0) a^(2)) [ q + (Q)/(2sqrt2)]`
and force along `DA , F_(2) = F_(AD) + F_(AC) cos 45^(@)`
=` (1)/(4 pi in_(0)) (Qq)/(a^(2)) + (1)/(4pi in_(0)) * (Q^(2))/(2a^(2)) (1)/(sqrt2)`
`= (Q)/(4 pi in_(0) a^(2)) [ a + (Q)/(2 sqrt2)]`
`therefore` Resultant electric force `F_(A) = sqrt(F_(1)^(2) - F_(2)^(2)) = (Q)/(4 pi in_(0) a^(2)) [ q + (Q)/(2sqrt2)] sqrt2 = (Q)/(8 pi in_(0) a^(2)) [2 sqrt2q + Q]`
The resultant force acts along the extended line CA.