Four point charges `+2nC,-3nC,+4nC,-5nC` are placed at the vertices A,B,C,D of a square ABCD of side 0.2 m respectively. Calculate the electric field intensity at a point of intersection of the diagonals.

Explanation

The angles between `E_(1)` and `E_(2),E_(3)` and `E_(4)` and `E_(5)` and `E_(6), theta=90^(@)`
`Ac=(sqrt(2))(0.2)m`

`E_(1)=9xx10^(9)xx(2xx10^(-9))/((0.2sqrt(2))^(2))=18/2xx1/(10^(-2))=9xx10^(2)Vm^(-1)`
`E_(2)=4.5xx10^(2)xx3=13.5xx10^(2)Vm^(-1)`
`E_(3)=4.5xx10^(2)xx4=18.0xx10^(2)Vm^(-1)`
`E_(4)=4.5xx10^(2)xx5=22.5xx10^(2)Vm^(-1)`
`E_(5)=E_(3)-E_(1)=9xx10^(2)Vm^(-1)`
`E_(6)=E_(4)-E_(2)=9xx10^(2)Vm^(-1)`
`E=10^(2)xx9sqrt(1^(2)+1^(2)+0)=12.72 Vm^(-1)`
`tan alpha=(E_(6)sin 90)/(E_(5)+E_(6)cos 90)=(E_(6))/(E_(5))=1 :. alpha=45^(@)` w.r.t `E_(5)` or `E_(6)`