Four point charges `q_(A)=2 muC, q_(B)=-5 muC, q_(C)=2 muC` and `q_(D)=-5 muC` are located at the corners of square ABCD of side 10 cm. What is the force on a charge of `1 muC` placed at the centre of the square?

Explanation

Let the centre of the square is at O.
The charge placed on the centre is `muC`
`AB=BC=CD=DA=10 cm, AC=sqrt(2)xx10=10sqrt(2)` cm
`AO=BO=CO=DO=(10sqrt(2))/2=5sqrt(2)` cm
Force between q and `q_(A)` :
`F_(A)=1/(4pi epsi_(0)) (q q_(A))/((OA)^(2))=(9xx10^(9)xx1xx10^(-6)xx2xx10^(-6))/((5sqrt(2)xx10^(-2))^(2))=(9xx2xx10^(-3))/(25xx10^(-4))`
`:. F_(A)=3.6 N` (direction towards O to C)
Force between q and `q_(C)` :
`F_(C)=1/(4pi epsi_(0)) (q q_(C))/((OC)^(2))=(9xx10^(9)xx1xx10^(-6) xx2xx10^(-6))/((5sqrt(2)xx10^(-2))^(2))=(9xx2xx10^(-3))/(25xx2xx10^(-4))`
`:. F_(C)=3.6 N` (direction towards O to A)
Force between q and `q_(B)` :
`F_(B)=1/(4pi epsi_(0)) (q q_(B))/((OB)^(2))=(9xx10^(9)xx1xx10^(-6)xx2xx10^(-6))/((5sqrt(2)xx10^(-2))^(2))`
`:. F_(B)=3.6 N` (direction towards O to B)
Force between q and `q_(0)` :
`F_(D)=1/(4pi epsi_(0))(q q_(D))/((OD)^(2))=(9xx10^(9)xx1xx10^(-6)xx2xx10^(-6))/((5sqrt(2)xx10^(-2))^(2))`
`:. F_(D)=3.6 N` (direction towards O to D)
Here we observe that, `F_(A)=-F_(C)` and `F_(D)=-F_(B)`
`:.` The net resultant force on `1 muC` is
`F=F_(A)+F_(B)+F_(C)+F_(D)`
`=-F_(C)+F_(B)+F_(C)-F_(B)`
`=0`.