The point charges `q_(A)=3 muC` and `q_(B)=-3 muC` are located 20 cm apart in vaccum.
What is the electric field at the midpoint O of the line AB joining the two charges?

Explanation

Given `q_(A)=3 muC=3xx10^(-6) C, q_(B)=-3 muC=-3xx10^(-6) C`
From fog. `AO=OB=10` cm `=0.1 m`
Electric field at midpoint 'O' due to `q_(A)`
at A is `E_(A)=1/(4pi epsi_(0)) q_(A)/((AO)^(2))=(9xx10^(9)xx3xx10^(-6))/((0.1)^(2))`
`=(27xx10^(3))/(0.1xx0.1)=2.7xx10^(-6) N//C`
The direction of `E_(A)` is A to O.
Electric field at midpoint 'O' due to `q_(B)` at B is
`E_(B)=1/(4pi epsi_(0)) (qB)/((OB)^(2))=(9xx10^(9)xx3xx10^(-6))/((0.1)^(2))=(27xx10^(3))/(0.1xx0.1)=2.7xx10^(6) N//C`
The direction of `E_(B)` is O to B.
Now we see that `E_(A)` and `E_(B)` are in same direction. So, the resultant electric field at O is E. Hence,
`E=E_(A)+E_(B)=2.7xx10^(6)+2.7xx10^(6)=5.4xx10^(6)N//C`
The direction of E will be from O to B or toward B.