Two point charges `q_(A)=3muC` and `q_(B)=-3muC` are located 20 cm apart in vacuum.
What is the electric field at the mid - point O of the line AB joining the two charges ?
First of all, calculate the electric fields at mid - point due to both charges and then find the resultant electric field by vector addition.

Explanation

Given, `AB=20cm`

`AO=OB=10cm=0.1m`
`q_(A)=3muC=3xx10^(-6)C`
`q_(B)=-3muC=-3xx10^(-6)C`
The electric field at a point due to a charge q,
`E=(1)/(4piepsilon_(0))*(q)/(r^(2))`
where, r is the distance between charge and the point.
electric field due to `q_(A)` at O is `E_(A)`
`E_(A)=(1)/(4piepsilon_(0))*(q_(A))/((AO)^(2))=(9xx10^(9)xx3xx10^(-6))/((0.1)^(2))`
`=(27xx10^(3))/((0.1)^(2))=(27xx10^(3))/(0.1xx0.1)=2.7xx10^(6)" N/C"`
The direction of `E_(A)` is from A to O, i.e. towards O or towards OB as the electric field is always directed away from positive charge.
electric field due to `q_(B)` at O is `E_(B)`
`E_(B)=(1)/(4piepsilon_(0))*(q_(B))/((OB)^(2))`
`E_(B)=(9xx10^(9)xx3xx10^(-6))/((0.1)^(2))=(27xx10^(3))/(0.1xx0.1)`
`=2.7xx10^(6)" N/C"`
The direction of `E_(B)` is from O to B, i.e. towards O or towards OB as the electric field is always directed towards the negative charge.
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 (resultant electric field) will be from O to be or towards B.