Two point charges `2 muC` and `-2 muC` are placed at point A and B as shown in figure. Find out electric field intensity at points C and D. [All the distances are measured in meter].

Explanation

Electric field at point C (`E_(A), E_(B)` are magnitudes only and arrows represent directions).
Electric field due to positive charge is away from it while due to negative charge, it is towards the charge. It is clear that `E(B) gt E_(A)`.
`:. E_("Net")=(E_(B)-E_(A))` towards negative X-axis
`=(K(2muC))/((sqrt(2))^(2))-(K(2 muC))/((3sqrt(2))^(2))` towards negative X-axis `=8000 (-hat(i)) N//C`
Electric field at point D :
Since magnitude of charges are same and also `AD=BD`
So, `E_(A)=E_(B)`
Vertical components of `vec(E)_(A)` and `vec(E)_(B)` cancel each other while horizontal components are in the same direction.
So, `E_("net")=2E_(A) cos theta =(2.K (2 muC))/2^(2) cos 45^(@)`
`=(K xx 10^(-6))/sqrt(2)=9000/sqrt(2) hat(i) N//C`.