Two point charges `e_(1)` and `e_(2)` are separated by a distance `d` but the electric field is not `0` on any point joining the two points. What can be concluded from this?

Explanation

In the first case, let the distance of point P from `e_(1)` is x and from `e_(2)` is (d - x)

Electric field at P due to `e_(1), E_(1)=(e_(1))/(x^(2))`
Electric field at P due to `e_(2), E_(2)=(e_(2))/((d-x)^(2))`
If `e_(1)` and `e_(2)` are oppositely charged, then the directions of `E_(1)` and `E_(2)` at point P will be the same and their resultant field,
`E=(e_(1))/(x^(2))+(e_(2))/((d-x)^(2))`
In this case, E will never be zero.
In the second case, let the distance of point P from `e_(1)` is x and from `e_(2)` is (x - d)
In this case, `E_(1)=(e_(1))/(x^(2)) and E_(2)=(e_(2))/((x-d)^(2))`
If `e_(1) and e_(2)` are oppositely charged. `E_(1) and E_(2)` will act along the opposite directions.
`therefore` Resultant electirc field, `E= (e_(2))/((x-d)^(2))-(e_(1))/(x^(2))`
In this case, if `e_(1)=e_(2)` then `E_(1) and E_(2)` will always be unequal.
E will never be zero.
Therefore, if the electric field is not zero at any point, then `e_(1) and e_(2)` are equal and opposite in nature, i.e., `e_(1)= -e_(2)` .