Two charges `+4e` and `+e` are at a distance `x` apart. At what distance,a charge `q` must be placed from charge `+e` so that is in equilibrium

Explanation

Remember if `Q_(1)` and `Q_(2)` are of same nature (means both positive or both negative) then the third charge should be put between (not necessarily at mid-point) `Q_(1)` and `Q_(2)` on the straight line joining `Q_(1)` and `Q_(2)`. But if `Q_(1)` and `Q_(2)` are of opposite nature, then the third charge will be put outside and close to that charge which is lesser in magnitude.

Here you can see `Q_(1)` and `Q_(2)` are of same nature so third charge q will be kept in between at a distance x from `Q_(1)` (as shown in figure). Hence, q will be at a distance `(3-x)` from `Q_(2)`. Since q is in equilibrium, so net force on it must be zero. You can see, the forces applied by `Q_(1)` and `Q_(2)` on q are in oppsite direction, so just balance their magnitude.
Force on q by `Q_(1)=(kQ_(1)q)/(x^(2))` and that by `Q_(2)=(kQ_(2)q)/((3-x)^(2))`
Now, `(kQ_(1)q)/(x^(2))=(kQ_(2)q)/((3-x)^(2))" or "(Q_(1))/(x^(2))=(Q_(2))/((3-x)^(2))" or "(4)/(x^(2))=(1)/((3-x)^(2))`
Take the square root: `(2)/(x)=(1)/((3-x))`
Or, `6-2x=x` (after cross multiplication) or, x = 2 m. So, q will be placed at a distance 2 m from `Q_(1)` and at 1 m from `Q_(2)`.
If `Q_(1)Q_(2)gt 0`
`x_(1)=(r_(0))/(sqrt((q_(2))/(q_(1)))+1)" "(x_(1)" is distance from "q_(1)" between "q_(1)" and "q_(2))`
If `Q_(1)Q_(2)lt 0`
`x_(1)=(r_(0))/(sqrt((q_(2))/(q_(1)))-1)`
`x_(1)` in this case is not between the charges.