The proton charges `+4q` and `+q` are at a distance 3m apart. At what point between the charges, a third charge `+q` must be placed tokeep it in equilibrium ?

Explanation

Remember if `Q_(1)` and`Q _(2)` are of same nature ( means both positive or 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 the charge which is lesser in magnitude.

Here you can see `Q_(1)` and `Q_(2)` are of same nature so third charge be kept in between ata distance x from `Q_(1)` . Hence q will be at a distance ( 3-x) from `Q_(2)`. Since Q is in equlibrium, so net force on it must be zero. You can see , the forces applied by `Q_(1)` and `Q_(2)` on q are in opposite direction, so just balanacetheir magnitude .
Forceon 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` ( aftercross multiplication )or `, x=2` m. So q will be placed at a distance 2m from `Q_(1)` and at 1m 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 changes.