(a) Two point charges `Q` and `16 Q` are fixed at separation `d`. Where should a third point charge be placed between them so that it experiences no force I.e. in equilibrium ?
(b) In the previous problem , if we replace `Q` by `- Q` , where should the third charge be placed on the joining the charges so that it is equilibrium ?
(c ) Three point charges `Q , q` and `Q` are placed at `x = 0 , x = d//2` and `x = d`. Find `q` in terms of `Q` so that all charges are in equilibrium.
(d) Two point charges `9 Q` and `25 Q` are placed at separation `d`. Where should a third charge `q` be placed between them so that all charges are in equilibrium ? Also find the value of `q` and its nature.

Explanation

(a) Let the third charge `q` be placed at distance `x` from `Q`.
Electric force on `q`:
Due to `Q , F_(1) = (1)/(4 pi in_(0)) (Qq)/(x^(2))` , towards right
Due to `16 Q , F_(2) = (1)/(4 pi in_(0)) (16 Qq)/(( d - x)^(2))` , towards left
Since net force on `q` is zero
`F_(1) = F_(2)`
`(1)/(x^(2)) = (16)/((d - x)^(2))`
Taking square root on both sides , we get
`(1)/(x) = (4)/((d - x)) rArr x = (d)/(5)`
Hence , the third charge be placed at distance `d//5` from `Q` or at `4d//5` from `16 Q`.
(b) The third charge `q` cannot be placed between the charges , because the force on `q` due to `- Q` and `16 Q` will be in same direction , hence net force will not be zero.
The third charge can be placed either left of `- Q` or right of `16 Q`. Let `q` is placed `x` left of `-Q`.
Force on `q`:
Due to `-Q , F_(1) = (1)/(4 pi in_(0)) (Qq)/(x^(2))` , towards right
Due to `16 Q , F_(2) = (1)/(4 pi in _(0)) (Qq)/(x^(2))` , towards left
`F_(1) = F_(2)`
`(1)/(x^(2)) = (16)/((d + x)^(2)) rArr (d +x)/(x) = 4 rArr x = (d)/(3)`
Let `q` be placed `x` right of `16 Q`.
`(-Q ,q) : F_(1) = (1)/(4 pi in_(0)) ( 16 Qq)/(x^(2))` , towards right
`F_(1) = F_(2)`
`(1)/((d +x)^(2)) = (16)/(x^(2)) rArr (x)/( d + x) = 4 rArr x = -(4d)/(3)`
i.e. the third charge be placed at `4d//3` left of `16 Q` or at `d//3` left of `-Q` . (c ) All charges are in equilibrium i.e. net force on each charge is zero. To find value of `q`, put net force on `Q` (at (`A` or `C`) equal to zero.
Force on `Q ("at" A)`
Due to `q , F_(1) = (1)/(4 pi in_(0)) (Qq)/((d//2)^(2))` , towards left
Due to `Q , F_(2) = (1)/( 4 pi in_(0)) (Qq)/(d^(2))` , towards left
Net force on `Q` is zero i.e.
`F_(1) + F_(2) = 0`
`4 Qq + Q^(2) = 0 rArr q = -(Q)/(4)`
(d) All charges are in equilibrium i.e. net force on each is equal to zero.
Force at `q`:
`(1)/( 4 pi in_(0)) ( 9 Qq)/(x^(2)) = (1)/(4 pi in_(0)) ( 25 Qq)/((d - x)^(2))`
`(9)/(x^(2)) = (25)/((d - x)^(2)) rArr 3(d - x) = 5x rArr x = (3d)/(8)`
Force at `9 Q`:
`(1)/(4 pi in_(0)) ( 9 Qq)/(x^(2)) + (1)/( 4 pi in_(0)) (9 Q . 25 Q)/(d^(2)) = 0`
`(q)/((3d//8)^(2)) = -( 25 Q)/(d^(2)) rArr q = -(225)/(64) Q`