Question

At which point along the x axis will `q_2` experience no net force given `d=2 m` and relation between charges 1 and 3 is `3q_1=q_3`?

Answer 1

Let charge `q_1` be placed at origin. `:.q_3` is located at `2hati`.
Let `q_2` be at `xhati` to meet the required condition.

Force on `q_2` due to `q_1` using Coulombs' Law

`vecF_(q_1to q_2)=k_e(q_1q_2)/x^2hati`

Similarly force on `q_2` due to `q_3`

`vecF_(q_3to q_2)=-k_e(q_3q_2)/(d-x)^2hati`

For net force on `q_2=0`, sum of above two forces must be zero. We have

`vecF_(q_1to q_2)+vecF_(q_3to q_2)=0=k_e(q_1q_2)/x^2hati+(-k_e(q_3q_2)/(d-x)^2hati)`
`=>(q_1)/x^2=(q_3)/(d-x)^2`

Inserting value of charges `3q_1=q_3`

`(q_1)/x^2=(3q_1)/(d-x)^2`
`=>1/x^2=(3)/(d-x)^2`
`=>3x^2-(d-x)^2=0`
`=>(sqrt3x)^2-(d-x)^2=0`
`=>[sqrt3x+(d-x)][sqrt3x-(d-x)]=0`
`=>[d+x(sqrt3-1)][x(sqrt3+1)-d]=0`

Two roots are

`[d+x(sqrt3-1)]=0and [x(sqrt3+1)-d]=0`
`x=-d/(sqrt3-1)and x=d/(sqrt3+1)`

Rationalizing denominators we get

`x=-d/2(sqrt3+1)and x=d/2(sqrt3-1)`

Inserting given value of `d` we get

`x=-(sqrt3+1)\ m`
`x=(sqrt3-1)\ m`

Both solutions are valid if we consider only the magnitude of the net force. When we consider direction as well we find that only valid solution is when `q_2` is placed between the two charges `q_1and q_3`.

Hence, `x=(sqrt3-1)\ m`