Question

A point charge of 5.9 μC is placed at the origin (x1 = 0) of a coordinate system, and another charge of –2.7 μC is placed placed on the x-axis at x2 = 0.23 m (Coulomb's law)?

A: Where on the x-axis can a third charge be placed in meters so that the net force on it is zero?

B: What if both charges are positive; that is, what if the second charge is 2.7 μC?

Answer 1

(A) Let the third charge `q` be placed at `x_3`
Coulomb's Force on charge `q` due to charge at the origin

`F_(5.9->q)=k(5.9q)/x_3^2` .....(1)

Coulomb's Force on charge `q` due to charge at the `x_2`

`F_(-2.7->q)=k(-2.7q)/(x_3-0.23)^2` ......(2)

For the net force to be zero, magnitude of both the forces must be equal.

`|k(5.9q)/x_3^2|=|k(-2.7q)/(x_3-0.23)^2|`
`=>(5.9)/x_3^2=(2.7)/(x_3-0.23)^2`
`=>(5.9)(x_3-0.23)^2=(2.7)x_3^2`
`=>5.9x_3^2-2.714x_3+0.31211=2.7x_3^2`
`=>3.2x_3^2-2.714x_3+0.31211=0`

Solving this quadratic with the help of in-built graphics tool we get
two solutions as

`x_3=0.14\ m and 0.71\ m`

The direction of both forces must be opposite to each other.
`=>q` can either be a positive or a negative charge. And this condition is met only if the third charge is placed at `x_3=0.71\ m`

(B) In such a case (2) becomes

`F_(2.7->q)=k(2.7q)/(x_3-0.23)^2`

The solutions for `x_3` remain same. However, the requirement for the direction of both to be opposite would be met only if the third charge was placed between the two charges.

Therefore, only valid solution would be `x_3=0.14\ m`