Question

Question #54e33

Answer 1

We draw a free-body diagram for one of the charged balls, say left. This is shown in the figure below.

All the forces acting on the charged ball are in equilibrium.

One force is tension `T` in the silk thread. The other force is weight `mg` of ball acting downward and third is electrostatic force of repulsion `F_"elec"` between the two charged balls.

Equating the magnitude of vertical component of tension with weight of the ball we get
`Tcostheta=mg` .....(1)
and equating the magnitude of horizontal component of tension with the magnitude of electrostatic force we get
`Tsintheta=F_"elec"=k_eq^2/x^2` .....(2)
where `x` is the distance between the two charged balls.

Dividing (2) by (1) we get
`tan theta=k_eq^2/x^2/ mg` ....(3)

It is given that the angle between the silk threads is small (`2theta` in our figure). We know that for small angles `theta`

`tanthetaapproxsintheta` ....(4)

From the geometry of the figure we see that
`sin theta=(x/2)/l=x/(2l)` .....(5)
Using (4) and (5) equation (3) reduces to
`x/(2l)~~k_eq^2/(x^2 mg)`

Substituting the value of Coulomb's constant `k_e` in terms of electric permittivity of free space `epsilon_@`and solving for `x` we get
`x^3=1/(4pi epsilon_@)xx2lq^2/( mg)`
`=>x=((q^2l)/(2pi epsilon_@mg))^(1/3)`