As shown in the figure, a configuration of two equal point charges $$q_0 = +2\mu C$$ is placed on an inclined plane. Mass of each point charge is 20 g. Assume that there is no friction between charge and plane. For the system of two point charges to be in equilibrium (at rest) the height $$h = x \times 10^{-3}$$ m. The value of x is _______
Take $$\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9$$ N m$$^2$$ C$$^{-2}$$, $$g = 10$$ m s$$^{-2}$$

For equilibrium along incline, electrostatic repulsion must balance component of weight along plane.

Inclination:

$$\theta=30^{\circ}$$

For each charge,

$$mg\sin⁡30^{\circ\ }$$

must equal Coulomb repulsion.

Mass

$$m=20g=0.02kg$$

So

$$mg\sin30^{\circ}=0.02(10)\cdot\frac{1}{2}$$

=0.1N

Electrostatic force:

$$F=\frac{kq_0^2}{r^2}$$

where r is separation between charges along incline.

Set equilibrium:

$$\frac{kq_0^2}{r^2}=0.1$$

Given

$$q_0=2\times10^{-6}C$$

and

$$k=9\times10^9$$

So

$$\frac{(9\times10^9)(4\times10^{-12})}{r^2}$$

$$\frac{36\times10^{-3}}{r^2}$$

$$r^2=0.36$$

$$r=0.6m$$

Now from geometry, the vertical height shown is

$$h=r\sin⁡30^{\circ\ }$$

$$=0.6\cdot\frac{1}{2}$$

=0.3m

$$=300\times10^{-3}m$$