Suppose a uniformly charged wall provides a uniform electric field of $$2 \times 10^4 \text{ N C}^{-1}$$ normally. A charged particle of mass 2 g being suspended through a silk thread of length 20 cm and remain stayed at a distance of 10 cm from the wall. Then the charge on the particle will be $$\frac{1}{\sqrt{x}} \mu C$$ where $$x$$ = ______ [use $$g = 10 \text{ m s}^{-2}$$]

Given

$$E=2\times10^4N/C$$

$$m=2g=2\times10^{-3}kg$$

l=20cm

Particle stays 10 cm from wall, so horizontal displacement is

x=10cm

From geometry,

$$\sin\theta=\frac{10}{20}=\frac{1}{2}$$

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

For equilibrium,

$$T\sin\theta=qE$$

$$T\cos\theta=mg$$

Dividing,

$$\tan\theta=\frac{qE}{mg}$$

So

$$q=\frac{mg\tan\theta}{E}$$

Substitute,

$$q=\frac{(2\times10^{-3})(10)\cdot\frac{1}{\sqrt{3}}}{2\times10^4}$$

$$=\frac{10^{-6}}{\sqrt{3}}C$$

$$=\frac{1}{\sqrt{3}}\mu C$$