A small bob of mass 100 mg and charge $$+10$$ $$\mu$$C is connected to an insulating string of length 1 m. It is brought near to an infinitely long non-conducting sheet of charge density '$$\sigma$$' as shown in figure. If string subtends an angle of 45° with sheet at equilibrium the charge density of sheet will be:

(Given, $$\epsilon_0 = 8.85 \times 10^{-12}$$ F/m and acceleration due to gravity, $$g = 10$$ m/s$$^2$$)

For an infinite non-conducting sheet, electric field is

$$E=\frac{\sigma}{2\varepsilon_0}$$

Force on charged bob due to field:

$$F_e=qE$$

$$=q\frac{\sigma}{2\varepsilon_0}$$

At equilibrium forces balance.

Weight downward:

mg

Electric force horizontal.

Tension along string.

Resolving tension:

$$T\cos45^{\circ}=mg$$

$$T\sin45^{\circ}=F_e$$

Dividing,

$$\tan45^{\circ}=\frac{F_e}{mg}$$

Since

$$\tan45^{\circ}=1$$

$$F_e=mg$$

So

$$q\frac{\sigma}{2\varepsilon_0}=mg$$

Thus

$$\sigma=\frac{2\varepsilon_0mg}{q}$$

Now

Mass:

$$100mg=100\times10^{-6}kg=10^{-4}kg$$

Charge:

$$q=10\mu C=10^{-5}C$$

Substitute:

σ

$$=\frac{2(8.85\times10^{-12})(10^{-4})(10)}{10^{-5}}$$

$$=17.7\times10^{-11}$$

$$=1.77\times10^{-10}\ \text{C/m}^2$$