In Fig. 23-49, a small, non-conducting ball of mass$m=1.0mg$and charge $q=2.0\times {10}^{-8}C$ (distributed uniformly through its volume) hangs from an insulating thread that makes an angle $\mathrm{\theta }={30}^{\mathrm{o}}$with a vertical, uniformly charged non-conducting sheet (shown in cross-section). Considering the gravitational force on the ball and assuming the sheet extends far vertically and into and out of the page, calculate the surface charge density s of the sheet.

1. Mass of the ball,$1mg$
2. Charge on the ball,role="math" localid="1657350162514" $q=2\times {10}^{-8}C$
3. The angle of the insulation thread,role="math" localid="1657350181635" $\mathrm{\theta }={30}^{\mathrm{O}}$

Using the electrostatic force value in the equations of the horizontal and vertical components of the free body diagram, we can get the expression of the electric field. Again, using the value of the electric field of a non-conducting sheet in this calculated expression, we can get the value of the required surface charge density.

Formula:

The electrostatic force of a charged particle,$F=qE$ (1)

The force due to the gravity, $F=mg$ (2)

The electric field of a non-conducting shell,$E=\frac{\sigma }{2{\epsilon }_0}$ (3)

The forces acting on the ball are shown in the diagram to the right. The electric field produced by the plate is normal to the plate and points to the right. Since the ball is positively charged, the electric force on it also points to the right. The tension in the thread makes the angle with the vertical. Since the ball is in equilibrium the net force on it vanishes.

The sum of the horizontal components from the free body diagram using equation (i) for force yields

$qE-T\mathrm{sin\theta }=0........................\left(4\right)$

The sum of the vertical components yields,

$T\mathrm{cos\theta }-mg=0.................\left(5\right)$

The expression of the tension from equation (4) is given as:

$T=\frac{qE}{\mathrm{sin\theta }}$

This value is substituted into equation (5) to get the following value:

$qE=\mathrm{T}\mathrm{tan\theta }$

As the tension of the body is equal to the body weight, so using equation (2) and equation (3), the surface density of the sheet is given as follows:

$$\begin{gathered} \frac{q\sigma }{2{\epsilon }_o}=mg\mathrm{tan\theta } \\ \sigma =\frac{2{\epsilon }_{o}mg\mathrm{tan\theta }}{q} \\ \sigma =\frac{298.85\times {10}^{-12}{C}^2/N.m^2\left)\right(1.0\times {10}^{-6}kg\left)\right(9.8,m/s^2)\tan {30}^o}{2.0\times {10}^{-8}C} \\ \sigma =5.0\times {10}^{-9}C/m^2 \end{gathered}$$

Hence, the value of the surface charge density is $5.0\times {10}^{-9}C/M^2$ .