An electric field $\overrightarrow{E}=100,000\hat{i}N/C$causes the 5.0 g point charge in FIGURE P22.67 to hang at a $20°$angle. What is the charge on the ball?

Electric field $=\overrightarrow{E}=100,000\hat{i}N/C$

Mass of the charge $=m=5.0g=5.0\times {10}^{-3}kg$

Angle$=\theta =20°$

Let us first draw the free body diagram of the charged sphere to understand the forces acting on it.

Shown below is the free body diagram of the charged sphere.

Since the system is at rest, the net force acting on the body is zero.

Let us resolve the forces in X and Y directions, and solve them separately.

Considering the vertical forces, we get,

$$\begin{gathered} T\cos \theta -F_g=0 \\ T\cos \theta =F_g.....\left(1\right) \end{gathered}$$

Considering the horizontal forces, we get,

$$\begin{gathered} F_e-T\sin \theta =0 \\ T\sin \theta =F_e.....\left(2\right) \end{gathered}$$

Dividing equation (2) by (1), we get,

$$\begin{gathered} \frac{T\sin \theta }{T\cos \theta }=\frac{F_e}{F_g} \\ \tan \theta =\frac{F_e}{F_g} \end{gathered}$$

Where,

localid="1648621647402" $F_e=qE=$Electrostatic force

localid="1648621737300" $F_g=mg=$force due to gravity or weight

By substituting the values in the above equation, we get,

$$\begin{gathered} \tan 20=\frac{qE}{mg} \\ q=\frac{\tan 20\times mg}{E} \\ q=\frac{0.364\times 5.0\times {10}^{-3}\times 9.81}{100000} \\ q=1.8\times {10}^{-7}C=0.18\mu C \end{gathered}$$

Hence, the charge on the ball is calculated to be$0.18\mu C$.