An electric field $\overrightarrow{E}=200,000\hat{i}N/C$causes the point charge in FIGURE P22.68 to hang at an angle. What is $\theta$?

Electric field $=\overrightarrow{E}=200,000\hat{i}N/C$

Mass of the charge $=m=2.0g=2.0\times {10}^{-3}kg$

Amount of charge$=25nC=25\times {10}^{-9}C$

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,

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

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

$$\begin{gathered} \tan \theta =\frac{qE}{mg} \\ \tan \theta =\frac{25\times {10}^{-9}\times 200000}{2.0\times {10}^{-3}\times 9.81} \\ \tan \theta =0.2548 \\ \theta =14.29° \end{gathered}$$

Hence, the required angle is$14.29°$.