Two charges of - 4 μC and + 4 μC are placed at the points A ( 1 , 0 , 4 ) m and B ( 2 , - 1 , 5 ) m located…

Given the electric field is

$\begin{array}{rcl}\overrightarrow{E}& =& 0.20 \hat{\mathrm{i}} \frac{\mathrm{V}}{\mathrm{cm}}\\ & =& 20\hat{\mathrm{i}} \frac{\mathrm{V}}{\mathrm{m}}\end{array}$

The formula for calculating the torque is

$\overrightarrow{\tau }=\overrightarrow{p}\times \overrightarrow{E} ...\left(1\right)$

where,

$\begin{array}{rcl}\overrightarrow{p}& =& q\overrightarrow{l}\\ & =& 4\times \left[\left(2\hat{\mathrm{i}}+\left(-1\right)\hat{\mathrm{j}}+5\hat{\mathrm{k}}\right)-\left(1\hat{\mathrm{i}}+0\hat{\mathrm{j}}+4\hat{\mathrm{k}}\right)\right]\\ & =& 4\left(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\right) \mathrm{\mu C} \mathrm{m}\end{array}$

Thus, from equation (1), it follows that

$$\begin{gathered} \overrightarrow{\tau }=(4\hat{\mathrm{i}}-4\hat{\mathrm{j}}+4\hat{\mathrm{k}})\times {10}^{-6}\times \left(20\hat{\mathrm{i}}\right) \mathrm{N} \mathrm{m} \\ =(8\hat{\mathrm{k}}+8\hat{\mathrm{j}})\times {10}^{-5} \mathrm{N} \mathrm{m} \\ \Rightarrow \left|\overrightarrow{\tau }\right|=\sqrt{8^2+8^2}\times {10}^{-5} \mathrm{N} \mathrm{m} \\ =8\sqrt{2}\times {10}^{-5} \mathrm{N} \mathrm{m} \end{gathered}$$

Hence, $\alpha =2$.