Two charges of - 4μ C and + 4μ C are placed at the points A ( 1, 0,...

Given the electric field is

\begin{cases}\overset{\to}{E} & = & 0 . 20\overset{^}{i}\dfrac{V}{cm} \\ & = & 20\overset{^}{i}\dfrac{V}{m}\end{cases}

The formula for calculating the torque is

$\overset{\to}{τ} = \overset{\to}{p} \times \overset{\to}{E} . . . \left(1\right)$

where,

\begin{cases}\overset{\to}{p} & = & q\overset{\to}{l} \\ & = & 4 \times \left[\left(2\overset{^}{i} + \left(- 1\right)\overset{^}{j} + 5\overset{^}{k}\right) - \left(1\overset{^}{i} + 0\overset{^}{j} + 4\overset{^}{k}\right)\right] \\ & = & 4\left(\overset{^}{i} - \overset{^}{j} + \overset{^}{k}\right)\mu Cm\end{cases}

Thus, from equation (1), it follows that

\overset{\to}{τ} = ( 4\overset{^}{i} - 4\overset{^}{j} + 4\overset{^}{k} ) \times {10}^{- 6} \times ( 20\overset{^}{i} ) Nm = ( 8\overset{^}{k} + 8\overset{^}{j} ) \times {10}^{- 5}Nm ⇒ \left|\overset{\to}{τ}\right| = \sqrt{{8}^{2} + {8}^{2}} \times {10}^{- 5}Nm = 8\sqrt{2} \times {10}^{- 5}Nm

Hence, $\alpha = 2$.