Question 248 of 311

Two charges of $-4\mu C$ and $+4\mu C$ are placed at the points $A(1,0,4)m$ and $B(2,-1,5)m$ located in an electric field $\vec{E}=0.20\hat{i}V{cm}^{-1}$. The magnitude of the torque acting on the dipole is $8\sqrt{\alpha }\times {10}^{-5}Nm$, where $\alpha =$_________.

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Answer: 2

Given the electric field is
$\begin{matrix}\vec{E} & = & 0.20\hat{i}\dfrac{V}{cm} \\ & = & 20\hat{i}\dfrac{V}{m}\end{matrix}$
The formula for calculating the torque is
$\vec{\tau }=\vec{p}\times \vec{E}...\left(1\right)$
where,
$\begin{matrix}\vec{p} & = & q\vec{l} \\ & = & 4\times \left[\left(2\hat{i}+\left(-1\right)\hat{j}+5\hat{k}\right)-\left(1\hat{i}+0\hat{j}+4\hat{k}\right)\right] \\ & = & 4\left(\hat{i}-\hat{j}+\hat{k}\right)\mu Cm\end{matrix}$
Thus, from equation (1), it follows that
$\vec{\tau }=(4\hat{i}-4\hat{j}+4\hat{k})\times {10}^{-6}\times (20\hat{i})Nm =(8\hat{k}+8\hat{j})\times {10}^{-5}Nm \Rightarrow \left|\vec{\tau }\right|=\sqrt{{8}^{2}+{8}^{2}}\times {10}^{-5}Nm =8\sqrt{2}\times {10}^{-5}Nm$
Hence, $\alpha =2$.