Question

A point charge $q$ is placed at the origin. Let ${\overrightarrow{E}}_A,{\overrightarrow{E}}_B$ and ${\overrightarrow{E}}_C$ be the electric fields at three points, $A(1,2,3)$, $B(1,1,–1)$ and $C(2,2,2)$ due to charge $q$. Then

$\left(i\right){\overrightarrow{E}}_A\perp {\overrightarrow{E}}_B$
$\left(ii\right)\left|\begin{array}{c}{\overrightarrow{E}}_B\end{array}\right|=4\left|\begin{array}{c}{\overrightarrow{E}}_C\end{array}\right|$

Select the correct alternative.

• A. Only $\left(i\right)$ is correct.
• B. Only $\left(ii\right)$ is correct.
• C. Both $\left(i\right)$ and $\left(ii\right)$ are correct.
• D. Both $\left(i\right)$ and $\left(ii\right)$ are wrong.

Answer 1

The correct option is C Both $\left(i\right)$ and $\left(ii\right)$ are correct.

Let $O$ be the origin, then, ${\overrightarrow{E}}_A$ will be along $OA$.
Similarly, ${\overrightarrow{E}}_B$ along $OB$.

So, $\overrightarrow{OA}=(1-0)\hat{i}+(2-0)\hat{j}+(3-0)\hat{k}=1\hat{i}+2\hat{j}+3\hat{k}$

Also, $\overrightarrow{OB}=1\hat{i}+1\hat{j}-1\hat{k}$

Now, $\overrightarrow{OA}.\overrightarrow{OB}=1.1+2.1+3\times -1=0$

Therefore, ${\overrightarrow{E}}_A$ is perpendicular to ${\overrightarrow{E}}_B$.

Further, $E_B=\frac{kq}{(1-0{)}^2+(1-0{)}^2+(-1-0{)}^2}=\frac{kq}{3}$

$E_C=\frac{kq}{(2-0{)}^2+(2-0{)}^2+(2-0{)}^2}=\frac{kq}{12}$

$\therefore \left|\begin{array}{c}{\overrightarrow{E}}_B\end{array}\right|=4\left|\begin{array}{c}{\overrightarrow{E}}_C\end{array}\right|$

Hence, both $\left(i\right)$ and $\left(ii\right)$ are correct.