Question

A point charge $$q$$ is placed at origin. Let ${\overrightarrow{E}}_A$ , ${\overrightarrow{E}}_B$ and ${\overrightarrow{E}}_C$ be the electric field at three points $A(1,2,3)$, $B(1,1,-1)$ and $C(2,2,2)$ due to charge $$q$$ . Consider the following statement and choose the correct alternative.
(i). ${\overrightarrow{E}}_A\mathrm{\perp }{\overrightarrow{E}}_B$
(ii). $|{\overrightarrow{E}}_B|=4|{\overrightarrow{E}}_C|$
(a). Only [i] is correct
(b). Only [ii] is correct
(c). Both [i] and [ii] is correct
(d). Both [i] and [ii] are wrong

Answer 1

- Hint: You can start by defining coulomb’s force and electric field. Then use the equation $\overrightarrow{E}={\displaystyle \frac{Kq}{|\overrightarrow{r}{|}^2}}\times {\displaystyle \frac{{\overrightarrow{r}}_{}}{|r|}}$ to find the electric field at points $$A$$ , $$B$$ and $$C$$ . Then find out if $${\overrightarrow{E}}_A\mathrm{\perp }{\overrightarrow{E}}_B$$ by calculating if $${\overrightarrow{E}}_A.{\overrightarrow{E}}_B=0$$ . Then compare the values of $|{\overrightarrow{E}}_B|$ and $$|{\overrightarrow{E}}_C|$$ to find out if $|{\overrightarrow{E}}_B|=4|{\overrightarrow{E}}_C|$. Use this method to reach the solution.

Complete step-by-step answer:
The force that a charge experiences in the presence of another charge is known as electrostatic force. This force is also known as Coulomb force, after the name of its discoverer Charles-Augustin de Coulomb.
The charge experiences Coulomb’s force due to the formation of an electric field around the other charge. Imagine a fisherman entrapping a fish in its net and pulling it, this is somewhat how electric fields trap a charge.

The electric field at any point is given by
$\overrightarrow{E}={\displaystyle \frac{Kq}{|\overrightarrow{r}{|}^2}}\times {\displaystyle \frac{{\overrightarrow{r}}_{}}{|r|}}$
The electric field on point $A$ is
${\overrightarrow{E}}_A={\displaystyle \frac{Kq}{|{\overrightarrow{r}}_{OA}{|}^2}}\times {\displaystyle \frac{{\overrightarrow{r}}_{OA}}{|r_{OA}|}}$
$\Rightarrow {\overrightarrow{E}}_A={\displaystyle \frac{Kq}{{(\sqrt{14})}^2}}\times {\displaystyle \frac{(\hat{i}+2\hat{j}+3\hat{k})}{\sqrt{14}}}$
$\Rightarrow {\overrightarrow{E}}_A={\displaystyle \frac{Kq(\hat{i}+2\hat{j}+3\hat{k})}{14\sqrt{14}}}$
The electric field at point $B$ is
${\overrightarrow{E}}_B={\displaystyle \frac{Kq}{|{\overrightarrow{r}}_{OB}{|}^2}}\times {\displaystyle \frac{{\overrightarrow{r}}_{OB}}{|r_{OB}|}}$

$\Rightarrow {\overrightarrow{E}}_B={\displaystyle \frac{Kq}{{(\sqrt{3})}^2}}\times {\displaystyle \frac{(\hat{i}+\hat{j}-\hat{k})}{\sqrt{3}}}$
$\Rightarrow {\overrightarrow{E}}_B={\displaystyle \frac{Kq(\hat{i}+\hat{j}-\hat{k})}{3\sqrt{3}}}$
The electric field at point $C$ is
${\overrightarrow{E}}_C={\displaystyle \frac{Kq}{|{\overrightarrow{r}}_{OC}{|}^2}}\times {\displaystyle \frac{{\overrightarrow{r}}_{OC}}{|r_{OC}|}}$
$\Rightarrow {\overrightarrow{E}}_C={\displaystyle \frac{Kq}{{(\sqrt{12})}^2}}\times {\displaystyle \frac{(2\hat{i}+2\hat{j}+2\hat{k})}{\sqrt{12}}}$
$\Rightarrow {\overrightarrow{E}}_C={\displaystyle \frac{Kq(\hat{i}+\hat{j}+\hat{k})}{12\sqrt{3}}}$
Let’s find out what the value of ${\overrightarrow{E}}_A.{\overrightarrow{E}}_B$ is
$${\overrightarrow{E}}_A.{\overrightarrow{E}}_B=\left({\displaystyle \frac{Kq}{14\sqrt{14}}}\right)\left({\displaystyle \frac{Kq}{3\sqrt{3}}}\right)(1\hat{i}+2\hat{j}+3\hat{k})(\hat{i}+\hat{j}-\hat{k})$$
$$\Rightarrow {\overrightarrow{E}}_A.{\overrightarrow{E}}_B=\left({\displaystyle \frac{Kq}{3\sqrt{3}}}\right)\left({\displaystyle \frac{Kq}{12\sqrt{3}}}\right)(1+2-3)$$
$$\Rightarrow {\overrightarrow{E}}_A.{\overrightarrow{E}}_B=0$$
$$\because {\overrightarrow{E}}_A.{\overrightarrow{E}}_B=0$$
$$\therefore {\overrightarrow{E}}_A\mathrm{\perp }{\overrightarrow{E}}_B$$
The magnitude of electric field at point $$A$$ and point $$B$$ are
$$|{\overrightarrow{E}}_B|={\displaystyle \frac{Kq}{3\sqrt{3}}}$$ (Equation 1)
And $$|{\overrightarrow{E}}_C|={\displaystyle \frac{Kq}{12\sqrt{3}}}$$ (Equation 2)
Dividing equation 1 by equation 2, we get
$${\displaystyle \frac{|{\overrightarrow{E}}_B|}{|{\overrightarrow{E}}_C|}}={\displaystyle \frac{{\displaystyle \frac{Kq}{3\sqrt{3}}}}{{\displaystyle \frac{Kq}{12\sqrt{3}}}}}$$
$${\displaystyle \frac{|{\overrightarrow{E}}_B|}{|{\overrightarrow{E}}_C|}}=4$$
$$|{\overrightarrow{E}}_B|=4|{\overrightarrow{E}}_C|$$
As we can see that both options i and ii are correct.
Hence, option (c). is the correct choice.

Note: In this type of problem, we view the electric field as a vector quantity. A vector quantity is a quantity that has both a magnitude and a direction. In most of the problems concerning the electric field, we mostly deal with the magnitude of the electric field, but in questions such as these we have to consider the direction of the electric field in all 3-dimensions and use the related equations i.e. $\overrightarrow{E}={\displaystyle \frac{Kq}{|\overrightarrow{r}{|}^2}}\times {\displaystyle \frac{{\overrightarrow{r}}_{}}{|r|}}$.