Question

A point charge $q$ is placed at origin. Let $E_{A^{\prime }}{E}_B$ and $E_C$ be the electric fields at three points $A$ $(1,2,3),B(1,1,-1)$ and $C(2,2,2)$ respectively due to the charge $q$. Then, the relation between them is
1. $E_A\perp E_B$
2. $E_A\Vert E_C$
3. $\left|E_B\right|=4\left|E_c\right|$
4. $\left|E_B\right|=8\left|E_c\right|$

• A. 1, 4 are correct
• B. 2, 4 are correct
• C. 1, 3 are correct
• D. 2, 3 are correct

Answer 1

Answer: (C)

In vector form, we write expressions of field at $A,B$ and $C$
$E_A=\left\{\frac{kq}{{\left(1^2+2^2+3^2\right)}^{3/2}}\right\}(\hat{i}+2\hat{j}+3\hat{k})$
$E_B=\frac{kq}{{\left(1^2+1^2+(-1{)}^2\right)}^{3/2}}\cdot (\hat{i}+\hat{j}-\hat{k})$
$E_C=\frac{kq}{{\left(2^2+2^2+2^2\right)}^{3/2}}\cdot (2\hat{i}+2\hat{j}+2\hat{k})$
So, $E_A=\frac{kq}{14^{3/2}}(\hat{i}+2\hat{j}+3\hat{k})$
$E_B=\frac{kq}{3^{3/2}}(\hat{i}+\hat{j}-\hat{k})$
$E_C=\frac{kq}{12^{3/2}}(2\hat{i}+2\hat{j}+2\hat{k})$
As, $E_A\perp E_B=0$ and $E_B=4\left|E_C\right|$