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(C) The electric field due to a point charge $q$ at position vector $\vec{r}$ is given by $\vec{E} = \frac{kq}{r^3} \vec{r}$.For point $A(1, 2, 3)$,$\

Vedclass · Accepted Answer

$1$ and $3$ are correct

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(C) The electric field due to a point charge $q$ at position vector $\vec{r}$ is given by $\vec{E} = \frac{kq}{r^3} \vec{r}$.For point $A(1, 2, 3)$,$\vec{r}_A = \hat{i} + 2\hat{j} + 3\hat{k}$,$r_A = \sqrt{1^2+2^2+3^2} = \sqrt{14}$. So,$\vec{E}_A = \frac{kq}{14^{3/2}}(\hat{i} + 2\hat{j} + 3\hat{k})$.For point $B(1, 1, -1)$,$\vec{r}_B = \hat{i} + \hat{j} - \hat{k}$,$r_B = \sqrt{1^2+1^2+(-1)^2} = \sqrt{3}$. So,$\vec{E}_B = \frac{kq}{3^{3/2}}(\hat{i} + \hat{j} - \hat{k})$.For point $C(2, 2, 2)$,$\vec{r}_C = 2\hat{i} + 2\hat{j} + 2\hat{k}$,$r_C = \sqrt{2^2+2^2+2^2} = \sqrt{12} = 2\sqrt{3}$. So,$\vec{E}_C = \frac{kq}{(12)^{3/2}}(2\hat{i} + 2\hat{j} + 2\hat{k}) = \frac{kq}{8 \cdot 3^{3/2}} \cdot 2(\hat{i} + \hat{j} + \hat{k}) = \frac{kq}{4 \cdot 3^{3/2}}(\hat{i} + \hat{j} + \hat{k})$.Check $1$: $\vec{E}_A \cdot \vec{E}_B \propto (1)(1) + (2)(1) + (3)(-1) = 1 + 2 - 3 = 0$. Thus,$E_A \perp E_B$ is correct.Check $3$: $|E_B| = \frac{kq}{3^{3/2}} \sqrt{1^2+1^2+(-1)^2} = \frac{kq}{3^{3/2}} \sqrt{3} = \frac{kq}{3}$.$|E_C| = \frac{kq}{4 \cdot 3^{3/2}} \sqrt{1^2+1^2+1^2} = \frac{kq \sqrt{3}}{4 \cdot 3^{3/2}} = \frac{kq}{4 \cdot 3} = \frac{kq}{12}$.Therefore,$|E_B| = 4|E_C|$ is correct.

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