Question

A point charge Q is placed at origin O. Let $\overrightarrow{E_A}$,$\overrightarrow{E_B}$ and $\overrightarrow{E_C}$ represent electric fields at A, B and C respectively. If coordination of A,B and C are respectively (1,2,3) m,(1,1,-1) m and (2,2,2) m then

• A. $\overrightarrow{E_A}\perp \overrightarrow{E_B}$
• B. $\overrightarrow{E_A}\parallel \overrightarrow{E_B}$
• C. $\left|\overrightarrow{E_B}\right|\parallel 4\left|\overrightarrow{E_C}\right|$
• D. $\left|\overrightarrow{E_B}\right|\parallel 8\left|\overrightarrow{E_C}\right|$

Answer 1

The correct option is C $\left|\overrightarrow{E_B}\right|\parallel 4\left|\overrightarrow{E_C}\right|$

$\overrightarrow{E_a}=\frac{h_E}{r{a}^3}=\frac{h_E}{1^2+2^2+3^2}(1\hat{i}+2\hat{j}+3\hat{k})\overrightarrow{E_a}=\frac{h_E}{{14}^{3/2}}(1\hat{i}+2\hat{j}+3\hat{k})\overrightarrow{E_b}=\frac{ha}{r_b^2}\overrightarrow{r_b}={\frac{h_E}{(1^2+1^2+1^2)}}^{3/2}(1\hat{i}-1\hat{j}+1\hat{k})=\frac{h_E}{3^{3/2}}(1\hat{i}-1\hat{j}+1\hat{k})\overrightarrow{E_c}=\frac{h_E}{r_c^2}\overrightarrow{r_c}=\frac{h_E}{(2^2+2^2+2^2{)}^{3/2}}(2\hat{i}+2\hat{j}+2\hat{k})$

$=\frac{h_E}{{12}^{3/2}}(2\hat{i}+2\hat{j}+2\hat{k})$

Now

$\overrightarrow{E_a}.\overrightarrow{E_b}=\frac{h_E}{{14}^{3/2}}(1\hat{i}+2\hat{j}+3\hat{k})=\frac{h_E}{3^{3/2}}(1\hat{i}-1\hat{j}+1\hat{k})\Rightarrow \overrightarrow{E_a}.\overrightarrow{E_b}=\left(\frac{h_E}{{14}^{3/2}}\right)\left(\frac{h_E}{3^{3/2}}\right)(1-2+3)\ne 0$

Thus$\overrightarrow{E_a}$ and $\overrightarrow{E_b}$ are perpendicular to each other

$|E_c|=\frac{h_E}{{12}^{3/2}}(2^2+2^2+2^2{)}^{1/2}=\frac{h_E}{{12}^{3/2}}(12{)}^{1/2}\Rightarrow |\overrightarrow{E_c}|=\frac{h_E}{12}\overrightarrow{|}{E}_b|=\frac{ha}{3^{3/2}}(1^2+1^2+1^2{)}^{1/2}=\frac{{}_E}{3}=4\times \frac{h_E}{12}=4|\overrightarrow{E_c}|$

So, $|\overrightarrow{E_b}|=4|E_c|$