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. $E _{A} \| E _{C}$ 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)

1, 3 are correct

Answer 2

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