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Question: Answered & Verified by Expert
A point charge $q$ is placed at origin. Let $\mathbf{E}_{A^{\prime}} \mathbf{E}_B$ and $\mathbf{E}_C$ 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. $\mathbf{E}_A \perp \mathbf{E}_B$
2. $\mathbf{E}_A \| \mathbf{E}_C$
3. $\left|\mathbf{E}_B\right|=4\left|\mathbf{E}_C\right|$
4. $\left|\mathbf{E}_B\right|=8\left|\mathbf{E}_C\right|$
PhysicsElectrostaticsAP EAMCETAP EAMCET 2018 (22 Apr Shift 2)
Options:
  • A 1 , 4 are correct
  • B 2, 4 are correct
  • C 1, 3 are correct
  • D 2, 3 are correct
Solution:
1433 Upvotes Verified Answer
The correct answer is: 1, 3 are correct
In vector form, we write expressions of field at $A, B$ and $C$
$$
\begin{aligned}
& \mathbf{E}_A=\left\{\frac{k q}{\left(1^2+2^2+3^2\right)^{3 / 2}}\right\}(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \\
& \mathbf{E}_B=\frac{k q}{\left(1^2+1^2+(-1)^2\right)^{3 / 2}} \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}) \\
& \mathbf{E}_C=\frac{k q}{\left(2^2+2^2+2^2\right)^{3 / 2}} \cdot(2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})
\end{aligned}
$$
So,
$$
\begin{aligned}
& \mathbf{E}_A=\frac{k q}{14^{3 / 2}}(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \\
& \mathbf{E}_B=\frac{k q}{3^{3 / 2}}(\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}) \\
& \mathbf{E}_C=\frac{k q}{12^{3 / 2}}(2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})
\end{aligned}
$$
As, $\mathbf{E}_A \perp \mathbf{E}_B=0$ and $\mathbf{E}_B=4\left|\mathbf{E}_C\right|$.

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