Question

The magnitude and direction of the electric field at point $P$ can be best represented by

• A. $\frac{3kq}{x^2}(\hat{i}+\hat{j})$
• B. $\frac{\sqrt{2}kq}{x^2}(\hat{i}+\hat{j})$
• C. $\frac{3\sqrt{2}kq}{x^2}(-\hat{i}-\hat{j})$
• D. $\frac{3\sqrt{2}kq}{x^2}(\hat{i}+\hat{j})$

Answer 1

Answer: (D)

${\overrightarrow{E}}_P=E\prime cos45^{\circ }\hat{i}+E\prime sin45^{\circ }\hat{j}$
$=E\prime \left(\frac{\hat{i}}{\sqrt{2}}+\frac{\hat{j}}{\sqrt{2}}\right)=\frac{k3q}{{\left(\frac{x}{\sqrt{2}}\right)}^2}\left(\frac{\hat{i}}{\sqrt{2}}+\frac{\hat{j}}{\sqrt{2}}\right)$
$=\frac{6kq}{x^2}\left(\frac{\hat{i}}{\sqrt{2}}+\frac{\hat{j}}{\sqrt{2}}\right)=\frac{3\sqrt{2}kq}{x^2}(\hat{i}+\hat{j})$