Question

The magnitude and direction of electric field at point 'P' can be best represented by-

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

Answer 1

Answer: (D)

$\frac{3\sqrt{2}kq}{x^{2}}(\widehat{i} + \widehat{j})$

Answer 2

$\overrightarrow { \mathrm { E } } _ { P } = E ^ { \prime } \cos 45 ^ { \circ } \hat { \mathrm { i } } + E ^ { \prime } \sin 45 ^ { \circ } \hat { \mathrm { j } }$ = $E ^ { \prime } \left( \frac { \hat { \mathrm { i } } } { \sqrt { 2 } } + \frac { \hat { \mathrm { j } } } { \sqrt { 2 } } \right)$ =

$\frac { \mathrm { k } 3 \mathrm { q } } { \left( \frac { \mathrm { x } } { \sqrt { 2 } } \right) ^ { 2 } } \left( \frac { \hat { \mathrm { i } } } { \sqrt { 2 } } + \frac { \hat { \mathrm { j } } } { \sqrt { 2 } } \right)$= =