A charge of 10–10C is placed at the origin. The electric field at (1, 1) m due to it (in NC

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A charge of 10^–10C is placed at the origin. The electric field at (1, 1) m due to it (in NC^–1) is

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$20 \bar{j}$

$\bar{i} + \bar{j}$

$\frac{0 .45}{\sqrt{2}} (\bar{i} + \bar{j})$

$4 .5 \sqrt{2} (\bar{i} + \bar{j})$

answer is B.

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Detailed Solution

$\begin{array}{l} q = 10^{- 10} c, \bar{r} = i + j \Rightarrow | r | = \sqrt{2}; \bar{E} = \frac{1}{4 {πε}_{0}} \cdot \frac{q}{r^{3}} \times \bar{r} \\ = 9 \times 10^{9} \times \frac{10^{- 10}}{2 \sqrt{2}} (i + j) = \frac{0 .45}{\sqrt{2}} (i + j) \end{array}$