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A charge of $10^{- 10} C$ is placed at the origin. The electric field at (1, 1) m due to it (in $N C^{- 1}$ ) is

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

$\frac{0.45}{\sqrt{2}} (\bar{i} + \bar{j})$

$20 \bar{j}$

$4.5 \sqrt{2} (\bar{i} + \bar{j})$

answer is B.

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Detailed Solution

$\vec{E} = \frac{kq}{r^{2}} \hat{r} = \frac{kq}{r^{3}} \vec{r}$

$\vec{r} = \hat{i} + \hat{j}$

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