A point charge $q = - 8.0 n C$ is located at the origin. The electric field $(i n N C^{- 1})$ vector at the point $x = 1.2 m, y = - 1.6 m,$ as shown in figure, is

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$- 14.4 \hat{i} - 10.8 \hat{j}$

$- 10.8 \hat{i} + 14.4 \hat{j}$

$- 14.4 \hat{i} + 10.8 \hat{j}$

$- 10.8 \hat{i} - 14.4 \hat{j}$

answer is C.

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Detailed Solution

$\vec{E} = \frac{1}{4 π ε_{0}} \frac{q}{r^{2}} \hat{r}$

$\hat{r} = \frac{\vec{r}}{r} = \frac{(1.2 - 0) \hat{i} + (- 1.6 - 0) \hat{j}}{\sqrt{{(1.2)}^{2} + {(1.6)}^{2}}} = \frac{1}{2} (1.2 \hat{i} - 1.6 \hat{j})$

$\vec{E} = 9 \times 10^{9} \times \frac{(- 8 \times 10^{- 9})}{2^{2}} \times [\frac{1}{2} (1.2 \hat{i} - 1.6 \hat{j})]$

$= - 10.8 \hat{i} + 14.4 \hat{j}$

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