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Electric potential in an electric field is given as $V = {Kr}^{- 1} .$ (K belong constant), if position vector $\vec{r} = 2 \hat{i} + 3 \hat{j} + 6 \hat{k}$ then electric field will be :

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$\frac{(2 \hat{i} + 3 \hat{j} + 6 \hat{k}) K}{243}$

$\frac{(2 \hat{i} + 3 \hat{j} + 6 \hat{k}) K}{343}$

$\frac{(3 \hat{i} + 2 \hat{j} + 6 \hat{k}) K}{243}$

$\frac{(6 \hat{i} + 2 \hat{j} + 3 \hat{k}) K}{343}$

answer is B.

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Detailed Solution

$V = \frac{1}{4 {πε}_{0}} \frac{q}{r} = k, here k = \frac{q}{4 {πε}_{0}}$

$\vec{E} = \frac{q \vec{r}}{4 {πε}_{0} r^{3}} = \frac{(2 \hat{i} + 3 \hat{j} + 6 \hat{k}) k}{{(2^{2} + 3^{2} + 6^{2})}^{3 / 2}}$

$\vec{E} = \frac{(2 \hat{i} + 3 \hat{j} + 6 \hat{k}) k}{343}$

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