Electric potential at any point is $V = - 5 x + 3 y + \sqrt{15} z,$ then the magnitude of the electric field is

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$4 \sqrt{2}$

$5 \sqrt{2}$

$7$

$3 \sqrt{2}$

answer is D.

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Detailed Solution

$E_{x} = - \frac{d V}{d x} = - (- 5) = 5; E_{y} = - \frac{d V}{d y} = - 3$
and $E_{z} = - \frac{d V}{d z} = - \sqrt{15}$
$E_{n e t} = \sqrt{E_{x}^{2} + E_{y}^{2} + E_{z}^{2}} = \sqrt{{(5)}^{2} + {(- 3)}^{2} + {(- \sqrt{15})}^{2}} = 7$

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