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Potential Gradient and Relation between Electric Field and Potential

Electric pote...

Question

Electric potential is given by $V = 6 x - 8 x y^{2} - 8 y + 6 y z - 4 z^{2}$
Then electric force acting on a 2C point charge placed at origin will be?

A

2 N

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B

6 N

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C

8 N

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D

20 N

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Solution

The correct option is D 20 N
$E_{x} = - \frac{d V}{d x} = - (6 - 8 y^{2}), E_{y} = - \frac{d V}{d y} = - (- 16 x y - 8 + 6 z), E_{z} = - \frac{d V}{d z} = - (6 y - 8 z)$
At origin $x = y = z = 0$ so, $E_{x} = - 6, E_{y} = 8$ and $E_{z} = 0$
$\Rightarrow E = \sqrt{E_{x}^{2} + E_{y}^{2}} = 10 N / C$
Hence force $F = Q E = 2 \times 10 = 20 N$

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