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Q.

Electric potential at a point in space is given as $V = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$ The electric field magnitude at point (1, 1, 1) will be

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a

3 N/C

b

$2 \sqrt{3} N / C$

c

$\sqrt{3} N / C$

d

$3 \sqrt{3} N / C$

answer is B.

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Detailed Solution

$E_{x} = - \frac{\partial V}{\partial x} = \frac{1}{x^{2}} E_{y} = - \frac{\partial V}{\partial y} = \frac{1}{y^{2}}$

$E_{y} = - \frac{\partial V}{\partial z} = \frac{1}{z^{2}}$

${\vec{E}}_{z} = + \frac{1}{x^{2}} \hat{i} + \frac{1}{y^{2}} \hat{j} + \frac{1}{z^{2}} \hat{k}$

at (1, 1, 1)

$\vec{E} = + \hat{i} + \hat{j} + \hat{k} | \vec{E} | = \sqrt{3} N / C$

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Electric potential at a point in space is given as V=1x+1y+1z The electric field magnitude at point (1, 1, 1) will be