The electric potential $V$ at any point $x, y, z$ (all the coordinates are in metres) in space is given by $V = 4 x^{2}$ volt. The electric field at the point $(1 m, 0, 2 m)$ in $V / m$ is:

8 along negative X-axis

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8 along positive X-axis

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16 along negative X-axis

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16 along positive X-axis

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Solution

The correct option is A 8 along negative X-axis
Given $V = 4 x^{2}$ volt
i.e, electric potential changes only along x-axis.

$∴ E_{x} = - \frac{d V}{d x} = - \frac{d}{d x} (4 x^{2}) = - 8 x$

Electric field at point $(1 m, 0, 2 m)$ is
$E_{x} = - 8 \times 1 = - 8 V / m$

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