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The electric potential V at any point $(x, y, z)$ in space is given by $V = 4 x^{2}$ volt. The electric field at $(1, 0, 2) m$ in $V m^{- 1}$ is

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16, along positive Z-axis.

16, along negative X-axis

8, along positive X-axis

8, along negative X-axis

answer is A.

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Detailed Solution

$E = - \frac{d V}{d r} = - \frac{d}{d x} (4 x^{2}) = - 8 x$ $= - 8 (1) = - 8 V m^{- 1}$

Negative sign indicates $\vec{E}$ is along negative direction of X-axis.

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