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

Sol. As electric field E is related to potential V through the relation

$\large E = – \frac{dV}{dr} $

$\large E_x = – \frac{dV}{dx} = – \frac{d}{dx}(4 x^2)$

E_{x} = – 8 x

$\large E_y = – \frac{dV}{dy} = – \frac{d}{dx}(4 x^2)$

E_{y} = 0

Similarly , E_{z} = 0

$\large \vec{E} = E_x \hat{i} + E_y \hat{j} + E_z \hat{k} $

$\large \vec{E} = -8x \hat{i} $

i.e., it has magnitude 8 V/m and is directed along negative x-axis.

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