The potential $V$ is varying with $x$ and $y$ as $V\, = \,\frac{1}{2}\,\left( {{y^2} - 4x} \right)\,volt.$ The field at ($1\,m, 1\,m$ ) is
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$E_{x}=\frac{-\partial V}{\partial x}=-\frac{1}{2}(-4)=2$
$\mathrm{E}_{\mathrm{y}}=\frac{-\partial \mathrm{V}}{\partial \mathrm{y}}=-\frac{1}{2}(2 \mathrm{y})=-\mathrm{y}$
$x=1, y=1$
$\mathrm{E}_{\mathrm{x}}=2, \mathrm{E}_{\mathrm{y}}=-1$
$\overrightarrow{\mathrm{E}}=2 \hat{i}-\hat{j}$
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