Potential V is varying with x and y as V, = ,frac{1}{2},left( {{y^2} - 4x} right),volt. field

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2. Electric Potential and Capacitance

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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

A

$2\hat i\, + \,\hat j\,\,V/m$

B

$-2\hat i\, + \,\hat j\,\,V/m$

C

$2\hat i\, - \,\hat j\,\,V/m$

D

$-2\hat i\, + \,2\hat j\,\,V/m$

Solution

$E_{x}=\frac{-\partial V}{\partial x}=-\frac{1}{2}(-4)=2$

$\mathrm{E}_{\mathrm{y}}=\frac{-\partial \mathrm{V}}{\partial \mathrm{y}}=-\frac{1}{2}(2 \mathrm{y})=-\mathrm{y}$

$x=1, y=1$

$\mathrm{E}_{\mathrm{x}}=2, \mathrm{E}_{\mathrm{y}}=-1$

$\overrightarrow{\mathrm{E}}=2 \hat{i}-\hat{j}$

Standard 12

Physics

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