Potential V is varying with x and y as V = frac{1}{2}({y^2} - 4x),volts field at (1,m,,1

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1. Electric Charges and Fields

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The potential $V$ is varying with $x$ and $y$ as $V = \frac{1}{2}({y^2} - 4x)\,volts$ 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

$\mathrm{E}_{\mathrm{x}}=\frac{-\partial \mathrm{V}}{\partial \mathrm{x}}=-\frac{1}{2}(-4)=2$

${{\rm{E}}_{\rm{y}}} = \frac{{ – \partial {\rm{V}}}}{{\partial {\rm{y}}}} = – \frac{1}{2}(2{\rm{y}}) = – {\rm{y}}$

$x=1, y=1$

$E_{x}=2, E_{y}=-1$

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

Standard 12

Physics

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