Potential in x-y plane is given as V = 5(x^2 + xy), volts . electric field at the point (1, -2)

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1. Electric Charges and Fields

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Potential in the $x-y$ plane is given as $V = 5(x^2 + xy)\, volts$. The electric field at the point $(1, -2)$ will be

A

$3j\, V/m$

B

$-5j\, V/m$

C

$5j \,V/m$

D

$-3j \,V/m$

Solution

$\mathrm{E}_{\mathrm{x}}=-\frac{\partial \mathrm{V}}{\partial \mathrm{x}}=-(10 \mathrm{x}+5 \mathrm{y})=-10+10=0$

$\mathrm{E}_{\mathrm{y}}=-\frac{\partial \mathrm{V}}{\partial \mathrm{x}}=-5 \mathrm{x}=-5$

$\therefore \overrightarrow{\mathrm{E}}=-5 \hat{\mathrm{j}} \mathrm{\,V} / \mathrm{m}.$

Standard 12

Physics

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