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Home  >>  AIMS  >>  Class12  >>  Physics  >>  Electric Charges and Fields
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Potential in the x-y plane is given as $V = 5(x^2 + xy)\; volts$. Find the electric field at the point (1, –2).

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$E_x=-\large\frac{dV}{dx} $$= -(10 x +5y)$
$\qquad= -10+10=0$
$E_r=-\large\frac{dV}{dx} $$= -5 x=-5$
$\overrightarrow{E}=-5 \hat j V/m$
answered Jun 8, 2014 by meena.p

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