Equipotential surfaces are shown in figure. T | vedclass

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Question

(c) Using $dV = - \overrightarrow E .d\overrightarrow r $
$&rArr;$ $\Delta V = - E.\Delta r\cos 	heta $
$&rArr;$ $E = \frac{{ - \Delta V}}{{\Delta

Vedclass · Accepted Answer

$200 \,Vm^{-1}$ at an angle $120&deg;$ with $X$-axis

Vedclass · Answer

(c) Using $dV = - \overrightarrow E .d\overrightarrow r $
$&rArr;$ $\Delta V = - E.\Delta r\cos 	heta $
$&rArr;$ $E = \frac{{ - \Delta V}}{{\Delta r\cos 	heta }}$
$&rArr;$ $E = \frac{{ - (20 - 10)}}{{10 	imes {{10}^{ - 2}}\cos 120^\circ }}$
$ = \frac{{ - 10}}{{10 	imes {{10}^{ - 2}}( - \sin 30^\circ )}} = \frac{{ - {{10}^2}}}{{ - 1/2}} = 200\,V/m$
Direction of $E$ be perpendicular to the equipotential surface i.e. at $120&deg;$ with $x-$axis.

Equipotential surfaces are shown in figure. Then the electric field strength will be

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