Equipotential surfaces are shown in figure. Then the electric field strength will be
Diffcult
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(c) Using $dV = - \overrightarrow E .d\overrightarrow r $
$⇒$ $\Delta V = - E.\Delta r\cos \theta $
$⇒$ $E = \frac{{ - \Delta V}}{{\Delta r\cos \theta }}$
$⇒$ $E = \frac{{ - (20 - 10)}}{{10 \times {{10}^{ - 2}}\cos 120^\circ }}$
$ = \frac{{ - 10}}{{10 \times {{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 $120°$ with $x-$axis.
$⇒$ $\Delta V = - E.\Delta r\cos \theta $
$⇒$ $E = \frac{{ - \Delta V}}{{\Delta r\cos \theta }}$
$⇒$ $E = \frac{{ - (20 - 10)}}{{10 \times {{10}^{ - 2}}\cos 120^\circ }}$
$ = \frac{{ - 10}}{{10 \times {{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 $120°$ with $x-$axis.
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