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2. Electric Potential and Capacitance
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Equipotential surfaces are shown in figure. Then the electric field strength will be
A$100 \,Vm^{-1}$ along $X$-axis
B$100 \,Vm^{-1}$ along $Y$-axis
C$200 \,Vm^{-1}$ at an angle $120°$ with $X$-axis
D$50 \,Vm^{-1}$ at an angle $120°$ with $X$-axis
Solution
(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.
Standard 12
Physics
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