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Some equip-potential surfaces are shown in figures. What can you say about the magnitude and direction of the electric field

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Electric field is perpendicular to the equipotential surface and in the direction of decreasing potential
The electric field will be at angle making an angle $120^{\circ}$ with x-axis.
Magnitude of electric field $E \cos 120^{\circ}=-\large\frac{(20-10)}{(20-10) \times 10^{-2}}-$$E . \large\frac{1}{2} =-\large\frac{-10}{0.10}$
=> $E= 200 V/m$
Direction of electric field will be radially outward, similar to a point charge kept at the centre, i.e $V= \large\frac{kq}{r}$
When $V=60 V=\large\frac{kq}{(0.1)}$
Hence potential at any distance from the center.
$V(r) =\large\frac{6}{r}$
Hence $E= -\large\frac{dV}{dr} = \bigg( \large\frac{6}{r^2} \bigg) $$V/m$
answered Jun 8, 2014 by meena.p

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