The variation of potential with distance $R$ from a fixed point is as shown below. The electric field at $R = 5\,m$ is......$volt/m$
Medium
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(a)Intensity at $5\,m$ is same as at any point between $B$ and $C$ because the slope of $BC$ is same throughout (i.e., electric field between $B$ and $C$ is uniform). Therefore electric field at $R = 5$ $m$ is equal to the slope of line $BC$ hence by $E = \frac{{ - dV}}{{dr}}$;
$E = - \frac{{(0 - 5)}}{{6 - 4}} = 2.5\,\frac{V}{m}$
At $R = 1\,m$, $E = - \frac{{(5 - 0)}}{{(2 - 0)}} = - 2.5\,\frac{V}{m}$
and at $R = 3\,m$ potential is constant so $E = 0$.
$E = - \frac{{(0 - 5)}}{{6 - 4}} = 2.5\,\frac{V}{m}$
At $R = 1\,m$, $E = - \frac{{(5 - 0)}}{{(2 - 0)}} = - 2.5\,\frac{V}{m}$
and at $R = 3\,m$ potential is constant so $E = 0$.
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