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2. Electric Potential and Capacitance
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The variation of potential with distance $R$ from a fixed point is as shown below. The electric field at $R = 5\,m$ is......$volt/m$

A
$2.5$
B
$ - 2.5$
C
$2/5$
D
$ - 2/5$
Solution

(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 $E = 0$.
Standard 12
Physics