Question: Q. 6. Two closely spaced equipotential surfaces $A$ and $B$ with potentials $V$ and $V+\delta V$, (where $\delta V$ is the change in $V$ ), are kept $\delta l$ distance apart as shown in the figure. Deduce the relation between the electric field and the potential gradient between them. Write the two important conclusions concerning the relation between the electric field and electric potentials.
U] [Delhi Comptt. I, II, III 2014]
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Solution:
Ans. Work done in moving a unit positive charge along distance $\delta l$
$$ \begin{aligned} \left|\mathrm{E}{l}\right| \delta l & =V{A}-V_{B} \ & =V-(V+\delta V)=-\delta V \end{aligned} $$
$$ \Rightarrow \quad\left|E_{l}\right|=-\frac{\delta \mathrm{V}}{\delta l} $$
(i) Electric field is in the direction in which the potential decreases the steepest.
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(ii) The magnitude of electric field is given by the change in the magnitude of potential per unit displacement, normal to the equipotential surface at the point.
[CBSE Marking Scheme 2014] 1/2
