Derive the relation between electric field and electric potential… - Vidyadip

Question

Derive the relation between electric field and electric potential. Also write two important conclusions concerning this relation.

✓

Answer

→ As shown in the fig., two equipotential surfaces $A$ and $B$ are very close to each other. Magnitudes of electric potentials on them are V and $V +\delta V$ respectively.
→ Here, $\delta V$ is change in electric potential in the direction of electric field $\vec{E}$.
→ Point $P$ is present on surface $B$. And the perpendicular distance from surface A to point P is $\delta l$.
→ The amount of work done in taking a unit positive charge on the perpendicular line from surface $B$ to surface $A$ is equal to $|\overrightarrow{ E }|$ $\delta l$ . This work is equal to the electric potential difference between surfaces A and B , which is $V _{ A }- V _{ B ^{\prime}}$
$\therefore|\overrightarrow{ E }| \delta l=\Delta V = V _{ A }- V _{ B }$
$\therefore|\overrightarrow{ E }| \cdot \delta l= V -( V +\delta V )$
$=-\delta V$
$\therefore \quad|\overrightarrow{ E }|=-\frac{8 V}{\delta l}$
→ Here, $\delta V$ is negative, so taking $-\delta V$ in place of $\delta V$,
$
|\overrightarrow{E}|=\frac{\delta V}{8 l}
$

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