Question
Derive the equation (formula) showing the relationship between electric field and electric potential.
✓
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 $|\vec{E}| \delta l$. This work is equal to the electric potential difference between surfaces $A$ and $B$, which is $V_A-V_B$.
$\begin{array}{l}
\therefore|\overrightarrow{ E }| \delta l=\Delta V = V _{ A }- V _{ B } \\
\therefore|\overrightarrow{ E }| \cdot \delta l= V -( V +\delta V ) \\
=-\delta V \\
\therefore \quad|\vec{E}|=-\frac{\delta V }{\delta l} \\
\end{array}$
→Here, $\delta V$ is negative, so taking $-\delta V$ in place of $\delta V$,
$|\overrightarrow{ E }|=\frac{\delta V }{\delta l}$
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