Question
Derive the equation $($formula$)$ showing the relationship between electric field and electric potential.
✓
Answer
$\rightarrow$ 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.
$\rightarrow$ Here, $\delta V$ is change in electric potential in the direction of electric field $\vec{E}$.
$\rightarrow$ Point $P$ is present on surface $B$.
And the perpendicular distance from surface $A$ to point $P$ is $\delta l$.
$\rightarrow $ 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 $V_A-V_B$.
$\therefore|\overrightarrow{ E }| \delta l=\Delta V = V _{ A }- V _{ B }$
$\therefore|\overrightarrow{ E }| \cdot \delta l= V -( V +\delta V )$
$=-\delta V$
$\therefore \quad|\vec{E}|=-\frac{\delta V }{\delta l}$
$\rightarrow$ 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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