Relation between Electric Field Intensity and Potential (or $E = -\frac{dV}{dr}$)}
Show that electric field intensity is given by the negative gradient of electric potential. What does the negative sign indicate in this expression?
Consider two points A and B in the electric field $\vec{E}$ due to a point charge $+Q$ placed at O. Assume that points A and B are very close to each other so that electric field intensity between A and B is uniform.
Let $q_0$ be the positive test charge placed at A. The force acting on charge $q_0$ in the electric field $\vec{E}$ is given by $\vec{F} = q_0 \vec{E}$. The direction of force $\vec{F}$ is along the direction of $\vec{E}$ i.e., away from the charge $+Q$.
Work done to move the test charge from point A to point B through distance $dr$ is given by
$dW= \vec{F} \cdot d\vec{r}
$(\because \vec{F} = q_0 \vec{E})$
$dW= q_0 \vec{E} \cdot d\vec{r}$
(\because \cos 180^\circ = -1)
$dW= q_0 E dr \cos 180^\circ
$dW = (-) q_0 E dr$
or
$ \frac{dW}{q_0} = -E dr$
According to the definition of potential difference,
$V = \frac{dW}{q_0}$
$\therefore$ eqn. (i) becomes
$dV = \frac{dW}{q_0}$
or
$dV = -E dr$
or
$E = -\frac{dV}{dr}$...(ii)
The quantity $\left(\frac{dV}{dr}\right)$ is called the electric potential gradient (the rate of change of potential with distance).
The negative sign shows that electric field intensity is in the direction of decreasing electric potential gradient.
Thus, we conclude that
1. electric field is in the direction in which the electric potential decreases.
2. the magnitude of electric field is equal to the electric potential gradient.
3. SI unit of potential gradient is volt/metre (V m$^{-1}$), which is also the unit of electric field.
4. The relation $E = -dV/dr$ is valid for non-uniform field also.
5. The SI unit of electric field intensity is same as that of potential gradient i.e. V m$^{-1}$.
$1~\text{V m}^{-1} = 1~\text{N C}^{-1}$
6. Electric potential is a scalar quantity, while electric potential gradient is a vector quantity.