# Intutive relation between Electric Field and Potential

I get the mathematical relationship between Electric Field and Potential i.e.

$$\vec{E} = -\frac{dV}{dr}\hat{r}$$

I also get that what it means is, if we travel through an electric field along a straight line and measure $V$ as we go, rate of change of $V$ with distance that we observe, when changed in sign, is the component of $\vec{E}$ in that direction.

Is there any intuitive way of understanding it.?

As an idea?

As you ride down a hill on a bicycle the force accelerating you down the hill ("field strength") is larger when the hill has a larger gradient ("potential gradient") with the minus sign being present because the direction of the force is in the direction of decreasing height, the gradient is negative.

What is electric potential?

The electrostatic field is a conservative force field (actually force per unit charge). As you may know, the work done in bringing a charge from infinity to a point in such a field will be path-independent. It depends on the end points only. This helps us to write the electrostatic field as the negative gradient of a scalar function, called electrostatic potential:

$$\vec{E}=-\nabla V$$

where $\displaystyle{\nabla=\left(\frac{\partial}{\partial{x}}\hat{x}+\frac{\partial}{\partial{y}}\hat{y}+\frac{\partial}{\partial{z}}\hat{z}\right)}$ is the gradient operator. The gradient operator acting on a vector gives you the direction of maximum rate of change of the vector.

The direction of electric field and electric potential: The work done

Electrostatic potential $V$ is the work done by an external agent in moving a unit positive charge from infinity to a point in the electrostatic field of another charge, without acceleration. The work done is stored as the potential energy of the system.

The negative sign indicates that the work done is in the opposite direction of the electric field from a positive charge. If the electric field is due to a negative charge, the work is done by the field and is hence negative.

If a charge $Q$ is to be brought through a distance $d\vec{r}$ from infinity against the electric field $E$, then the work done by the agency is

\begin{align} W & = -\int_\infty^{r=a} \vec{F}.d\vec{r}\\ & = -Q\int_\infty^{r=a}\vec{E}.d\vec{r} \end{align}

Which is stored as the potential energy of the system. Now, the potential energy per unit charge is

\begin{align} \frac{W}{Q} & = -\int_\infty^{r=a}\vec{E}.d\vec{r}\\ & = V(a)-V(\infty) \end{align}

Usually the potential at infinity is taken as a reference point and is zero. Hence the potential at a point $a$ is

$$V(a)=-\int_\infty^{r=a}\vec{E}.d\vec{r}$$

and the inverse process, i.e., the electric field at that point can be derived from the potential as follows:

$$\vec{E}=-\left[\frac{dV}{d\vec{r}}\right]_{r=a}$$

This means that the electric field at a point is the negative gradient of the electric potential at that point.

...Is there any intuitive way of understanding it?

The negative sign emphasizes the direction that the electric field points in a direction from higher potential to lower potential. This is a known concept. When you connect a battery to a load, the electric field points in a direction from the higher potential (positive terminal) to lower potential (negative potential), which indicates the direction of current flow.