## Potential Gradient; Electric Field from Electric Potential

The work done by an electric field $dW$ on a test charge $q_0$ between two points in an electric field $\vec E$ through a displacement $d\vec l$ is $dW = {q_0}\vec E \cdot d\vec l $ and therefore,

\[dV = -\vec E \cdot d\vec l \tag{1} \label{1}\]

In terms of x, y and z-components of $\vec E$ and $d\vec l$, $\vec E = {E_x}\hat i + {E_y}\hat j + {E_z}\vec k $ and $d\vec l = dx\hat i + dy\hat j + dy\hat k$, and

\[\begin{align*} dV &= - ({E_x}\hat i + {E_y}\hat j + {E_z}\vec k ) \cdot (dx\hat i + dy\hat j + dy\hat k)\\ or,\quad dV &= - ({E_x}dx + {E_y}dy + {E_z}dz) \end{align*}\]

Now we use partial derivative of $dV$ with respect to $x$, $y$ and $z$ to find the corresponding component of the electric field. Here it is assumed that you know what the partial derivative is. The partial derivative of $dV$ with respect to $x$ is

\[{E_x} = - \frac{{\partial V}}{{dx}}\]

Similarly the y and z-components of the electric field are

\[\begin{array}{c} {E_y} = - \frac{{\partial V}}{{dy}}\\ {E_z} = - \frac{{\partial V}}{{dz}} \end{array}\]

In terms of vector notation the electric field vector in terms of its components is

\[\begin{align*} \vec E &= - \left( {\hat i\frac{{\partial V}}{{dx}} + \hat j\frac{{\partial V}}{{dy}} + \hat k\frac{{\partial V}}{{dz}}} \right) \tag{2} \label{2}\\ {\rm{or,}}\quad \vec E &= - \vec {\nabla {\kern 1pt} } V \tag{3} \label{3} \end{align*}\]

Here $\vec {\nabla \kern 1pt} $ is the gradient operator called *grad* which in this case represents the gradient of electric potential $V$. The gradient operator is $\vec {\nabla {\kern 1pt} } = \left( {\hat i\frac{\partial }{{dx}} + \hat j\frac{\partial }{{dy}} + \hat k\frac{\partial }{{dz}}} \right)$. The quantity $\vec \nabla V$ is called potential gradient.

It's easier to determine electric field from electric potential. The direction of electric field is in the direction of decreasing potential.

If the electric field is radial or the charge distribution has spherical symmetry which produces radial electric field, $\vec E \cdot d\vec l = Edl\cos \alpha = Edr$ and $dr$ is the radial displacement. And from Eq.\eqref{1},

\[E = - \frac{{dV}}{{dr}} \tag{4} \label{4}\]

## Equipotential Surfaces

The surface on which the potential is the same at every point on the surface is equipotential surface. Since the potential is the same at every point on the surface, the change in potential on an equipotential surface is zero. If an element charge moves through a small displacement $d\vec s$ on an equipotential surface within electric field $\vec E$, the change in potential is $dV = -\vec E \cdot d\vec s = 0$. Therefore, the electric field must be perpendicular to an equipotential surface.

The equipotential surfaces are drawn in Figure 1 and Figure 2 in a way that the potential differences between the adjacent pairs are the same. Note that the equipotential surfaces are three dimensional. The equipotential surfaces in Figure 2 are closer on the right side of positive charge or the left side of negative charge where the electric field is relatively stronger than on the right side of negative charge or the left side of positive charge, that is the electric field is stronger between the charges and greater work is done over a relatively small displacement.

On the right side of negative charge or left side of positive charge, the electric field is relatively weaker and therefore the equipotential surfaces are farther apart. This concludes that the electric field may not necessarily be the same over an equipotential surface.

## Corona Discharge

We first know the relationship between the electric field and electric potential on the surface of a charged conducting sphere. Consider a sphere of radius $R$ and total charge $q$. If you enclose the conducting sphere by a Gaussian sphere of radius $r$ ($r > R$) and apply Gauss's law, you will find the electric field on the Gaussian surface as:

\[\begin{align*} EA=&\frac{q}{{{\epsilon }_{0}}} \\ \text{or,}\quad E=&\frac{kq}{{{r}^{2}}} \end{align*}\]

Which is the same as the electric field of a point charge. So, the total charge $q$ on the sphere acts as if all the charge were concentrated at the centre of the sphere. And the electric potential is also the same as that of a point charge which is

\[V=\frac{kq}{{{r}^{{}}}}\]

On the surface of the conductor ($r=R$), the electric field is $E=kq/R^2$ and the electric potential is $V=kq/R$. So,

\[{{V}_{\text{s}}}=R{{E}_{\text{s}}}\]

The subscript $\text{s}$ in the above equation represents the *surface*. If you keep the radius $R$ constant and increase the charge in the conducting sphere, the electric potential also increases. But there is a limit on how much charge you can add in the sphere for a certain radius. If you keep on adding the charge on the sphere, the electric potential becomes maximum at a certain point at which the air is ionized and becomes conductive. And you keep on adding the charge any further, the electric discharge through the air takes place. So the maximum potential to the conducting sphere of a given radius is limited, that is any further adding of charge results the electric discharge. The maximum electric field at which the air is ionized is about $3.0\times {{10}^{6}}\text{V/m}$

The maximum potential of the sphere decreases by decreasing the radius of the sphere and even adding the small amount of charge can reach the limit and further adding causes electric discharge. Therefore the conducting sphere can withstand large electric potential if the radius of the conducting sphere is large enough; it means you can add even more charge on it until the electric discharge through the air. We have the relationship $E_s = V_s/R$, and therefore the sharp edges or sharp point in a conductor has very less radius of curvature and even small electric potential can cause sufficiently large electric field which can cause the ionization of the air and then results the electric discharge through the air also called corona discharge. The blue glow through the sharp edges are called *corona*.