# Electric Potential

Electric potential is the electric potential energy per unit charge. We know the potential energy of test charge $q_0$ and point charge $q$ from the previous article and therefore the electric potential is

\[V = \frac{{\left( {k\frac{{q{q_0}}}{r}} \right)}}{{{q_0}}} = k\frac{q}{r} \tag{4}\]

The above equation shows that the electric potential is independent of $q_0$. The SI unit of electric potential or simply potential is J/C or *volt* written as V.

Now the electric potential due to the collection of charges $q_1$, $q_2$, $q_3$, $q_4$ etc. at a point say $p$ at distance $r_1$, $r_2$, $r_3$ and $r_4$ respectively is the sum of the potentials due to individual charges.

\[\begin{align*} {V_{{\rm{total}}}} &= k\left( {\frac{{{q_1}}}{{{r_1}}} + \frac{{{q_2}}}{{{r_2}}} + \frac{{{q_3}}}{{{r_3}}} + \frac{{{q_4}}}{{{r_4}}}} \right)\\ {\rm{or,}}\quad {V_{{\rm{total}}}} &= k\sum\limits_{n = 1}^n {\frac{{{q_n}}}{{{r_n}}}} \tag{5} \end{align*}\]

If the electric charge is uniformly distributed over a surface or throughout a volume, we can make a small element of charge $dq$ and determine the electric potential at any point inside the electric field. The total electric potential due to the surface or volume of charge is the sum of electric potentials due to all element charges.

\[V = k\int {\frac{{dq}}{r}} \tag{6}\]

## Potential Difference

In Figure 2 the work done by the electric force from point $a$ to point $b$ is the potential energy at $a$, $U_a$ minus the potential energy at $b$, $U_b$.

\[{W_{{\rm{ab}}}} = {U_a} - {U_b} = \frac{{kq{q_0}}}{{{r_a}}} - \frac{{kq{q_0}}}{{{r_b}}}\]

In terms of *per unit* basis the work done per unit charge from point $a$ to $b$ is $W_\text{ab}$ divided by $q_0$

\[\frac{{{W_{{\rm{ab}}}}}}{{{q_0}}} = k\frac{q}{{{r_a}}} - k\frac{q}{{{r_b}}} = {V_a} - {V_b}\]

So the work done by the electric force per unit charge from the initial position at $a$ to the final position at $b$ is the electric potential at $a$ minus the electric potential at $b$. The difference $V_a - V_b$ is called the potential difference of point $a$ with respect to point $b$ which is the work done but in *per unit* basis. If the potential of point $a$ with respect to point $b$ is $V_{ab}$,

\[{V_{ab}} = k\frac{q}{{{r_a}}} - k\frac{q}{{{r_b}}} = {V_a} - {V_b}\tag{7} \label{7}\]

In the below examples we consider continuous charge distribution of various charge distributions and find the electric potential.

## Electric Potential of a Line of Charge

Here we determine the electric potential of a line of charge. It's always easier to start with a coordinate system whenever necessary. The line of charge has total charge $q$ and length $2l$ along y-axis as shown in Figure 1. We determine the electric potential on the x-axis at point $p$ at a distance $x$ from the origin. Note that the middle point of the line of charge is at the origin our coordinate system.

Consider a small element of charge $dq$ at $y$ which has length of $dy$. The linear charge density is the charge per unit length and therefore $dq/dy = q/2l$ which implies $dq = (q/2l)dy$. The charge $dq$ is at distance $\sqrt {{x^2} + {y^2}}$ from the point $p$. To find the electric potential due to the line of charge we integrate the small electric potential $dV$ due to the charge $dq$ from $-l$ to $l$. The electric potential at the point $p$ due to the element charge $dq$ is,

\[dV = k\frac{{dq}}{{\sqrt {{x^2} + {y^2}} }} = k\frac{q}{{2l}}\frac{{dy}}{{\sqrt {{x^2} + {y^2}} }}\]

Now we integrate $dV$ from $-l$ to $l$ to get the total electric potential due to the line of charge:

\[V = k\frac{q}{{2l}}\int_{ - l}^l {\frac{{dy}}{{\sqrt {{x^2} + {y^2}} }}} = k\frac{q}{{2l}}\ln \left( {\frac{{\sqrt {{l^2} + {x^2}} + l}}{{\sqrt {{l^2} + {x^2}} - l}}} \right)\]

Note that electric potential is a scalar quantity so we don't need to take the account of any components of electric potential.

## Electric Potential of an Infinite line of charge

We already know the electric field of an infinite line of charge from electric field calculations of uniform charge distributions. We determine the electric potential using that electric field which is $E = k\frac{{2\lambda }}{x}$ with the distance $x$ replaced by $r$. And $\lambda $ is the linear charge density which is the charge per unit length.

Now consider a point $a$ at distance $r_1$ where the potential is $V_a$ and another point $b$ at distance $r_2$ where the potential is $V_b$ as shown in Figure 2. The potential difference between these points is

\[\begin{align*} {V_a} - {V_b} &= \int_{{r_1}}^{{r_2}} {\vec E \cdot d\vec r } = \int_{{r_1}}^{{r_2}} {E{\mkern 1mu} dr} \\ &= 2k\lambda \int_{{r_1}}^{{r_2}} {\frac{1}{r}dr} = 2k\lambda \left( {\ln {r_2} - \ln {r_1}} \right)\\ {\rm{or,}}\quad {V_{ab}} &= 2k\lambda \ln \frac{{{r_2}}}{{{r_1}}} \end{align*}\]

Note that the dot product $\vec{E}\cdot d\vec{r}$ has value $E\,dr$ because the displacement is perpendicularly outward from the line of charge. For convenience you can put the point $b$ at infinity and the potential at that point is zero. In this way you can get the value of the potential at the point $a$:

\[{V_a} = 2k\lambda \ln \frac{\infty }{{{r_1}}} = \infty \]

You may not have guessed that the result would be $\infty $ at the point $a$. Now what we do to remove this error is we take the point $b$ at a finite distance, say $r_0$ and make the potential zero at that distance. It's because you can make any point a zero potential. For example, take an example of gravitational potential energy. When you are at the bottom of a hill, you think the surface where you are standing to be at zero potential energy and apply $mgy$ to get the potential energy of a body of mass $m$ at height $y$ from the ground. Now you climb the hill and still you think the surface where you are standing to be at zero potential energy and again apply $mgy$ to get the potential energy. Now you just recently made the two different points (at the bottom and at the top of the hill) to be at zero potential energy.

All this means is we can take any point to be in zero potential and get the potential of another point with respect to that point. Now the above equation for the potential at the point $a$ with respect to the zero potential at $b$ is:

\[{V_a} = 2k\lambda \ln \frac{{{r_0}}}{{{r_1}}}\]

So the electric potential at any point $a$ at a perpendicular distance $r$ from the infinite line of charge can be obtained by taking any other point such as $b$ at a finite distance ${{r}_{0}}$ to be in zero electric potential, $V = {V_a} = 2k\lambda \ln \frac{{{r_0}}}{r}$. The same result is obtained for the infinite charged conducting cylinder. We use Gaussian surface to enclose the cylinder and all the charge on the cylinder is the same as the charge concentrated along its axis, the same as the infinite line of charge.

## Electric Potential of a Ring of Charge

Consider a ring of total charge $q$. The centre of the ring is ths same as the origin of our coordinate system and x-axis is perpendicular to the plane of the ring. We find the electric potential at point $p$ at distance $x$ from the centre of the ring on x-axis (see Figure 3). The radius of the ring is $a$. We first divide the ring of charge into infinitesimal elements of charge $dq$ and sum up the electric potentials due to all elements by integration.

The distance between the point $p$ and the element of charge $dq$ is ${\sqrt {{x^2} + {a^2}} }$. The small electric potential due to the charge $dq$ at the point $p$ is:

\[dV = k\frac{{dq}}{{\sqrt {{x^2} + {a^2}} }}\]

The total electric potential due to the ring of charge is

\[V = k\frac{1}{{\sqrt {{x^2} + {a^2}} }}\int {dq} = k\frac{q}{{\sqrt {{x^2} + {a^2}} }}\]

## Electric Potential of a Disk of Charge

Now we determine the electric potential of a disk of charge with total charge $q$ uniformly distributed in the disk. We find the electric potential at distance $x$ from the center of the disk on the x-axis of our coordinate system at point $p$. The plane of the disk is perpendicular to the x-axis (see Figure 4).

From the previous example we know the electric potential of a ring of charge. In this example we can divide the disk of charge into infinitesimal ring elements of charge $dq$ and thickness $dr$ where we consider that $r$ is the radius of a ring element and $a$ is the radius of the disk of charge. You can find the electric potential of the disk of charge by integrating the electric potential due to one ring element within limits $0$ to $a$. The area of the ring element is $2\pi r dr$ and the area of the disk of charge is $\pi a^2$. The ratio of the small charge $dq$ to the total charge $q$ is the same as the ratio of the small area $2\pi rdr$ to the total area $\pi a^2$ and therefore,

\[\begin{align*} \frac{{dq}}{q} &= \frac{{2\pi rdr}}{{\pi {a^2}}}\\ {\rm{or,}}\quad dq &= \frac{{2qrdr}}{{{a^2}}} \end{align*}\]

The electric potential of the disk of charge is determined by integration:

\[\begin{align*} V &= k\int_0^a {\frac{{dq}}{{\sqrt {{x^2} + {r^2}} }}} \\ {\rm{or,}}\quad V &= \frac{{2kq}}{{{a^2}}}\int_0^a {\frac{{rdr}}{{\sqrt {{x^2} + {r^2}} }}} \\ {\rm{or,}}\quad V &= \frac{{2kq}}{{{a^2}}}(\sqrt {{x^2} + {a^2}} - \left| x \right|) \end{align*}\]