Electric Potential

Do you know why water flows when there is a slope? The reason turns out to be the difference in the heights of the two regions, which means there is a difference in the potential energy, causing the water to flow so that a state of equilibrium is attained. For a detailed analysis see this.

But what is potential? Well, we have already seen in our mechanics wikis about its definition and we know that it's a form of energy.

Water flowing from a region of higher potential to a region of lower potential $^\text{[1]}$

But it takes many different forms in physics, and this wiki is concerned about the electrical definition of the term potential and its applications. As potential is form of energy, which is a scalar quantity, calculations are easier than those involving forces which is a vector. Basically electric potential is defined as the work done in moving a point charge from one point to another point under a constant electric field, and we find the formula to be $V=W/Q$.

Motivation decided: Energy is a scalar and enables easier calculations than sticking to forces.

Gravitational Potential

Let us begin our journey about electric potential energy by relating it to the gravitational potential energy because the electric potential is very closely related to the gravitational potential. We know from Newton's inverse square law that

$$\vec{F_g}=-\dfrac{GMm}{r^2}\hat r.$$

We had also seen that the potential energy gained by an object when raised from a point $h_A$ to $h_B$ through a height of $\Delta h$ was equivalent to the work done against gravity to bring the object to that point:

$$W_g=\displaystyle\int \vec{F_g} \cdot\vec{ds}=-\displaystyle\int_{h_A}^{h_B} mg \cdot dh = -mg\Delta h.$$

Now let us try to define a function that gives us the gravitational potential difference between two points. But how do we define it? Well, we will just assume that the gravitational field at a given point is $\frac{F_g}{m}=g$. Then

$$V_g=-\displaystyle\int_{h_A}^{h_B} \dfrac{\vec{F_g}}{m}\cdot \vec{ds}=\displaystyle\int_{h_A}^{h_B} \vec{g}\cdot\vec{ds}.$$

Electric Potential

Now, let us try to co-relate the Gravitational potential to the topic of our concern. If you have read coulomb's law you would have come across this equation which gives us the force experienced between two charges:

$$\vec{F_e}=\dfrac{kq_1q_2}{r^2}\hat r$$

We can also define the same function for electric potential and find the electric potential difference, where $V_e$ is the potential difference function, which defines the negative work done in moving a test charge from a point $a$ to $b$:

$$V_e=-\displaystyle\int_{a}^{b} \dfrac{\overrightarrow{F_e}}{q}\cdot \vec{dl}=\displaystyle\int_{a}^{b} \vec{E}\cdot\vec{dl}.$$

We had previously defined the electric potential as the work done by the electric field in moving a point charge through a certain distance. And we also know that the electric field and work are defined as

$$\begin{aligned} \vec{E}=\dfrac{\overrightarrow{F}}{q},\ \ W=\overrightarrow F\times \vec{dl} \implies V&=\int_{a}^{b} \overrightarrow{E}\cdot\vec{dl}\\ &=\int_{a}^{b}\dfrac{\overrightarrow F}{q}\cdot\vec{dl}\\ &=\dfrac{W}{q}. \end{aligned}$$

There is an important relationship between electric field and potential. To understand this, make sure you have read Electric Fields and Electrostatics.

$$V_a - V_b = \int_a^b{\overrightarrow{E}.\overrightarrow{dl}}$$

If the electric field $\overrightarrow{E}$ at various points is known, we can use the above equation to calculate the potential differences between any two points. In some cases, we may need to find the electric potential at a point rather than potential difference. In that case, the potential is calculated as the work done in bringing a unit positive charge from infinity to the given point.

We have

$$V_a - V_{\infty} = \int_a^{\infty}{\overrightarrow{E}.\overrightarrow{dl}}.$$

We know that $V_{\infty} = 0$. So this equation can be written as

$$V_a = \large{-} \int_{\infty}^a{\overrightarrow{E}.\overrightarrow{dl}}.$$

In some cases, the potential at point is given and we may need to find electric field at that point. In such cases, we have the relation

$$\overrightarrow{E} = - \overrightarrow{\nabla}V.$$

This is read $“\overrightarrow{E}$ is the negative gradient of $V."$ The quantity $\overrightarrow{\nabla}V$ is called the potential gradient.

A conservative field is one that can be written as the gradient of a scalar potential $V$ as

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

The scalar potential of the electric field is known to be $V=\frac{kQ}{r}$. The gradient in spherical coordinates is $\nabla = {\partial \over \partial r}\hat{r} + {1 \over r}{\partial \over \partial \theta}\hat{\theta} + {1 \over r\sin\theta}{\partial \over \partial \varphi}\hat{\varphi}$, so

$$\nabla V ={\partial V \over \partial r}\hat{r} + {1 \over r}{\partial V \over \partial \theta}\hat{\theta} + {1 \over r\sin\theta}{\partial V \over \partial \varphi}\hat{\varphi}.$$

Evaluate each of the partial derivatives of the scalar potential equation, $V=\frac{kQ}{r}:$

$$\begin{aligned} \dfrac{\partial V }{\partial r} &=- \dfrac{kQ}{r^2}\\ \dfrac{\partial V }{\partial \theta} &= 0\\ \dfrac{\partial V }{\partial \varphi} &= 0. \end{aligned}$$

Using these partial derivatives with the definition for the gradient, only the $\hat{r}$ component survives, since the other components contain partial derivatives that are equal to 0:

$$\begin{aligned} \vec{E} &={\partial V \over \partial r}\hat{r} + {1 \over r}{\partial V \over \partial \theta}\hat{\theta} + {1 \over r\sin\theta}{\partial V \over \partial \varphi}\hat{\varphi}\\ &=-\dfrac{kQ}{r^2}\hat{r}. \end{aligned}$$

Note: The negative sign is the reason that the relationship between the electric field and scalar potential is actually written $\vec{E}=-\nabla V.$

The following examples are applications of the potential function:

In the $xy$-plane, the electric potential at a point $(x,y)$ is given by the relation $V(x,y) = x^{2}y^{3} + xy^{5}$. Find the electric field vector at the point $(2,1)$.

---

We are given that $V(x,y) = x^{2}y^{3} + xy^{5}$. Using the relation $\vec{E} =-\vec{\nabla}V,$

$$\begin{aligned} \vec{E} &= - \dfrac{\partial}{\partial x}\left[x^{2}y^{3} + xy^{5}\right] \hat{i} - \dfrac{\partial}{\partial y}\left[x^{2}y^{3} + xy^{5}\right] \hat{j} \\ &= -\left[2xy^{3} + y^{5}\right] \hat{i} - \left[3x^{2}y^{2} + 5xy^{4}\right] \hat{j}. \end{aligned}$$

Substituting the point $(x,y) = (2,1),$

$$\begin{aligned} \vec{E} &= -\left[2\cdot2\cdot(1)^{3} + (1)^{5}\right]\hat{i} - \left[3\cdot(2)^{2}\cdot(1)^{2} + 5\cdot(2)\cdot(1)^{4}\right]\hat{j} \\ &= -5\hat{i} - 22\hat{j} \\ \left|\vec{E}\right| &= \sqrt{25+484} \\&= \sqrt{509} \\ \Rightarrow \vec{E} &= \sqrt{509}\hat{r}, \end{aligned}$$

where $\hat{r} = -\dfrac{5}{\sqrt{509}}\hat{i} -\dfrac{22}{\sqrt{509}}\hat{j}.$

In three dimensional space, the electric field at a certain point is given by

$$\vec{E} = \dfrac{5}{y}\hat{i} - 5 \dfrac{x}{y^{2}}\hat{j} + 8\hat{j}.$$

If $V_{a}$ and $V_{b}$ are the electric potentials at the points $A=(1,1,1)$ and $B=(2,2,2),$ respectively, find $V_{b} - V_{a}$ given that the potential at the point $(0,1,0)$ is 5 volts.

---

Using the relation $V(x,y,z) = \displaystyle \int \vec{E}\cdot\vec{dl} + c,$

$$\begin{aligned} V(x,y,z) &= \displaystyle \int \left[\dfrac{5}{y}\hat{i} - 5 \dfrac{x}{y^{2}}\hat{j} + 8\hat{j}\right]\cdot\left[dx\hat{i} + dy\hat{j} + dz\hat{k}\right] + c \\ &= \displaystyle \int \left[5\dfrac{dx}{y} - 5\dfrac{x}{y^{2}}dy + 8dz \right] + c \\ &= \displaystyle \int \left[ 5\dfrac{ydx -xdy}{y^{2}}\right] + 8z + c \\ &= \displaystyle \int \left[5d\dfrac{x}{y}\right] + 8z + c \\ &= 5\cdot\dfrac{x}{y} + 8z + c\text{ (volts)}. \end{aligned}$$

Given that $V(0,1,0) = 5$, $5 = 0 + 0 + c \implies c = 5,$

$$\begin{aligned} V(x,y,z) &= 5\cdot\dfrac{x}{y} + 8z + 5 \\ V_{a} &= V_{1,1,1} = 5\cdot\dfrac{1}{1} + 8(1) + 5 = 5 + 8 + 5 = 18 \\ V_{b} &= V_{2,2,2} = 5\cdot\dfrac{2}{2} + 8(2) + 5 = 5 + 16 + 5 = 26 \\ V_{b} - V_{a} &= 26 -18 = 8 \text{ (volts)}. \end{aligned}$$

Potential due to a Point Charge

Let's compute the potential at any point A located at a distance of $r_a$ from the point charge Q.
Using our relation $\displaystyle V_a = \large{-} \int_{\infty}^a{\overrightarrow{E}\cdot \overrightarrow{dl}},$ we can easily solve it.

$\overrightarrow{E}$ for a point charge is given by $\displaystyle \overrightarrow{E} = \frac{kQ}{r^2}$. Plugging it in the equation $\displaystyle V_a = \large{-} \int_{\infty}^a{\overrightarrow{E}\cdot \overrightarrow{dl}},$ we get

$$\boxed{V_a = \dfrac{kQ}{r_a}}.$$

Potential between two charges

Potential inside conductors

Potential due to a dipole

To solve problems related to this topic, we will use Newton's Shell theorem. Although it is for gravitational fields, we can apply it here as well.

Potential inside and outside charged shell:

For potential in the case of a spherical shell, i.e. a spherical charge configuration with charge uniformly distributed over the surface of the body, we will consider 2 cases. The first case will discuss the potential due to the shell on and outside of it, whereas the second case will discuss it in the shell.

Consider a spherical shell of radius $R$, with charge $Q$ distributed uniformly on its surface with charge density $\sigma$. Now, according to the shell theorem, this entire charge can be considered at the center of this configuration, i.e. point 'O'. So, we have successfully converted our complex charged body into a simple configuration for which Coulomb's law can easily be applied.

• Case 1: So, according to Coulomb's law, the electric field due to a point charge at distance $r$ from it is given by $$\vec{E}=\dfrac { 1 }{ 4\pi { \varepsilon }_{ o } } \dfrac { Q }{ { r }^{ 2 } }\hat r,$$ and hence its potential is given by $$V(r)=\displaystyle\int\overrightarrow{ E }\cdot \vec{dr}= \displaystyle\int \dfrac{1}{4\pi\varepsilon_0} \dfrac{Q}{r}^{2} dr =\dfrac { -1 }{ 4\pi\varepsilon_0}\dfrac Qr.$$ And since the shell theorem holds true for our configuration, electric potential in this case is also given by $$V(r)=\displaystyle\int\vec{ E } \cdot \vec{dr} = \displaystyle\int { \dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \dfrac { Q }{ { r }^{ 2 } } } dr =\dfrac { -1 }{ 4\pi \varepsilon _{ 0 } } \dfrac { Q }{ r },$$ where $r$ would act as the distance of the center of the sphere from the point at which the electric potential is to be calculated. This can also be written as $$V(r)=\dfrac { -1 }{ 4\pi { \varepsilon }_{ 0 } } \dfrac { \sigma (4\pi { R }^{ 2 }) }{ r } =\dfrac { -1 }{ \varepsilon _{ 0 } } \dfrac { \sigma { R }^{ 2 } }{ r },$$ where $\sigma$ is the charge density for the shell.

• Case 2: For the second case, we will begin by proving a little fact that will bring us the famous result of having electric field inside a spherical shell to be zero. Consider a hypothetical spherical area inside the shell of a radius $r_{0}$. Now, since this hypothetical sphere does not include any charge in it, by Gauss's law, we can easily say that the electric flux and hence the electric field due to this area = $0$. This can be done for infinite hypothetical areas inside the shell, which gives us our final argument that "electric field inside a charged spherical shell = 0."
  $$
  Using this result, since electric field inside the shell is zero and potential at any point due to an external electric field is given by $\int { E\cdot dr }$, the potential inside the spherical shell turns out to be some constant value which would be the same as the potential of the surface of the shell, i.e. $${ V }_{\text{Inside}}={ V }_{\text{Surface}}=\dfrac { -1 }{ 4\pi { \varepsilon }_{ 0} } \dfrac { Q }{ R }.$$

Potential inside and outside charged sphere:

Let us try to divide this into $3$ cases. First, let's calculate the potential outside the sphere; second, let's calculate the potential on the surface; finally, let's calculate the potential inside the sphere. Let us have a sphere of radius $R$ with a uniform charge density $\sigma$ carrying a charge $Q$.

Potential of a solid sphere carrying a charge of Q

To calculate the potential at any point outside the conductor, we will figure out the potential difference between any two arbitrary points $a$ and $b$ situated at distances $r_a$ and $r_b$ $(r_a<r_b),$ and then we will assume that one of them is at infinity. So the potential difference between the two points will be

$$\begin{aligned} V_a - V_b &= \displaystyle\int_a^b\vec{E}\cdot \overrightarrow{dl}\\ &=\displaystyle\int_{r_a}^{r_b}\dfrac{Q}{4\pi\epsilon_0r^2}dr\\ &=\dfrac{Q}{4\pi\epsilon_0}\left. \dfrac 1r\right|_{r_a}^{r_b}\\ &=\dfrac{Q}{4\pi\epsilon_0}\left(\dfrac{1}{r_a}-\dfrac{1}{r_b}\right). \end{aligned}$$

Now, let us assume $r_b\rightarrow \infty,$ then $V_b=0$. Therefore the potential at any point outside the conductor will be

$$\boxed{V_{\text{Outside}}=\dfrac{Q}{4\pi\epsilon_0}\dfrac 1{r_a}}.$$

Now, we can deduce the formula for the second case, i.e. for the potential on the surface of the sphere, by substituting $r_a=R$, and we will arrive at this equation

$$\boxed{V_{\text{Surface}}=\dfrac{Q}{4\pi\epsilon_0}\dfrac 1R}.$$

Finally, we know that the electric field inside the sphere is zero because the charge is always collected at the surface, so $\vec E =0$ and hence the potential inside a uniformly charged solid sphere is

$$\begin{aligned} V_{\text{Any point inside}} - V_{\text{Surface}} &= \int_{\text{Any point inside}}^{\text{Surface}}{\overrightarrow{E}\cdot\overrightarrow{dl}} \\ &= 0 \\ V_{\text{Any point inside}} &= V_{\text{Surface}} \\ \Rightarrow V_{\text{Any point inside}} &= \dfrac{Q}{4\pi\epsilon_0}\dfrac 1R. \end{aligned}$$

Lightning:

We have all seen lightning. It is that spark in the sky when the weather isn't too nice. But trust me when I say that it is not as boring as it seems! We all know from the previously discussed topics that 2 bodies which have a potential difference will always cause charge to flow from a higher-potential region to a lower one. Well, that is what all lightning is!

During a storm, or in humid weather, the air between the clouds and between the clouds and the ground gets partially ionized, i.e. it allows charge to flow through it, unlike the neutral air we have in hot weathers. So, this causes a potential difference to be developed between the two surfaces, which further causes a flow of charges, and hence an electric shock is produced in the form of a zig-zag projectile. This type of sudden discharge occurs due to the reason that both the bodies possess a very high amount of charge, because of which they become very unstable, and hence they come to equilibrium by this process.

Now, the basic question that arises is, "Why zig-zag?"

The reason is simple. The path of least resistance is the path that is the motive of every single charge in the lightning strike. Moreover, the entire atmosphere varies with humidity, temperature, pressure and what not, as we move along it? This causes fluctuations in its resistance, and hence the path is never straight, and rather it is one of the millions of possibilities that a lightning surge could have taken!

Capacitors:

Read Capacitors and Series and Parallel Capacitors for information on its formulas.

Basically, a capacitor is an instrument that stores electric charge; it stores this charge using a simple principle. Let's see how a capacitor works. A cap (as we call it) just has two metal rods inside; they act as the charge storage for it.

These metal rods, when connected to two terminals of a battery, start accumulating the respective charges present in the terminal, i.e. the rod connected to the positive terminal accumulates positive charge and the rod connected to the negative terminal accumulates negative charge. Thus it creates a potential difference between the two rods, and hence the cap acts as a temporary battery.

Plates inside a capacitor

Let us now formulate some useful formulae. Suppose that we have a capacitor with two plates of equal area $A$ and that each plate stores a charge of $\pm Q$ and has uniform charge density $\sigma$. Let us say they are separated by distance $d$ with a dielectric between them having permittivity $\epsilon_0$. We can now say that the voltage between the plates is

$$\begin{aligned} V &=\int_0^d E\, dl =\int_0^d \dfrac{\sigma}{\epsilon_0}dl =\dfrac{Qd}{\epsilon_0A}\\\\\\ \text{but } V&=Ed\\\ \\ \Rightarrow Ed&=\dfrac{Qd}{\epsilon_0A}\\ E&=\dfrac{Q}{\epsilon_0A}. \end{aligned}$$

Now,

$$\dfrac{Q}{\epsilon_0A}=\dfrac{V}{d}\implies Q=\dfrac{\epsilon_0A}d V \implies \boxed{Q=CV}.$$

Equipotential surfaces and shielding apparatuses:

Well, you can easily decipher from the word equipotential surfaces that we are going to discuss surfaces which have equal potential.

A dipole consists of two point charges $+Q$ and $-Q$ a distance of $d$ apart. Place the first charge at the origin and the second at $r=-d$. The potential is simply the sum of the potential for each charge:

$$\phi=\frac {1}{4\pi\epsilon_0}\left(\frac {Q}{r}-\frac {Q}{|r+d|}\right).$$

More precisely, the electric field is just the sum of the electric fields made by the two point charges. An electric field is not spherically symmetric. The leading order contribution is governed by the combination $p=Qd$. This property is known as a dipole electric moment. It points from the negative charge to the positive. Dipole electric field $E=-\nabla\phi=\frac {1}{4\pi\epsilon_0}\left(\frac{3(p\cdot\vec{r})\vec{r}-p}{r^3}\right)+\cdots$. Sine $\phi$ is constant, and the surface of the conductor must be an equipotential. This implies that any $E=-\nabla\phi$ is perpendicular to the surface. Any component of the electric field that lies to the tangential to the surface would make the surface charges move. Sign of the electric field depends on where it sits in space. In some parts, the force will be attractive, in other parts repulsive.

[1] GIF taken from http://gifloop.tumblr.com/post/15018621452 through giphy.com

[2] Image taken from http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elepe.html Hyperphysics.

[3] Image from https://en.m.wikipedia.org/wiki/File:Lightningssequence2_animation.gif under Creative Commons licensing for reuse and modification.

[4] Image from https://commons.m.wikimedia.org/wiki/File:Parallelplatecapacitor.svg under Creative Commons licensing for reuse and modification.

[5] PDF file https://www.math.ksu.edu/~dbski/writings/shell.pdf as reference.

[6] J.D Jackson's classical electrodynamics.

[7] Fitzpatrick electromagnetism.

[8] A.Zangwill ED.