# 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 wiki's about its definition and we know that it's a form of energy.

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.

#### Contents

## Introduction

**Gravitational potential**

Let us begin our journey about Electric potential energy by relating it to the the gravitational potential energy because the electric potential is very closely related to the gravitational potential , we know, from Newton's inverse square law: \[\vec{F_g}=-\dfrac{GMm}{r^2}\hat r\] 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 \(\Delta h\) 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 give point is \(\frac{F_g}{m}=g\). And thus: \[V_g=-\displaystyle\int_{h_A}^{h_B} \dfrac{\vec{F_g}}{m}\cdot \vec{ds}=\displaystyle\int_{h_A}^{h_B} \vec{g}\cdot\vec{ds}\]

**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: \[\vec{E}=\dfrac{\overrightarrow{F}}{q}\quad \text{ and }\quad W=\overrightarrow F\times \vec{dl}\\\therefore V=\displaystyle\int_{a}^{b} \overrightarrow{E}\cdot\vec{dl}=\displaystyle\int_{a}^{b}\dfrac{\overrightarrow F}{q}\cdot\vec{dl}=\dfrac{W}{q}\]

A 9V battery has an electric potential difference of \(9~\mbox{V}\) between the positive and negative terminals. How much kinetic energy **in J** would an electron gain if it moved from the negative terminal to the positive one?

**Details and assumptions**

- The charge on the electron is \(-1.6 \times 10^{-19}~\mbox{C}\).
- You may assume energy is conserved (so no drag or energy loss due to resistance for the electron).

## Relationship between potential and E-field

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 above equation to calculate the 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.

\[V_a - V_{\infty} = \int_a^{\infty}{\overrightarrow{E}.\overrightarrow{dl}}\] 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 potential at point is given and we may need to find electric field at that point. In such cases, we have the relation :

\[\large \overrightarrow{E} = - \overrightarrow{\nabla}V\]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

Vas\(\vec{E}=\nabla V\)

The scalar potential of the electric field is known to be \(V=\dfrac{kQ}{r}\).

\(\vec{E}=\nabla V\)

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, \(V=\dfrac{kQ}{r}\).

\(\dfrac{\partial V }{\partial r} =- \dfrac{kQ}{r^2}\)

\(\dfrac{\partial V }{\partial \theta} = 0\)

\(\dfrac{\partial V }{\partial \varphi} = 0\)

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.

\(\vec{E}={\partial V \over \partial r}\hat{r} + {1 \over r}{\partial V \over \partial \theta}\hat{\theta} + {1 \over r\sin\theta}{\partial V \over \partial \varphi}\hat{\varphi}\)

\(\vec{E}=-\dfrac{kQ}{r^2}\hat{r}\)

**Note:** The negative sign is the reason that the relationship between the electric field and scalar potential is *actually* written

\(\vec{E}=-\nabla V\)

**Applications of the potential function**

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

Solution:We are given that, \( V(x,y) = x^{2}y^{3} + xy^{5} \). Using the relation, \( \vec{E} =-\vec{\nabla}V \)\[\vec{E} = - \dfrac{\partial}{\partial x}\left[x^{2}y^{3} + xy^{5}\right] \hat{i} - \dfrac{\partial}{\partial y}\left[x^{2}y^{3} + xy^{5}\right] \hat{j} \\ \vec{E} = -\left[2xy^{3} + y^{5}\right] \hat{i} - \left[3x^{2}y^{2} + 5xy^{4}\right] \hat{j} \]

Substituting the point \( (x,y) \equiv (2,1) \)

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

Where \( \hat{r} = -\dfrac{5}{\sqrt{509}}\hat{i} -\dfrac{22}{\sqrt{509}}\hat{j} \)

Question: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 \( V_{a} , V_{b} \) are the electric potentials at the points \( A (1,1,1) \) & \( B(2,2,2) \) find \( V_{b} - V_{a} \) , given that the potential at the point \( (0,1,0) \) is 5 Volts.

Solution:Using the relation,

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

\[\begin{align}V(x,y,z) &= \displaystyle \int \left[\dfrac{5}{y}\hat{i} - 5 \dfrac{x}{y^{2}}\hat{j} + 8\hat{j}\right]\cdot\left[dx\hat{i} + dy\hat{j} + dz\hat{k}\right] + c \\ V(x,y,z) &= \displaystyle \int \left[5\dfrac{dx}{y} - 5\dfrac{x}{y^{2}}dy + 8dz \right] + c \\ V(x,y,z) &= \displaystyle \int \left[ 5\dfrac{ydx -xdy}{y^{2}}\right] + 8z + c \\ V(x,y,z) &= \displaystyle \int \left[5d\dfrac{x}{y}\right] + 8z + c \\ V(x,y,z) &= 5\cdot\dfrac{x}{y} + 8z + c\text{ Volts} \end{align}\]Given that \( V(0,1,0) = 5 \), \(5 = 0 + 0 + c \rightarrow c = 5 \)

\[\begin{align} V(x,y,z) &= 5\cdot\dfrac{x}{y} + 8z + 5 \\ V_{a} = V_{1,1,1} &= 5\cdot\dfrac{1}{1} + 8(1) + 5 = 5 + 8 + 5 = 18 \\ V_{b} = V_{2,2,2} &= 5\cdot\dfrac{2}{2} + 8(2) + 5 = 5 + 16 + 5 = 26 \\ V_{b} - V_{a} &= 26 -18 = 8 \text{Volts} \end{align}\]

## Finding Electric Potential in different conditions

Potential due to a point charge

Let's compute the potential at any point

Alocated at a distance \(r_a\) from the point chargeQ.

Using our relation \( \displaystyle V_a = \large{-} \int_{\infty}^a{\overrightarrow{E}.\overrightarrow{dl}} \) we can easily solve it.\(\overrightarrow{E}\) for a point charge is given by \(\displaystyle \overrightarrow{E} = \dfrac{kQ}{r^2}\). Plugging it in the equation \( \displaystyle V_a = \large{-} \int_{\infty}^a{\overrightarrow{E}.\overrightarrow{dl}} \) we get. \[\boxed{V_a = \dfrac{kQ}{r_a}}\]

Potential between two charges

Potential inside conductors

Potential due to a dipole

## Potential with spherical symmetry

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

Potential inside and outside charged shellFor potential in the case of a spherical shell, i.e. a spherical charge configuration with charge

uniformlydistributed over the surface of the body, we will consider 2 cases for potential. The first case will discuss the potential due to the shell on and outside it, whereas the second case would discuss it in the shell.Consider a spherical shell of radius \(R\), with charge \(Q\) distributed uniformly on it's surface with a charge density \(\sigma\). Now, according to shell theorem, we can consider this entire charge to be considered at the centre 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 a distance \(r\) from it is given by: \[\vec{E}=\dfrac { 1 }{ 4\pi { \varepsilon }_{ o } } \dfrac { Q }{ { r }^{ 2 } }\hat r\] and hence, it's potential is given by:\[V(r)=\displaystyle\int\overrightarrow{ E } .\vec{dr}= \displaystyle\int \dfrac{1}{4\pi\varepsilon_0} \dfrac{Q}{r}^{2} dr\quad =\dfrac { -1 }{ 4\pi\varepsilon_0}\dfrac Qr\]

And since, shell theorem holds true for our configuration, electric potential in this case is also given by:

\[V(r)=\displaystyle\int\vec{ E } .\vec{dr} = \displaystyle\int { \dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \dfrac { Q }{ { r }^{ 2 } } } dr\quad =\dfrac { -1 }{ 4\pi \varepsilon _{ 0 } } \dfrac { Q }{ r } \]

where \(r\) would act as the distance of the centre of the sphere from the point where electric potential is to be calculated at.

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 bezero. 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, therefore by gauss' 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 out 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 \(\int { E.dr } \), therefore the potential inside the sphercial 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 sphereLet us try to divide thus into \(3\) cases, first let's calculate the potential outside the sphere, second let's calculate the potential on the surface, and finally the potential inside the sphere. Let us have a sphere of radius \(R\) with a uniform charge density \(\sigma\) carrying a charge \(Q\).

To calculate the potential at any point outside the conductor, we will figure out the

potential differencebetween any two arbitrary points \(a\) and \(b\) situated at a distance \(r_a\) 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:\[V_a - V_b = \displaystyle\int_a^b\vec{E}.\cdot \overrightarrow{dl}=\displaystyle\int_{r_a}^{r_b}\dfrac{Q}{4\pi\epsilon_0r^2}dr\]\[=\dfrac{Q}{4\pi\epsilon_0}\left. \dfrac 1r\right|_{r_a}^{r_b}=\dfrac{Q}{4\pi\epsilon_0}\left(\dfrac{1}{r_a}-\dfrac{1}{r_b}\right)\]

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 substituenting \(r_a=R\), 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, this is because the charge is always collected at the surface, thus \(\vec E =0\), and hence the potential inside a uniformly charged solid sphere is \[V_{\text{Any point inside}} - V_{\text{Surface}} = \int_{\text{Any point inside}}^{\text{Surface}}{\overrightarrow{E}.\overrightarrow{dl}} = 0 \\ \therefore V_{\text{Any point inside}} = V_{\text{Surface}} \\ \Rightarrow \boxed{ V_{\text{Any point inside}} = \dfrac{Q}{4\pi\epsilon_0}\dfrac 1R}\]

The above graph shows the electric potential inside or around a charged sphere as a function of \(r\) which denotes the distance from the center of the sphere. If the radius of the sphere is \(R_0,\) which of the following statements is correct?

a) The charged sphere is charged positively.

b) The intensity of the electric field inside the charged sphere is zero.

c) The intensity of the electric field at \(r=2R_0\) is twice as large as that at \(r=4R_0.\)

## Applications

LightningSo, 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 previous 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 a humid weather, the air between the clouds, and the clouds and the ground gets partially

ionizedi.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 azig-zagprojectile. This type of sudden discharge occurs due to the reason that both the bodies posses a very high amount of charge, because of which they become very unstable, 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 it's resistance, and hence the path is never straight, rather it is one of the millions of possibilities that a lightning surge could have taken!

Capacitors

Read capacitors, 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.

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

\[V=\int_0^d E\, dl=\int_0^d \dfrac{\sigma}{\epsilon_0}dl=\dfrac{Qd}{\epsilon_0A}\\ \text{but, }V=Ed;\quad\therefore Ed=\dfrac{Qd}{\epsilon_0A}\implies\boxed{ E=\dfrac{Q}{\epsilon_0A}}\] Now, \[\dfrac{Q}{\epsilon_0A}=\dfrac{V}{d}\therefore Q=\dfrac{\epsilon_0A}d V \implies \boxed{Q=CV}\]

Equipotential surfaces and shielding apparatusesWell, you can easily decipher from the word equipotential surfaces that we are going to discuss about surfaces which have equal potential.

A dipole consists a two point charges \(+Q\) and \(-Q\) a distance \(d\) apart. Placing first charge at the origin and the second at \(r=-d\) . The potential is simply the sum of the potential for each charge \[\phi=\frac {1}{4\pi\epsilon_0}(\frac {Q}{r}-\frac {Q}{|r+d|})\] More precisely, electric field is just the sum of the electric fields made by the two point charges. Electric field is not spherically symmetric. The leading order contribution is governed by the combination \(p=Qd\). This property is known as dipole electric moment. It points from the negative charge to the positive. Dipole electric field \(E=-\nabla\phi=\frac {1}{4\pi\epsilon_0}(\frac{3(p\cdot\vec{r})\vec{r}-p}{r^3})+\cdots\). Sine \(\phi\) is constant, 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.

## References

[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:Lightnings*sequence*2_animation.gif under Creative Commons licensing for reuse and modification.

[4] Image from https://commons.m.wikimedia.org/wiki/File:Parallel*plate*capacitor.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.

**Cite as:**Electric Potential.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/electric-potential/