# Electrostatics

Ever wondered how lightning occurs? Or why you see sparks when you remove your sweater in a dark room? Then you are in the right place. **Electrostatics** is the branch of physics which can help explain these amazing wonders. It is very vital because it can be used to explain natural electrostatic phenomena and help us learn how to solve basic problems in this field. So what next? Let us explore this world of interesting phenomena!

#### Contents

## Introduction

The charges observed in the universe are of only two types: positive and negative. These charges interact in different ways; in general, we can say that *like charges repel* each other and *unlike charges attract* each other. To be specific, they interact in the following ways:

$\text{When two charged bodies are brought together} = \begin{cases} \text{positive - positive} && \text{ repel } \\ \text{negative - negative} && \text{ repel } \\ \text{positive - negative} && \text{ attract}. \\ \end{cases}$

Note: Charges are conserved, i.e. charges can neither be created nor destroyed.

Let us now understand one of the most basic concept of electrostatics: quantization of charges.

Quantization of Charges:One of the most basic principle that many people fail to understand is the quantization of charges, according to which a charge can only be an

integer multipleof the elementary charge $e = 1.6\times 10^{-19}\text{C}$. Mathematically, we can say$Q=ne,\text{ where } n=\pm 1,\pm 2,\pm 3,\pm 4,\ldots.$

But this is very simple because we know that the charges, which are nothing but protons (+ve) or electrons (-ve), are indivisible. Hence a charge can't be a fraction, because protons can't be further divided.

But if we consider quantum physics, protons can be further divided into things called quarks, which have fractional values of $e$, more specifically in the ratio of one-thirds. But they are always found in a colony (as protons).Proton (a quantum model):

Now that we have understood that a charge can never be a fraction, we'll try to get familiar with the force between two charges with the help of the famous Coulomb's law.

Coulomb's Law:This law states that the force between two charges is directly proportional to the product of the charges and inversely proportional to the square of their separation. Therefore, we have

$F\propto \dfrac{q_1q_2}{r^2}.$

Now, we introduce a constant of proportionality $k$ (Coulomb's constant):

$F=k\dfrac{q_1q_2}{r^2}, \text{ where }k = \dfrac{1}{4\pi\epsilon_0}\approx9\times 10^9.$

Here, the quantity $\epsilon_0$ is known as the

permittivity of free space. This formula gives us the magnitude of the force exerted between the two charges.For more details visit Coulomb's law.

Electric Fields and Field Lines:Electric field lines are the imaginary curves in an electric field along which a point charge would move if it is placed in it. The field lines are different for a positive charge and a negative charge. For a positive charge, the electric field lines extend outward infinitely, and for a negative charge, the electric field lines appear to come inwards.

But when two charges interact with each other, things are different. We can classify it into two cases: first when two charges are unlike, and second when they are like. The image below can help you understand it:

But if we want to quantize the electric field, we do it this way, mathematically,

$\begin{aligned} \vec{E} &= \dfrac{\vec{F}}{q}\\ &=\dfrac 1{4\pi\epsilon_0}\dfrac{q}{r^2}. \qquad (\text{Hence the unit of electric field is Newtons/Coulomb}.) \end{aligned}$

For more information, visit field lines and field strength.

## Superposition

Main article: superposition of electric fields.

Every charged particle or a system of charged particles in the universe creates an electric field in the space around itself. This field can be calculated with the help of Coulomb's law. But, what would happen if there were two or more charges in space? How would the electric field in the region around them be affected? The principle of superposition helps us understand this phenomenon.

The Principle of Superposition:It states that every charge in space creates an electric field at a point independent of the presence of other charges in that medium. The resultant electric field is a vector sum of the electric field due to individual charges.

Charges in a Straight Line:Two charges $q_{1}$ and $q_{2}$ are kept at the endpoints of a rod $AB$ of length $L = 2\text{ m}$ in vacuum. What is the magnitude of electric field at the center of the rod due to these 2 charges: $q_{1} = +10^{-4} C, \quad q_{2}= +10^{-5} C\, ?$

$$

$\dfrac{1}{4\pi\epsilon_{0}} = 9 \times 10^{9}\text{ Nm}^2/\text{C}^2$

According to the principle of superposition, each charge creates its own electric field independent of the other charge. Let the electric fields by $q_{1}$ and $q_{2}$ be $E_{1}$ and $E_{2},$ respectively. Using Coulomb's law, we have

$\begin{aligned} |E_{1}| &= \dfrac{q_{1}}{4\pi\epsilon_{0}(1)^{2}} = 9 \times 10^{9} \times 10^{-4} = 9 \times 10^{5} \\ |E_{2}| &= \dfrac{q_{2}}{4\pi\epsilon_{0}(1)^{2}} = 9 \times 10^{9} \times 10^{-5} = 9 \times 10^{4}. \end{aligned}$ Now, $E_{1}$ and $E_{2}$ are oppositely directed. The angle between them is $\theta = \pi$ radians. Therefore, $\begin{aligned} |E_\text{net}| &= \sqrt{E_{1}^{2} + E_{2}^{2} + 2E_{1}E_{2}\cos\theta} \\ &= \sqrt{E_{1}^{2} + E_{2}^{2} - 2E_{1}E_{2}} \\ &= | |E_{1}| - |E_{2}| |\\ \Rightarrow E_\text{net} &= 9 \times 10^{5} - 9 \times 10^{4} \\ &= 81 \times 10^{4} \text{ (N/C)}. \end{aligned}$

Charges in a Triangle:Three charges $q_{1}, q_{2}$ and $q_{3}$ are placed at the vertices of an equilateral triangle of side length $a$ in vacuum. Find the magnitude of electric force experienced by charge $q_{3}$.

Data: $q_{1} = +10 \mu \text{C},\ q_{2} = -10 \mu \text{C},\ q_{3} = +2 \mu \text{C},\ a = 3\text{ meters},\ \dfrac{1}{4\pi\epsilon_{0}} = 9 \times 10^{9}$

Let us consider the interaction of the charge $q_{3}$ with each of the two charges separately.

Let the electric field by the charges $q_{1}$ and $q_{2}$ be $E_{1}$ and $E_{2},$ respectively.

The magnitudes of $E_{1}$ and $E_{2}$ can be calculated using Coulomb's law: $\begin{aligned} \left|E_{1}\right| &= \dfrac{1}{4\pi\epsilon_{0}}\cdot\dfrac{10\cdot 10^{-6} }{3^{2}} = 10^{9}\cdot 10^{-5} = 10^{4}\\ \left|E_{2}\right| &= \dfrac{1}{4\pi\epsilon_{0}}\cdot\dfrac{10\cdot 10^{-6} }{3^{2}} = 10^{9} \cdot 10^{-5} = 10^{4}. \end{aligned}$ Using the principle of superposition, we have $\begin{aligned} \vec{E}_\text{resultant} &= \vec{E_{1}} + \vec{E_{2}} \\ \left|E_\text{resultant}\right| &= \sqrt{\left|E_{1}\right| ^{2} + \left|E_{2}\right|^{2} + 2\left|E_{1}\right|\left|E_{2}\right| \cos\dfrac{2\pi}{3}} \\ &= \sqrt{10^{8} + 10^{8} - 10^{8}} = \sqrt{10^{8}} = 10^{4} \text{ (N/C)}. \end{aligned}$ Let the force acting on $q_{3}$ be $F.$ Then $\begin{aligned} \vec{F} &= q_{3} \times \vec{E}_\text{resultant} \\ \left|F\right| &= q_{3} \left|E_\text{resultant}\right| \\ &= 2 \times 10^{-6} \times 10^{4} = 2\cdot 10^{-2} = 0.02 \text{ (N)}.\ _\square \end{aligned}$

Three equal charges are placed in such a way that they form a right triangle with sides $3\text{ cm}, 4\text{ cm}, 5\text{ cm}.$ Each charge has a magnitude of $2\times 10^{-6}\text{C}$.

Find the net force acting on the charge at the right angle.

$$

**Details and assumptions:**

Take $\dfrac 1{4\pi\epsilon_0} = 9\times 10^9\text{ Nm}^2/\text{C}^2$.

Round your answer to the nearest integer.

Try solving this problem by applying the concepts of charges in a straight line and triangle.

## Charge Distribution in Conductors

Charge distribution between plates can be figured out using the following basic principles:

Electric Field inside a Conductor

At any point inside a conductor, the electric field equals zero. This phenomenon can be explained using the fact that all the charges present on a conductor move away towards the outer surface of the conductor.Conservation of Local Charge

Charges present on an isolated conductor that are conserved, i.e. the net charge on the conductor, remain constant.Potential of a Conductor

The surface of a conductor is an equipotential, since the electric field is zero and the conductor behaves like an equipotential surface.

Charges in a Sphere:Let us investigate the charge distribution in a hollow sphere. First, lat us see how the charges are distributed on a sphere; by Newton's shell theorem, the charge distributed over a sphere can be considered as a point charge at the center of the sphere with equal magnitude. Thus, we can consider the electric field to be pointing outward/inward depending on the charge the sphere is carrying.

Charge Distribution between Two Charged Plates:Consider the below situation:

There are two charged plates with initial charge $Q_1$ and $Q_2,$ respectively. To find the charges inside and outside respectively of both the charges, we use the above-mentioned rules along with Gauss' law.

Using Rule 1, net electric field inside both the plates should be zero, i.e. $\vec E = 0:$ $\begin{aligned} \oint_\gamma \vec E\cdot \mathrm d\vec a &= {1\over \epsilon_0}Q_\text{encl}\\ 0=\oint_\gamma 0~\mathrm d\vec a &= {1\over \epsilon_0}Q_\text{encl}\\ \Rightarrow Q_\text{encl} &= 0. \end{aligned}$ But the enclosed charge amounts to $q_2+q_3$. Thus $q_2 = -q_3 =q_\text{in}\implies \boxed{q_2 = q_\text{in},~q_3 = -q_\text{in}}.$

Now, at point $P,$ net charge is zero. The charge due to a large charged plate is given by $\vec E = {q\over 2\epsilon_0}\hat r.$ So, net field at $P$ is $\begin{aligned} \vec {E_1}+\vec {E_2}+\vec {E_3}+\vec {E_4} &= {q_1\over 2\epsilon_0}+{q_{in}\over 2\epsilon_0}-{q_{in}\over 2\epsilon_0} - {q_4\over 2\epsilon_0} \\ &= 0\\\\ &\Rightarrow \boxed{q_1 = q_4=q_\text{out}}. \end{aligned}$ So, the situation becomes as follows:

Now, using Rule 2 on both of the charges, we have $\begin{aligned} q_\text{out}+q_\text{in} = Q_1 &\Rightarrow\boxed{q_\text{out} = \dfrac{Q_1+Q_2} 2}\\ q_\text{out}-q_\text{in}=Q_2 &\Rightarrow\boxed{q_\text{in} = \dfrac{Q_1-Q_2}2}.\ _\square \end{aligned}$

Charge distribution between two charged plates of which either/both of them is/are earthed

## Rearrangement of Charges under the Influence of Electric Field

[Anyone interested can do it]

What happens to the charges in an object when an electrified body is brought towards it? Charges present in an object can rearrange themselves under the influence of an electric field, depending upon the nature of the field.

PolarizationWhen an uncharged object is placed in an electric field, its charges rearrange themselves. The electrons move away from the direction of the field, while the positive charges are displaced towards the electric field. This effect is known as polarization. The charges align themselves in the following manner:

Dielectrics

Capacitors

$(A)$ is the distribution of electrical charges in an object, being influenced by a nearby electrified body $(B).$ What are the object $(A)$ and the type of charge on $(B),$ respectively?

In the above diagram,## Applications

**Electrostatics** is the field of physics and especially electrodynamics that has many examples that can be seen in real life. Out of all of them, *lightning* and the *Van de Graaff generator* are a couple, one of which is natural while the other is one of the most ingenious human inventions ever.

We are going to discuss both of them in detail below, and hopefully show you the utter beauty of science, especially physics.

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 previously discussed topics that 2 bodies which have a potential difference will always cause charges 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 between the clouds and the ground gets partially

ionized, i.e. it allows the 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 because both the bodies posses a very high amount of charges, 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, but rather it is one of the millions of possibilities that a lightning surge could have taken!

The Van De Graaff Generator

So, the next application on our list is the

amazingVan de Graff generator. According to me, it is one of the best inventions by the human mind, but its principle is pretty simple.So, picture this: you want to create a very high potential barrier or a very high potential difference between an object and its surroundings. If you would go the traditional way of accumulating the charge from the outside of the object, it would get

reallyharder as you go on, as you'd have to face increasing amount repulsions. To overcome this problem, and to create large potential differences, that could be used in various fields; for example, in accelerating a subatomic particle, we use this device.The working of Van de Graaff generator is pretty cool, to be honest. What happens is that we use the principle of the motion of electric charge from a higher potential to lower potential. In the above diagram, you can see a pointed object

Dat the bottom and a pointed objectEat the top. These are known as brushes and they help in collecting or depositing charges.The brush

Ddeposits a positive charge on an insulated and rotating sheet, which then transports this charge to the brushE, which in turn lets the charge flow through it, and deposits it on the larger sphere in the end. In this way, we can create a potential difference of millions of volts, and only stop when we reach the breakdown voltage of the environment surrounding the apparatus.Now the question arises, "Why won't we face the similar repulsions as before?"

That's a perfectly logical assumption to make, but the problem is that, now, we are transporting the charge not against the charge on the bigger sphere (as all its charge is on its outer surface). Moreover, with a little more mathematical manipulation, you can clearly show that the potential of the bigger sphere will

alwaysbe greater than the inner pulleys, in case of a positive charge.

In the end, if you are even a teensy bit interested, I'd highly recommend you to check out this collection of electrostatic experiments and lectures by a professor at MIT, who's really good at what he does. (Trust me! :P)