Equation of continuity for current density

We know the equation of continuity is

$\overrightarrow{\nabla}. \overrightarrow{J}+ \frac{\partial \rho}{\partial t}=0 \qquad(1)$

According to Maxwell's first differential equation

$\overrightarrow{\nabla}. \overrightarrow{D}=\rho \qquad(2)$

From equation $(1)$ and equation$(2)$

$\overrightarrow{\nabla}. \overrightarrow{J}+ \frac{\partial }{\partial t}(\overrightarrow{\nabla}. \overrightarrow{D})=0$

$\overrightarrow{\nabla}. (\overrightarrow{J}+ \frac{\partial \overrightarrow{D} }{\partial t}) =0 $ Where the term →
$(\overrightarrow{J}+ \frac{\partial \overrightarrow{D} }{\partial t})$ → solenoidal vector and it is also regarded as total current density for time varying electric field.
$D$ → The displacement vector
$\frac{\partial \overrightarrow{D} }{\partial t}$ → Displacement current density

The above equation is known as the "Equation of continuity for current density".

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Equation of continuity of electromagnetic wave

Definition:

The mathematical representation of the law of conservation of charge in differential form is called the "continuity equation".

Mathematical representation of Equation of continuity:

If $\overrightarrow{J}$ is the current density of a closed surface $\overrightarrow{S}$ then the current through a closed surface is

$i=\oint_{S} \overrightarrow{J}. \overrightarrow{dS} \qquad(1)$

Let $V$ be the volume enclosed by the surface $S$. So the total charge in this volume-

$q=\oint_{V} \rho. dV \qquad(2)$

By the law of conservation of charge i.e. "Charge can neither be created nor destroyed". If some charge flows out from the volume per unit time giving rise to current density, the charge in the volume decreases at the same rate. So the current

$i=-\frac{\partial q}{\partial t}$

$i=-\frac{\partial}{\partial t} (\oint_{V} \rho. dV) \qquad (from \: equation(2) \: )$

$i=-\oint_{V} \frac{\partial \rho}{\partial t} dV \qquad(3)$

from equation $(1)$ and equation $(3)$

$\oint_{S} \overrightarrow{J}. \overrightarrow{dS}= -\oint_{V} \frac{\partial \rho}{\partial t} dV \qquad(4)$

According to Gauss's divergence theorem-

$\oint_{S} \overrightarrow{J}. \overrightarrow{dS}= \oint_{V} (\overrightarrow{\nabla}. \overrightarrow{J}) dV \qquad(5)$

From equation $(4)$ and equation $(5)$

$\oint_{V} (\overrightarrow{\nabla}. \overrightarrow{J}) dV = -\oint_{V} \frac{\partial \rho}{\partial t} dV$

$\oint_{V} (\overrightarrow{\nabla}. \overrightarrow{J}) dV + \oint_{V} \frac{\partial \rho}{\partial t} dV =0 $

$\oint_{V} [(\overrightarrow{\nabla}. \overrightarrow{J}) + \frac{\partial \rho}{\partial t}] dV =0 $

$\overrightarrow{\nabla}. \overrightarrow{J} + \frac{\partial \rho}{\partial t} =0 $

This equation is known as the equation of continuity and it is based on the conservation of charge.

For the study state $\frac{\partial{\rho}}{\partial{t}}=0$

$\overrightarrow{\nabla}. \overrightarrow{J}=0$

This means that in the steady state, there is no source or sink of current.

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Derivation of Maxwell's forth equation

Maxwell's fourth equation is the differential form of Ampere's circuital law.

$\overrightarrow{\nabla} \times \overrightarrow{H} = \overrightarrow{J} + \frac{\partial{\overrightarrow{D}}}{\partial{t}}$

Derivation:

According to Ampere's circuital law

$\oint_{l} \overrightarrow{B}. \overrightarrow{dl}=\mu_{0} i \qquad(1)$

According to Stroke's theorem-

$\oint_{l} \overrightarrow{B}.\overrightarrow{dl}=\oint_{S} (\overrightarrow{\nabla} \times \overrightarrow{B}). \overrightarrow{dS} \qquad(2)$

From equation$(1)$ and equation$(2)$

$\oint_{S} (\overrightarrow{\nabla} \times \overrightarrow{B}). \overrightarrow{dS} = \mu_{0} i \qquad(3)$

where $i=\oint_{S} \overrightarrow{J}. \overrightarrow{dS} \qquad(4)$

So from equation$(3)$ and equation$(4)$

$\oint_{S} (\overrightarrow{\nabla} \times \overrightarrow{B}). \overrightarrow{dS} = \mu_{0} \oint_{S} \overrightarrow{J}. \overrightarrow{dS}$

$\oint_{S} (\overrightarrow{\nabla} \times \overrightarrow{B}). \overrightarrow{dS} - \mu_{0} \oint_{S} \overrightarrow{J}. \overrightarrow{dS}=0$

$\oint_{S} [(\overrightarrow{\nabla} \times \overrightarrow{B}) - \mu_{0} \overrightarrow{J}]. \overrightarrow{dS}=0$

$(\overrightarrow{\nabla} \times \overrightarrow{B}) - \mu_{0} \overrightarrow{J}=0$

$\overrightarrow{\nabla} \times \overrightarrow{B} = \mu_{0} \overrightarrow{J}$

We know that $\overrightarrow{B}=\mu_{0} \overrightarrow{H}$. So the above equation can be again written as-

$\overrightarrow{\nabla} \times \overrightarrow{H} = \overrightarrow{J}$

Modified Maxwell's fourth equation:

The modified Maxwell's fourth equation is the differential form of the modified Ampere's circuital law.

We know the modified Ampere's circuital law-

$\oint_{l} \overrightarrow{B}. \overrightarrow{dl}=\mu_{0} i + i_{d}$

Where $i_{d}$ - Displacement current

So the derivation of the modified Maxwell equation is similar to the above derivation. Therefore the modified Maxwell's fourth equation can be written as-

$\overrightarrow{\nabla} \times \overrightarrow{H} = \overrightarrow{J} + \overrightarrow{J_{d}} \qquad(1)$

Where $\overrightarrow{J_{d}}$ - Displacement current density

And the value of $\overrightarrow{J_{d}}$ is

$\overrightarrow{J_{d}}=\epsilon_{0} \frac{\partial{\overrightarrow{E}}}{\partial{t}}$

$\overrightarrow{J_{d}}=\frac{\partial{\overrightarrow{D}}}{\partial{t}} \qquad(\because \overrightarrow{D}=\epsilon_{0} \overrightarrow{E})$

Now substitute the value of $\overrightarrow{J_{d}}$ in equation $(1)$, then

$\overrightarrow{\nabla} \times \overrightarrow{H} = \overrightarrow{J} + \frac{\partial{\overrightarrow{D}}}{\partial{t}}$

This is modified by Maxwell's fourth equation.

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Derivation of Maxwell's third equation

Maxwell's third equation is the differential form of Faraday's law induction.i.e

$\overrightarrow{\nabla} \times \overrightarrow{E}=- \frac{\partial{\overrightarrow{B}}}{\partial{t}}$

Derivation:

According to Faraday's Induced law-

$e=-\frac{\partial{\phi_{B}}}{\partial{t}} \qquad(1)$

According to Gauss's law of magnetism-

$\phi_{B}=\oint_{S} \overrightarrow{B}.\overrightarrow{dS} \qquad(2)$

Now substitute the value of $\phi_{B}$ in equation $(1)$

$e=-\frac{\partial}{\partial{t}} \oint_{S} \overrightarrow{B}.\overrightarrow{dS}$

$e=-\oint_{S} \frac{\partial{\overrightarrow{B}}}{\partial{t}}.\overrightarrow{dS} \qquad(3)$

The line integral of the electric field around a closed loop is called electromotive force. Thus

$e=\oint_{l} \overrightarrow{E}.\overrightarrow{dl} \qquad(4)$

from equation $(3)$ and $(4)$

$\oint_{l} \overrightarrow{E}.\overrightarrow{dl}=-\oint_{S} \frac{\partial{\overrightarrow{B}}}{\partial{t}}.\overrightarrow{dS} \qquad(5)$

According to Stroke's Theorem-

$\oint_{l} \overrightarrow{E}.\overrightarrow{dl}=\oint_{S} (\overrightarrow{\nabla} \times \overrightarrow{E}).\overrightarrow{dS} \qquad(6)$

from equation $(5)$ and equation $(6)$

$\oint_{S} (\overrightarrow{\nabla} \times \overrightarrow{E}).\overrightarrow{dS}=-\oint_{S} \frac{\partial{\overrightarrow{B}}}{\partial{t}}.\overrightarrow{dS}$

$\oint_{S} [(\overrightarrow{\nabla} \times \overrightarrow{E})+ \frac{\partial{\overrightarrow{B}}}{\partial{t}}].\overrightarrow{dS}=0$

If the surface is arbitrary then-

$(\overrightarrow{\nabla} \times \overrightarrow{E})+ \frac{\partial{\overrightarrow{B}}}{\partial{t}}=0$

$\overrightarrow{\nabla} \times \overrightarrow{E}=- \frac{\partial{\overrightarrow{B}}}{\partial{t}}$

This is Maxwell's third equation.

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Derivation of Maxwell's second equation

Maxwell's second equation is the differential form of Gauss's law of magnetism.

As magnetic, monopoles do not exist in magnets and the magnetic field lines form closed loops. There is no source of the magnetic field from which the lines will either only diverge or only converge. Hence the divergence of the magnetic field is zero.

$\overrightarrow{\nabla}. \overrightarrow{B}=0$

Derivation-

According to Gauss's law of magnetism

$\oint_{S} \overrightarrow{B}. \overrightarrow{dS}=0 \qquad(1)$

Now apply the Gauss's divergence theorem-

$\oint_{S} \overrightarrow{B}. \overrightarrow{dS}= \oint_{v} \overrightarrow{\nabla}.\overrightarrow{B}.dV \qquad (2)$

from equation $(1)$ equation $(2)$

$\oint_{v} (\overrightarrow{\nabla}.\overrightarrow{B}).dV =0$

$\overrightarrow{\nabla}.\overrightarrow{B} =0$

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Derivation of Maxwell's first equation

Maxwell's first equation is the differential form of Gauss's law of electrostatics.i.e

$\overrightarrow{\nabla}.\overrightarrow{E}= \frac{\rho}{\epsilon_{0}} $

Derivation:

According to Gauss's law for electrostatic-

$\oint_{s} \overrightarrow{E}.\overrightarrow{dS}=\frac{q}{\epsilon_{0}} \qquad(1)$

For continuous charge distribution inside the surface-

$q=\oint_{v}\rho.dV$

Where
$\rho$→Charge density
dV→Small volume

Now substitute the value of $q$ in equation $(1)$ then

$\oint_{s}\overrightarrow{E}.\overrightarrow{dS}=\frac{1}{\epsilon_{0}} \oint_{v}\rho.dV \qquad(2)$

Now according to Gauss's divergence theorem-

$\oint_{s} \overrightarrow{E}.\overrightarrow{dS}= \oint_{v} \overrightarrow{\nabla}.\overrightarrow{E} dV \qquad (3)$

From equation$(2)$ and equation$(3)$, we can write the above equation-

$\oint_{v} \overrightarrow{\nabla}.\overrightarrow{E} dV= \frac{1}{\epsilon_{0}} \oint_{v}\rho.dV $

$\oint_{v} \overrightarrow{\nabla}.\overrightarrow{E} dV- \frac{1}{\epsilon_{0}} \oint_{v}\rho.dV=0 $

$\oint_{v} (\overrightarrow{\nabla}.\overrightarrow{E}- \frac{\rho}{\epsilon_{0}})dV=0 $

On solving the above equation-

$\overrightarrow{\nabla}.\overrightarrow{E}- \frac{\rho}{\epsilon_{0}}=0 $

$\overrightarrow{\nabla}.\overrightarrow{E}= \frac{\rho}{\epsilon_{0}} $

This is Maxwell's first equation.

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Conservative force and non conservative force

Conservative force: There are the following points that describe the conservative force-

1.) The conservative force depends only on the position of the particle and does not depend on the path of the particle.

2.) In a conservative force, the kinetic energy of the particle does not change between the positions.

Let us consider a particle is moving from position $A$ to position $B$ under the conservative force. If the kinetic energy of the particle at position $A$ and $B$ is $K_{i}$ ,$K_{f}$ respectively then for conservative force-

$K_{i}=K_{f}$

3.) In conservative force, the work done by the force in completing one round between any two positions is zero.

According to the work-energy theorem

$W=\Delta{K}$

$W=K_{f}-K_{i} \quad(1)$

For a conservative force, the kinetic energy of the particle does not change between the positions of the particle i.e., $ K_{f}=K_{i}$. So from equation $(1)$

$W=0$

Alternative Method:

The above statement can also be proven by the following method-

The work is done by the conservative force to move a particle from position $A$ to position $B$ = $W_{AB}$.

The work done by the conservative force to move a particle from position $B$ to position $A$ = $W_{BA}$

We know that the kinetic energy of the particle does not change between the positions, so the work done by the force between the positions will be equal and opposite. i.e

$W_{AB}=-W_{BA}$

$W_{AB}+W_{BA}=0$

So, from above, we can conclude that the net work done by force between the two positions in completing one round is zero.
4.) The central force is also known as the conservative force.
5.) The conservative force is always equal to the negative gradient of potential energy.

$\overrightarrow{F}=-\overrightarrow{\nabla}U$

$\overrightarrow{F}= -\frac{dU}{dr}$

Non-conservative force:

1.) The non-conservative force is path-dependent between the two positions and does not depend on the positions of the particle.

2.) The kinetic energy of the particle varies from one position to another due to friction.

3.) For a non-conservative force, the work done by the force in completing one round between two positions is not zero.

4.) This is not a central force.

5.) The non-conservative force does not equal the negative gradient of potential energy.

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Biot Savart's Law and Equation

Biot-Savart Law: Biot-Savart law was discovered in 1820 by two physicists Jeans-Baptiste Biot and Felix Savart. According to this law:

1. The magnetic field is directly proportional to the length of the current element.
   
   
   $dB \propto dl \qquad (1)$


2. The magnetic field is directly proportional to the current flowing in the conductor.
   
   
   $dB \propto i \qquad (2)$


3. The magnetic field is inversely proportional to the square of the distance between length of the current element $dl$ and point $P$ (This is that point where the magnetic field has to calculate).
   
   
   $dB \propto \frac{1}{r^{2}} \qquad (3)$


4. The magnetic field is directly proportional to the angle of sine. This angle is the angle between the length of the current element $dl$ and the line joining to the length of the current element $dl$ and point $P$.
   
   
   $dB \propto sin\theta \qquad (4)$

From equation $(1)$,$(2)$,$(3)$,$(4)$ :

$\qquad dB \propto \frac{i dl sin\theta}{r^{2}}$

Now replace the proportional sign with the constant i.e. $\frac{\mu_{0}}{4 \pi}$. Therefore the above given equation can be written as

$ dB = \frac{\mu_{0}}{4 \pi} \frac{i dl sin\theta}{r^{2}}$

The magnetic field at point $P$ due to entire conductor:-

$ B =\frac{\mu_{0}}{4 \pi} \int \frac{i dl sin\theta}{r^{2}}$

Case$(1)$: If $\theta=0^{\circ}$ then the magnetic field will be zero from the above equation i.e.

$B=0$

Case$(2)$: If $\theta=90^{\circ}$ then the magnetic field will be maximum from the above equation i.e.

$B =\frac{\mu_{0}}{4 \pi} \int \frac{i dl}{r^{2}}$.

The vector form of Biot-Savart magnetic field equation is:-

$ \overrightarrow{B} =\frac{\mu_{0} i}{4 \pi} \int \frac{ \overrightarrow{dl} \times \overrightarrow{r}}{r^{3}}$

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Applications of Lasers

Description of Application of Lasers:

There are widespread applications of lasers in various disciplines such as medicine, industries, astronomy, communication, chemistry, etc. Some of the laser applications are given below in short:

1.) Lasers in Medicine: Some of the applications of lasers in medical care such as in:

• Controlling haemorrhage.
• Treatment of the liver and lungs and for the elimination of moles and tumours developing on the skin tissues.
• Therapy and stomatology.
• Microsurgery for virtually painless treatment.
• Ophthalmology to reattach a detached retina.
• Penetration of blood vessels in the eye for treating glaucoma.
• Treatment of cancer.
• Dentistry etc.

2.) Lasers in Industries: Some of the industrial applications of lasers are as follows:

• Testing the quality of optical components such as lenses, prism, gratings etc.
• More accurate measurement of the sizes of physical quantities, precision length measurement.
• Gelling of extremely fine holes in various substances, such as in paper clips, teeth, diamonds, human hair, etc.
• Cutting of different types of hard materials.
• Technical motion picture photography.
• Detection of fingerprints.
• High-power laser eyes are useful in welding small metal points, such as in the field of electronics and microelectronics thermocouple welding to a substrate etc.
• Used in vaporizing materials for subsequent deposition on a substrate.
• Rock-crushing and boring tunnels.

3.) Lasers in astronomy: Lasers are useful in radio telescopes to exchange and extend their range of observation and in amplification of very faint radio signals from space. The application of lasers is to record the bursts of light and radiation waves from stars etc.

4.) Atmospheric Optics: Lasers are used for remote probing of the atmosphere including, the measurement of traces of pollutant gases, temperature, water, vapour concentration etc. Laser radar provides the distribution of atmospheric pollutants in different vertical sections.

5.) Lasers in Biology: Lasers are useful in micro Raman spectroscopic analysis for biological and biomedical samples available only in very small quantities. Argon-ion Laser is used to obtain scattering spectra of a wide range of biological materials.

6.) Laser in Ranging: Lasers are used in finding the accurate position of a distant object and also make it possible to determine the size and shape of your object and its orientation. Lasers are useful in the measurement of the velocity of moving objects. Laser fluouresensors are used to monitor remote environments.

7.) Lasers in Communication: Lasers are very useful in transmitting a large volume of signals over long distances. The communication capacity of typical light is about 10⁶ greater than that of typical microwaves. Due to the fact that optical frequencies are extremely large as compared to conventional radio waves and microwaves, a light beam acting as a carrier wave can carry more information in comparison to radio waves and microwaves. Lasers are thus more efficient in long-distance communication. Due to the coherence and monochromaticity of laser beams, these are used in signal modulation. Glass fibres are used for the transmission of light waves employing the principle of total internal reflection. Laser is useful in communication with Earth satellites, in rocketry, etc.

8.) Lasers in Chemistry: Lasers are used in chemistry in different ways such as in:

• Isotope separation for the enrichment of Uranium reactor fuel.
• The study of the nature of chemical bonds.
• Trace analysis of gases.
• Detection of the small number of atoms produced by the interaction of low-energy solar neutrinos.
• Microelectronic designing and fabrication.
• Triggering chemical and photochemical reactions
• Information regarding the presence of a trace of metals in various tissues etc.
• Thermo nuclear fusion reactor.

9.) Lasers in spatial frequency filtering: lasers are used in:

• Contrast enhancement of the image.
• Detecting random error in a periodic structure.
• Character recognition problems are used to identify certain objects of interest in optical images of forests in military defence.
• Removing the dot patterns of the images.

10.) Lasers in Holography: Although the principle of holography was laid down by Gabor in 1948, but holography gained practical importance after the development of the laser in 1960. Due to high monochromaticity and spatial coherence properties lasers are used in holography. Holography is a method of recording information from a three-dimensional object in such a way that a three-dimensional image may subsequently be reconstructed. In holography, the photographic plate for holograms is illuminated by two laser beams simultaneously, one carrier beam and the other object beam. The hologram contains the detail of both the amplitude and phase of light received from different parts of the three-dimensional object. The holographic technique using a laser beam can be used for the examination of complicated shapes with diffusely reflecting surfaces which is not possible when ordinary light sources are used. The computer memories using laser as a source have very high storage capacity (~10¹⁰ bits/mm³) with rapid access and are easy to align and less subject to problems of vibration than other optical memories. A holographic memory records and reads out a large number of bits simultaneously.

11.) Lasers in the Military: Lasers are useful in various military applications such as:

• Range finders, forget accurate information about the range to improve the first hit probability.
• Beam rider, for the guidance of weapons by making the weapon remain within the beams, All the way to get the target.
• The simulator assimilates lasers and simulates all aspects of the firing situation to ensure that the enemy is destroyed.
• In communication with submarines or satellites.
• Used in cruise missiles towards the high value- targets such as ships, command bunkers, bridges dams etc.

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Significance of Compton effect

Description of Significance of Compton effect:

There are the following significance of the Compton effect→

1. The greatest significance of the Compton effect is that is to provide final and deciding proof for Planck-Einstein's visualization of the quantum nature of radiation. The particle nature of light was established after the discovery of the Compton effect.
2. The discovery of the Compton effect led to the formulation of quantum mechanics by W. Heisenberg and E. Schrodinger and provided the basis for the beginning of the theory of quantum electrodynamics.
3. It is most important to radiobiology, as it happens to be the most probable interaction of high energy x-ray with atomic nuclei in living beings and is applied in radiation therapy.
4. It is used to prove the wave function of electrons in the matter in the momentum representation.
5. It is the most effective in Gamma spectroscopy that gives rise to Compton edge, as it is possible for gamma rays to scatter out of the detectors used.
6. The Compton effect has played a significant role in diverse scientific areas such as nuclear engineering, experimental and theoretical nuclear physics, atomic physics, plasma physics, x-ray crystallography, etc.
7. The Compton effect provides an important research tool in some branches of medicine, including molecular chemistry, solid-state physics, etc.
8. The Compton effect has an appropriate application in the measurement of lungs density in living organisms.
9. The Compton effect is useful in putting large detectors in orbit above the earth's atmosphere.
10. The development of a high-resolution semiconductor radiation detector opened a new area for the application of Compton scattering.

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Coulomb′s Law and Applications

Coulomb’s Law: This law was first published by French physicist Charles-Augustin de Coulomb in the year 1785. According to Coulomb’s Law-

The electric force acting between the two point charges is directly proportional to the product of magnitude of the two charges and inversely proportional to square of the distance between these two charges. The electric force always acts along the line joining the charges.

Coulomb's force between the two charges

Let us consider two positive charges whose magnitudes $q_{1}$ and $q_{2}$ are placed at a distance $‘r’$. According to Coulomb’s Law (magnitude only):

$F\propto q_{1}q_{2} \qquad(1)$

$F\propto \frac{1}{r^{2}} \qquad(2)$

From equation$(1)$ and equation$(2)$, we can write as:

$F\propto \frac{q_{1}q_{2}}{r^{2}} \qquad(3)$

$F=\frac{1}{4\pi \varepsilon K} \frac{q_{1}q_{2}}{r^{2}} \qquad(4)$

Where
$\epsilon$= Permittivity of any medium,
$K$ = Dielectric constant

For air and vacuum:
$\epsilon= \epsilon_{0}$ and $K=1$

So Coulomb’s Law for Air and Vacuum:

$F=\frac{1}{4\pi \varepsilon_{0}}\frac{q_{1}q_{2}}{r^{2}} \qquad(5)$

Where $\varepsilon _{0}=8.0854 \times 10^{-12} \:\: C^{2}N^{-1}m^{-2}$

Then, the value of $\frac{1}{4\pi \varepsilon_{0}}=9\times10^{9} N-m^{2}/C^{2}$ So, From Equation$(5)$

$F=9\times 10^{9} \frac{q_{1}q_{2}}{r^{2}}$

This is the equation of Coulomb's Law that applies to the medium of air or vacuum. The above equation of Coulomb Law shows only the magnitude value of electrostatic force.

Properties of Coulomb's law:-

There are the following properties of Coulomb's law:-

1.) The Coulomb force is an action and reaction pair and follows Newton's third law.
2.) The Coulomb force is a conservative force.
3.) The Coulomb force is a central force i.e. it is always acting along the line joining between two charges.
4.) If the net force is zero then momentum will be conserved.
5.) If the center of mass is at rest and momentum is conserved then it follows the mass conservation law.
6.) The force between two charges is independent of the presence or absence of other charges but the net force increases on that particular charge.

Limitations of Coulomb's Law:

1.) This law does not apply to moving charges, i.e., it applies to static charges (charges at rest), and charges must be stationary relative to each other.
2.) It applies to charges of regular and smooth shape. It is very difficult to apply to irregular shapes.
3.) The charges must not overlap for example they must be distinct point charges.
4.) This law can not be directly applicable to calculate the charge on big planets.

Application of Coulomb's Law:

1.) Coulomb's law calculates the distance between the charges.
2.) Coulomb's law calculates the electrostatic force between the charges.
3.) Coulomb's law calculates the electrostatic force on a point charge due to the presence of several point charges. It is also known as the Superposition theorem of electrostatic force.

**What is $1$ Coulomb's?

Answer: When two unit charges are placed in a vacuum at one meter apart, then the force acting between charges is $9\times10^{9}$ $N$.This force is known as $1$ Coulomb.

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