Energy distribution laws of black body radiation

1.) Wein’s laws of Energy distributions→

A.) Wein's Fifth Power law→ The total amount of the energy emitted by a black body per unit volume at an absolute temperature in the wavelength range $\lambda$ and $\lambda + d\lambda$ is given as

$E\lambda \cdot d\lambda= \frac{A}{\lambda^{5}}f\left ( \lambda T \right ) \cdot d\lambda \qquad (1)$

Where $A$ is a constant and $f(\lambda T)$ is a function of the product $\lambda T$ and is given as

$ f\left ( \lambda T\right )=e^-\frac{hc}{\lambda kT}\qquad (2)$

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

$E_\lambda \cdot d\lambda = \frac{A}{\lambda ^{5}}e^\frac{-hc}{\lambda kT} \cdot d\lambda$

$E_\lambda \cdot d \lambda = A \lambda ^{-5} e^\frac{-hc}{\lambda kT} \cdot d \lambda$

Wien’s law energy distribution explains the energy distribution at the short wavelength at higher temperatures and fails for long wavelengths.

B.) Wein's Displacement law→ As the temperature of the body is raised the maximum energy shift toward the shorter wavelength i.e.

$\lambda_{m} \times T = Constant $

Where
$\lambda_m$- Wavelength at which the energy is maximum
$T$-Absolute temperature

Thus, if radiation of a particular wavelength at a certain temperature is adiabatically altered to another wavelength then temperature changes in the inverse ratio.

2.) Rayleigh-Jean’s law→ The total amount of energy emitted by a black body per unit volume at an absolute temperature T in the wavelength range $\lambda $ and $\lambda +d\lambda $ is given as

$E_{\lambda}.d\lambda = \frac{8\pi kt}{\lambda ^{4}}.d\lambda$

Where K– Boltzmann’s Constant which has valve $ 1.381\times 10^{23}\frac{J}{K}$

This law, explains the energy distribution at the longer wavelength at all temperatures and fails totally for the shorter wavelength.

Note→ The energy distribution curves of the black body show a peak while going towards the ultraviolet wavelength (shorter $ \lambda $) and then fall while Rayleigh-Jeans law indicates continuous rise only. This is the failure of classical physics.

3.) Stefan-Boltzmann Law→ The total amount of heat radiated by a perfectly black body per unit area per second is directly proportional to the fourth power of its absolute temperature $(T)$. i.e.

$E \propto T^{4}$

$E = \sigma T^{4}$

Where $\sigma$= Stefan’s Constant which has value $5.67\times 10^{-8} W-\frac{K^{4}}{m^{2}}$

It is a black body at absolute temperature $T$ is surrounded by another black body at absolute temperature $T_{0}$, The net amount of heat $E$ lost by the former per second per $cm^{2}$ is→

$E=\sigma (T^{4}-T_{0}^{^{4}})$

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Relation between group velocity and phase velocity

We know that phase velocity

$V_{p}=\frac{\omega }{k}$

$\omega =V_{p}.k \qquad(1)$

And group velocity

$V_{g}=\frac{d\omega}{dk} \qquad(2)$

Substitute the value of $\omega$ from equation$(1)$ in equation $(2)$

$V_{g}=\frac{d}{dk}(V_{p}.k)$

$V_{g}=V_{p}+k.\frac{dV_{p}}{dk}$

$V_{p}=V_{p}+k.\frac{dV_{p}}{d\lambda}.\frac{d\lambda }{dk} \qquad (3)$

But

$\lambda=\frac{2\pi }{k}$

The above equation can be obtain from following formula i.e. $k=\frac{2\pi}{\lambda }$

Now put the value of $\lambda$ in equation $(3)$

$V_{g}=V_{p}+k\frac{dV_{p}}{d\lambda}\frac{d}{dk}(\frac{2\pi }{k}$

$V_{g}=V_{p}+k\frac{dV_{p}}{d\lambda}(\frac{-2\pi }{k^{2}}$

$V_{g}=V_{p}-\frac{2\pi}{k}\frac{dV_{p}}{d\lambda }$

$V_{g}=V_{p}-\lambda\frac{dV_{p}}{d\lambda }$

Thus, the above equation represents the relation between group velocity and phase velocity.

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Electric Line of Force and its Properties

Electric Field:

That region around any charged particle (source-charge) in which other test charged particle experience an electric force (attraction or repulsion). This region is called electric field of that charged particle.

The size and magnitude of the test charged particle is very small i.e tends to zero which does not modify the electric field of source charged particle. The electric field can only experience, not be seen by eyes. The electric field is defined by imaginary lines which are called electric field lines.

Electric Field Lines:

Electric field lines are imaginary lines that describe the behavior of electric fields. The behavior of the electric field tells about the force acting on free test-charged particles. This force gives the motion to the free test charge particle which is placed in the path of the electric field line. So the test charged particle moves in the direction of the electric field line. Hence, the direction of electric force and electric field lines both are the same. Thus electric field line is also known as electric lines of force.

Electric Lines of Force:

An electric line of force is defined as an imaginary smooth curve drawn in electric field along which a free isolated test charged particle moves. The tangent drawn at any point on the electric line of force gives the direction of the force acting on a test charged particle placed at that point.

Properties of Electric Line of Force:

There are the following properties of the electric line of force:

1.) The direction of the electric line of force is outward for positively charged particles, and inward for negatively charged particles.

Electric field line of positive charge

Electric field line of negative charge

So when two opposite charge are placed near each other, the electric line of force start from a positive charge and end at a negative charge

Electric lines of force of unlike charges

When two same nature charges are placed near each other, the field produced by one charge is equal and opposite to the field produced by the other charges at the midpoint of the line joining the two charges. Therefore, at this point, the resultant electric field is zero. This point is called the neutral point.

Electric lines of force of like charges

2.) The tangent drawn at any point on the line of force gives the direction of the force acting on a test-charged particle at that point.

3.) Two electric lines of force can never intersect with each other because if they do so then at the point of intersection two tangents can be drawn which would mean that there are two directions of the force at that point which is impossible.

4.) When electric lines of force are closer to each other, the electric field is stronger in this region. And where The line of force are far away from each other then the electric field is weaker in this region. When the line of force are parallel and in a straight line to each other, the electric field is known as the uniform electric field.

5.) The electric line of force has a tendency to contract in length like a stretched elastic string for the opposite nature of the charged particle. It is due to the attraction force between charged particles.

6.) The electric line of force has a tendency to separate from each other in the direction perpendicular to their length for the same nature of charged particles. It is due to the repulsion force between charged particles.

7.) The electric lines of force can never form closed loops as a line can never start and end on the same charge.

8.) The electric line of force also gives us an indication of the equipotential surface (That surface which has the same potential on each point of the surface).

9.) The electric line of force always flows from higher potential to lower potential.

10.) In a region where there is no electric field, lines are absent so the electric field inside the conductor is zero. Electric lines of force end or start normally from the surface of a conductor.

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Electric Charge and its properties

What is an electric charge?

Electric charge is an intrinsic property of elementary particles (e.g., electrons, protons, neutrons, etc.) of any substance that gives rise to electric force between them when they get closer.

Types of charge:

There are two types of electric charge

1.) Positive Electric Charge like Proton
2.) Negative Electric Charge like Electron

The same nature of charged particles repels each other and the opposite nature of charged particles attracts each other. For example, electrons and electrons repel each other, and electrons and protons attract each other. Similarly, proton and proton repel each other.

Generally, any substance is electrically neutral because the number of electrons and protons are equal in it but according to the Free electron Model theory, these substances are classified into three categories. These are:

1.) Conductor
2.) Semiconductor
3.) Insulator

The electric force is generated in conductors and insulating substances but not in semiconductor substances because these semiconductor substances have four electrons in their valence shells.

Generation of Electric force in Conductor: The electric force is generated in conductors due to the transfer of electrons from one conducting substance to another conducting substance. There are various methods for the transfer of electrons in conductors, like contact, friction, and induction. Therefore, When electrons are transferred from one conductor to another conductor then the conductor has deficiency or excess of electrons, these conductors are called electrostatically charged.

Electrostatic charged Conductors:

 The charge on the conductors due to the transfer of electrons between the conductors is known as electrostatic charged conductor.

There are two types of electrostatic-charged conductors.

1.) Positive Electrostatic Charged Conductor:

When conductors have a deficiency of electrons i.e. the number of electrons is less than protons, these conductors are called positive electrostatic charged conductors.

2.) Negative Electrostatic Charged Conductor:

When conductors have excessive electrons i.e. number of electrons is greater than protons, these conductors are called negative electrostatic charged conductors.

Generation of Electric force in Insulator: Insulators are charged by induction this charging occurs due to the polarization of atoms of insulating substances. In the absence of an external electric field, the center of mass of the nucleus and electron cloud in an atom lies on the same point but when an insulating substance is placed in an external electric field, the center of mass of the nucleus and electron cloud in an atom is separated from each other at a very small distance. The center of mass of the electron cloud is shifted toward the positive side of the external electric field and the center of mass of the nucleus is shifted toward the negative side of the external electric field. thus these atoms are polarized. When insulators are electrically charged, these polarized atoms store the electric potential energy.

Basic Properties of Electric Charge:

1.) Additive Nature of Charge:

The total charge on any system is always equal to the algebraic sum of all the charges in the system.

If a system contains $'n'$ number of point charges, the total charge of a system will be the sum of $'n'$ number of point charges.

Let us consider, a system containing $'n'$ number of point charges like $q_{1},q_{2},q_{3}......q_{n}$, the total charge of the system will be

$Q=q_{1}+q_{2}+q_{3},.....+q_{n}$

$Q=\sum_{i}^{n}q_{i}$

2.) Conservation of Charge:

The total Charge of the isolated system is always conserved i.e.

The electric charges neither can be created nor destroyed i.e. The electric charges are generated due to transfer of electrons from one subtance to another substance.

3.) Quantization of Charge:

The electric charge is always an integer multiple of $‘e’$. It is known as the quantization of charge.

$Q=\pm ne$

Where $'n'$ is an integer number.

4.) The mass of charged particle depends on velocity:

The velocity of the charged particle is significant to the speed of light so the mass of the charged particle depends on the velocity. This variation can be found by following the formula:

$m=\frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

Where
$m_{0}$ - Rest mass of particle
$m$ - Relative mass of the particle

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Electric Potential and Derivation of electric potential at a point due to point charged particle

Electric Potential:

When a test-charged particle is brought from infinity to a point in the electric field then the work done per unit test charge particle is called electric potential. It is represented by $V$. It is a scalar quantity.

Let's consider a test-charged particle $q_{0}$ bring from infinity to at a point $P$ in the electric field. If the work done by test charged particle is $W$ then electric potential →

$V=\frac{W}{q_{0}}$

Unit of Electric Potential: $Joul/Coulomb$  OR  $N-m/Ampere-sec$

In MKS: $Kg-m^{2}-Ampere^{-1}-sec^{-3}$

Dimension of Electric potential: $[ML^{2}A^{-1}T^{-3}]$

The electric potential at a point due to point charged particle:

Let us consider, a source point charge $+q$ is placed in air and vacuum at point $O$.Let's take a point $P$ at distance $r$ from the source point charged particle. Here the test-charged particle $+q_{0}$ is brought from infinity to point $P$.If the test-charged particle moves a very small distance $dx$ from point $A$ to $B$ against the electrostatic force. So electrostatic force at point $A$ which is placed at a distance $x$ from point $O$ →

$F=\frac{1}{4\pi\epsilon{0}} \frac{qq_{0}}{x^{2}} \qquad(1)$    

Electric potential due to point charge

The work is done against the electrostatic force $\overrightarrow{F}$ to move small distance $dx$ from point $A$ to Point $B$

$ dW=\overrightarrow{F}\: \overrightarrow{dx}$

$ dW=F\: dx \: cos 180^{\circ}$

Here the angle between the electrostatic force and displacement is $180^{\circ}$. So work done

$ dW=- F\: dx$

$ dW=-\frac{1}{4\pi\epsilon{0}} \frac{qq_{0}}{x^{2}}\:dx \qquad \left\{ from\: equation \: (1) \right\}$

The total work is done in moving the charge $q_{0}$ from infinity to the point $P$ will be

$W=-\int_{0}^{W}{dW}$

Here negative sign shows that the work done from infinity to at point $P$ is stored in the form of potential energy between the charges.

$ W=-\int_{\infty}^{r}{\frac{1}{4\pi\epsilon{0}} \frac{qq_{0}}{x^{2}}\:dx }$

$ W=-\frac{qq_{0}}{4\pi\epsilon{0}} \int_{\infty}^{r}{\frac{dx}{x^{2}} }$

$W=-\frac{qq_{0}}{4\pi\epsilon{0}} \left [\frac{-1}{x} \right ]_{\infty}^{r}$

$ W=-\frac{qq_{0}}{4\pi\epsilon{0}} \left [-\frac{1}{r}-\frac{-1}{\infty} \right ]$

$W=\frac{qq_{0}}{4\pi\epsilon{0}} \left [\frac{1}{r} \right ]$

$ W=\frac{1}{4\pi\epsilon{0}} \left [\frac{qq_{0}}{r} \right ]$

Hence, the work is done to move a unit test charge from infinity to the point $P$, or the electric potential at point $P$ is →

$V=\frac{W}{q_{0}}$

$V=\frac{1}{4\pi\epsilon{0}} \frac{q}{r}$

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Construction and Working of Nuclear Reactor or Atomic Pile

A nuclear reactor is a device within which a self-sustaining controlled chain reaction is produced by fissionable material. it is thus a source of control energy that is utilized for several useful purposes. The reactor has some important part which is given below:

1. Fuel: The fassionable material such as Uranium-235 and Plutonium-239 known as fuel. These materials play an important role in operating the nuclear reactor.


2. Moderator: It slows down the neutrons to thermal energies through the elastic collision between its nuclei and fission neutrons. Thermal neutrons have a very high probability of fissioning Uranium-235 nuclei. Examples: heavy water graphite beryllium oxide. Heavy water is the best moderator.


3. Control Rods: These rods are used to control the fission rate in the reactor. these Rods are fixed in the reactor walls. These rods are made up of the material of cadmium and Boron. These materials are good absorbers of slow neutrons. Therefore when the rods are pushed into the reactor, the fission rate decreases, and when they are pulled out, the fission grows.


4. Coolant: The coolant is used to remove the heat from the reactor which is produced inside the reactor. For this purpose air-water or CO₂ is passed through the reactor.


5. Shield: various type of intense rays are emitted from the reactor, which may be injurious to the people working near the reactor. To protect them from this radiation, thick concrete walls are erected around the reactor.


6. Safety device: In case of an accident or other emergency, a special set of control rods, called " shut-off rods" enter the reactor automatically. They immediately absorb the neutrons so that the chain reaction stops entirely.

Construction:

It is made up of a large number of uranium rods which are placed in calculated geometrical lattice between layers of pure graphite ( moderator) blocks. To prevent oxidation of Uranium, the rods are covered by close-fitting aluminum cylinders. The control rods are so inserted within the lattice that they will be raised or lowered between the Uranium rods whenever necessary. The whole reactor is encircled by a concrete shield.

Working:

The actual operation of the reactor is started by pulling out the control rods so that they do not absorb many neutrons. Then, the stray neutrons, which are always present in the tractor, start fissioning the U-235 nuclei. In each fission, two or three fast neutrons are produced. These neutrons repeatedly strike the moderator and slowed down. Then, These neutrons start fissioning the U-235 nuclei. So, a chain-reaction of fission starts. The number of neutrons, which is produced in fissioning, is controlled by pushing the cadmium rods into the reactor. These rods absorb a number of the neutrons. Thus, the energy produced in the reactor is kept under control to avoid any explosion. The coolant is pond pumped through the reactor to carry away The Heat generated by the fission of uranium nuclei. the hot CO₂ passes through the heat exchanger and convert cold water into steam. This steam is used to operate turbines for generating electricity.

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Einstein’s Mass Energy Relation Derivation

Einstein’s Mass-Energy Relation: Einstein's mass energy relation gives the relation between mass and energy. It is also knows as mass-energy equivalence principle.

According to Newtonian mechanics, Newton’s second law

$f=\frac{dP}{dt}$

Where $P$ is the momentum of the particle. So put $P=mv$ in above equation:

$f=\frac{d}{dt}\left ( mv \right )\quad\quad (1)$

According to theory of relativity, mass of the particle varies with velocity so above equation $(1)$ can be written as:

$f=m \frac{dv}{dt}+v\frac{dm}{dt}\quad\quad (2)$

When the particle is displaced through a distance $dx$ by the applied force $F$. Then the increase in kinetic energy $dk$ of the particle is given by

$dk= Fdx\quad\quad (3)$

Now substituting the value of force $F$ in equation $(3)$

$dk =m\frac{dv}{dt}\cdot dx+v\frac{dm}{dt}\cdot dx \quad (4) $

$dk=mv\cdot dv +v^{2}\cdot dm \:\: (5) \: \left \{ \because \frac{dx}{dt}=v \right \}$

The variation of mass with velocity equation

$m=\frac{m_{\circ}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\quad\quad (6) $

Square both sides in above equation:

$m^{2}=\frac{m_\circ^{2}}{1-\frac{v^{2}}{c^{2}}}$

$m^{2}c^{2}-m^{2}v^{2}=m_\circ ^{2}c^{2}$

Differentiate the above equation which can be written as

$2mdm\cdot c^{2}- 2m \cdot dm \cdot v^{2}-2v\cdot dv\cdot m^{2}=0$

$c^{2}dm-v^{2}dm-vm\cdot dv$

$c^{2}dm=v^{2}dm+mv\cdot dv \quad\quad (7)$

Now substitute the value of $dk$ from equation $(5)$ in equation $(7)$. So above equation can be written as:

$dk = c^{2}dm$

Now consider that the particle is at rest initially and by the application of force it acquires a velocity $v$. The mass of body increase from ${m_{\circ}}$ to $m$. The total kinetic energy acquired by the particle is given by

$dk = \int_{m_\circ}^{m}c^{2}\cdot dm$

$k = c^{2}\left ( m-m_{\circ} \right )$

$k = mc^{2} - m_\circ c^{2}$

$k+m_\circ c^{2} = mc^{2}$

Where $k$ is the kinetic energy of the particle and $m_\circ c^{2}$ is the rest mass-energy of the particle. So The sum of these energies is equal to the total energy of the particle $E$. So

$E= m c^{2}$

Where $E$ is the total energy of the particle.

The above equation is called the mass energy equivalence equation.

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Concept of Simultaneity in Special Relativity

Concept of Simultaneity (Relative character of Time ): The interval aren't the same for two observes in relative motion. This cause an important incontrovertible fact that two events that appear to happen simultaneously to at least one observer are not simultaneous to another observer in relative motion.

Suppose two events occur (or two-time bombs explode) at different places $x_{1}$ and $x_{2}$ but at the same time $t_{0}$ with respect to an observer in a stationary frame (or on the ground). The situation of the different to an observer in moving frame $S'$ or to a pilot of a spaceship moving with velocity $v$ relative to stationary frame $S$ (or ground). To him, according to Lorentz transformation for time. The explosion at $x_{1}$ occurs at

$t'_{1} = \frac{t_{0} -x_{1}\frac{v}{c^{2}}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$ x_{2}$ occurs at

$t'_{2} = \frac{t_{0}- x_{2}\frac{v}{c^{2}}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

Hence the two events (explosions) that occur simultaneously to one observer in the stationary frame are separated to another observer in a moving frame by an interval of

$t'_{2}-t'_{1}=\frac{\left ( x_{1}-x_{2} \right )\frac{v}{c^{2}}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

Therefore, the principle of simultaneity in relativity is an absolute concept for two events. It depends on an observer or a frame of reference. The effect is not due to the time dilation. Hence, we conclude that there is no such thing as “absolute time” which is the same for all observers.

“Time is relative and it varies for all observers in relative motion.”

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Assumptions of Planck’s Radiation Law

Planck in 1900 suggested the correct explanation of the black body radiation curve. They gave the following assumption →

1. A chamber contains black body energy radiation and simple harmonic oscillators (atoms of Wall, i.e. Black lamp & Platinum coating inside wall, behave as oscillators or resonators) of molecular dimensions which can vibrate with all possible frequencies.


2. The frequency of energy radiation emitted by an oscillator is the same as the frequency of its vibration.


3. An oscillator cannot emit or absorb the energy in a continuous manner it can emit or absorb energy in a small unit (packet) called Quanta.

If an oscillator is vibrating with a frequency $ \nu $ it can only radiate in quanta of magnitude $h\nu $ i.e. “The oscillator can have only discrete energy value $E_{n}$ ” given by–

$E_{n}=nh\nu$

Where
$n$ – an integer
$h$– Planck ’s constant and the value is $6.626\times10^{-34} J-s$

The average energy of Planck’s oscillator of frequency $\nu$ -

$E_{\lambda}d\lambda = \frac{8\pi hc}{\lambda ^{5}} \frac{d\lambda }{(e^{\frac{hc}{\lambda kT}}-1)}$

$E_{\nu}d\nu= \frac{8\pi h\nu^{3}}{c^{3}}\frac{d\nu }{(e^{\frac{h\nu}{kt}}-1)}$

This assumption is most revolutionary in character. This implies that the exchange of energy between radiation and matter (Black lamp or platinum Coating ) cannot take place continuously but are limited to a discrete set of value $ 0, h\nu, 2h\nu, 3h\nu,------ nh \nu $.

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Heisenberg uncertainty principle

If the x-coordinate of the position of a particle is known to an accuracy of $\delta x$, then the x-component of momentum cannot be determined to an accuracy better than $\Delta P_{x}\approx \frac{\hbar }{\Delta x}$.

$\Delta P_{x}. \Delta x\approx \hbar$

The above inequality must be satisfied

$\Delta P_{x}. \Delta x\geqslant \hbar$

Where $\hbar $ - Planck’s Constant

This is the Uncertainty principle with macroscopic objects. Exact statement of the Uncertainty principle →

The product of the uncertainties in determining the position and momentum of the particle can never be smaller than the number of the order $\frac{\hbar }{2}$.

$\Delta P_{x}. \Delta x\geqslant \frac{\hbar}{2}$

Where $\delta x$ and $\delta P $ are defined as the root mean square deviation from their mean values.

The Uncertainty principle can also describe by the following formula →

$\Delta x.\Delta p_{x}\approx \frac{\hbar}{2}$

$\Delta x.\Delta p_{x}\geqslant \frac{\hbar}{2}$

$\Delta x.\Delta p_{x}\geqslant \frac{h}{4\pi }$

Expression for $y$ and $z$ component →

$\Delta y.\Delta p_{y}\geqslant \frac{h}{4\pi }$

$\Delta z.\Delta p_{z}\geqslant \frac{h}{4\pi }$

The uncertainty relation between energy and time →

$\Delta E.\Delta t\geqslant \frac{h}{4\pi }$

$\Delta E.\Delta t\geqslant \frac{\hbar }{2 }$

The uncertainty relation between momentum and Angular Position→

$\Delta L.\Delta \theta \geqslant \frac{h }{4\pi }$

$\Delta L.\Delta \theta \geqslant \frac{\hbar}{2}$

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The electric potential energy of a system of Charges

The Potential Energy of a system of two-point like charges→

When the system of two charged particles is configured, in which one charge is at rest of position and another is brought from infinity to near the first charge then the work done acquire by this charged particle is stored in the form of electric potential energy between these charges.

Derivation→

Let us consider, If two charge $q_{1}$ and $q_{2}$ in which one charge $q_{1}$ is at the rest of the position at point $P_{1}$ and another charge $q_{2}$ is brought from infinity to a point $P_{2}$ to configure the system then the electric potential at point $P_{2}$ due to charge particle $q_{1}$ →

$V=\frac{1}{4\pi\epsilon_{0}}\frac{q_{1}}{r}$

Where $r$ is the distance between the point $P_{1}$ and Point $P_{2}$

Here, Charge $q_{2}$ is moved in from infinity to point $P_{2}$ then the work required is →

$W=V q_{2}$

$W= \frac{1}{4\pi\epsilon_{0}}\frac{q_{1}q_{2}}{r}$

Since the electric potential at infinity is zero the work- done will also be zero. So total work-done from infinity to a point $P$ will be stored in the form of electric potential energy.

$U=W$

$ U= \frac{1}{4\pi\epsilon_{0}}\frac{q_{1}q_{2}}{r}$

The electric potential energy of a system of three-point-like charges→

To obtain the potential energy of a system of three charges. First, Obtain the work done between any two charges and then obtain the different work done for both those charges from the third charge, and then the total work done will be equal to electric potential energy.

Let us consider a system is made up of three charges $q_{1}$, $q_{2}$ and $q_{3}$ which are placed at point $P_{1}$,$P_{2}$ and $P_{3}$. Now the work done between two charges $q_{1}$ and $q_{2}$ is

$W_{1}=\frac{1}{4\pi\epsilon_{0}}\frac{q_{1}q_{2}}{r_{12}}$

Now, the charge $q_{3}$ is brought from infinity to the point $P_{3}$. Work has to be done against the forces exerted by $q_{1}$ and $q_{2}$ Therefore, The work done between charges of $q_{2}$ and $q_{3}$

$W_{2}=\frac{1}{4\pi\epsilon_{0}} \frac{q_{2}q_{3}}{r_{23}}$

Now, The work done between charges $q_{1}$ and $q_{3}$

$ W_{3}=\frac{1}{4\pi\epsilon_{0}} \frac{q_{1}q_{3}}{r_{13}}$

The total work done to make a system of three charges:

$ W=W_{1}+W_{2}++W_{3}$

Now substitute the value of $W_{1}$, $W_{2}$and $W_{2}$ in above equation i.e.

$ W=\frac{1}{4\pi\epsilon_{0}} \left[\frac{q_{1}q_{2}}{r_{12}}+\frac{q_{1}q_{3}}{r_{13}}+\frac{q_{2}q_{3}}{r_{23}}\right]$

This work is stored in the form of electric potential energy in the system.

$U=W$

$ U=\frac{1}{4\pi\epsilon_{0}} \left[\frac{q_{1}q_{2}}{r_{12}}+ \frac{q_{1}q_{3}}{r_{13}}+\frac{q_{2}q_{3}}{r_{23}}\right]$

Similarly, the Potential energy of a system of N point system i.e.

$ U=\frac{1}{4\pi\epsilon_{0}} \sum_{i=1}^{N}\sum_{j=1}^{N}\frac{q_{i}q_{j}}{r_{ij}}$

Here $i\neq j$

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Gauss's Law for Electric Flux and Derivation

Gauss's Law:

Gauss's law for electric flux is given by Carl Friedrich Gauss in 1813. He extended the work of Joseph-Louis Lagrange. This formula was first formulated in 1713 by Lagrange. Gauss's law stated that:

The electric flux passing normal through any closed hypothetical surface is always equal to the $\frac{1}{\epsilon_{0}}$ times of the total charge enclosed within that closed surface. This closed hypothetical surface is known as Gaussian surface.

Let us consider that a $+q$ coulomb charge is enclosed within the Gaussian's surface. Then according to Gauss's Law, the electric flux will be:

$\phi _{E}= \frac{q}{\epsilon_{0}}$

The electric flux of the electric field →

$\phi_{E}=\oint \overrightarrow{E}\cdot\overrightarrow{dA}$

Substitute this value of electric flux $\phi_{E}$ in the above formula so we get →

$\oint \overrightarrow{E}\cdot\overrightarrow{dA}=\frac{q}{\epsilon_{0}}$

Where $\epsilon_{0}$ → Permittivity of the free space

The above formula of Gauss's law is applicable only under the following two conditions:

1.) The electric field at every point on the surface is either perpendicular or tangential.
2.) The magnitude of the electric field at every point where it is perpendicular to the surface has a constant value.

Derivation of Gauss's law from Coulomb's law:

1.) When the charge is within the surface
2.) When the charge is outside the surface

1. When the charge is within the surface:

Let a charge $+q$ is placed at point $O$ within a closed surface of irregular shape. Consider a point $P$ on the surface which is at a distance $r$ from the point $O$. Now take a small element or area $\overrightarrow{dA}$ around the point $P$. If $\theta$ is the angle between $\overrightarrow{E}$ and $\overrightarrow{dA}$ then electric flux through small element or area $\overrightarrow{dA}$

$d\phi_{E}=\overrightarrow{E}\cdot\overrightarrow{dA}$

$d\phi_{E}=E\:dA\:cos\theta \qquad\quad\quad (1)$

When charge is inside the surface

According to Coulomb's law, the electric field intensity $E$  at point $P$.

$E=\frac{1}{4\pi\epsilon_{0}}\frac{q}{r^{2}}$

Now substitute the value of electric field intensity $E$ in equation $(1)$

$d\phi_{E}=\frac{q}{4\pi\epsilon_{0}}\frac{dA\:cos\theta}{r^{2}}$

but $\frac{dA\:cos\theta}{r^{2}}$ is the solid angle $d\omega$ subtended by $dA$ at point $O$. Hence the above equation can be written as

$d\phi_{E}=\frac{q}{4\pi\epsilon_{0}}d\omega$

So, The total flux $\phi_{E}$ over the entire surface can be found by integrating the above equation

$\oint d\phi_{E}= \frac{q}{4\pi\epsilon_{0}}\oint d\omega$

For entire surface solid angle $d\omega$ will be equal to $4\pi$ i.e. $d\omega=4\pi$

$\phi _{E}= \frac{q}{\epsilon_{0}}$

If the closed surface enclosed with several charges like $q_{1},q_{2},q_{3},.....-q_{1},-q_{2},-q_{3},.....$. Now each charge will contribute to the total electric flux $\phi_{E}$.

$\phi_{E}= \frac{1}{\epsilon_{0}}\left [ q_{1}+q_{2}+q_{3}...-q_{1}-q_{2}-q_{3}... \right ]$

Here $\quad q=q_{i}-q_{j}$

$\phi_{E}= \frac{1}{\epsilon_{0}}\sum_{i=1,j=1}^{n}(q_{i}-q_{j})$

$\phi_{E}= \frac{1}{\epsilon_{0}}\sum q$

Where $\sum q$ → Algebraic Sum of all the charges

2. When the charge is outside the surface:

Let a point charge $+q$ be situated at point $O$ outside the closed surface. Now a cone of solid angle $d\omega$ from point $O$ cuts the surface area $dA_{1}$, $dA_{2}$ at point $P$ and $Q$ respectively. The electric flux for an outward normal is positive while for inward normal is negative so

The electric flux at point $P$ through an area

$d\phi_{1}$= $-\left (\frac{q}{4\pi \epsilon_{0}} \right )d\omega$

The electric flux at point $Q$ through area

$d\phi_{2}$= $+\left (\frac{q}{4\pi \epsilon_{0}} \right )d\omega$

The Total electric flux will be sum of all the electric flux passing through areas of surface →

$\phi_{E}=d\phi_{1}+d\phi_{2}$

$\phi_{E}=-\left ( \frac{q}{4\pi \epsilon_{0}} \right )d\omega+\left ( \frac{q}{4\pi \epsilon_{0}} \right )d\omega $

$\phi_{E}=0$

The above equation is true for all cones from point $O$ through any surface, however irregular it may be-

The total electric flux over the entire surface due to an external charge is zero.

This verifies Gauss's law.

Application of Gauss's law:

There are following some important application given below:

1. Electric field intensity due to a point charge


2. Electric field intensity due to uniformly charged spherical Shell (for Thin and Thick)


3. Electric field intensity due to a uniformly charged solid sphere (Conducting and Non-conducting)


4. Electric field intensity due to uniformly charged infinite plane sheet (for Thin and Thick)


5. Electric field intensity due to uniformly charged parallel sheet


6. Electric field intensity due to charged infinite length wire

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Vector Form of Coulomb's Law

Derivation of vector form of Coulomb's law:

Let us consider, Two-point charges $+q_{1}$ and $+q_{2}$ are separated at a distance $r$ (magnitude only) in a vacuum as shown in the figure given below.

Let $\overrightarrow{F_{12}}$ is the force on charge $+q_{1}$ due to charge $+q_{2}$ and $\overrightarrow{F_{21}}$ is the force on charge $+q_{2}$ due to charge $+q_{1}$. Then

$\overrightarrow{F_{12}}=\frac{1}{4\pi \varepsilon _{0}}\frac{q_{1}q_{2}}{r^2}\:\:\hat{r_{21}}\qquad(1)$

Where $\widehat{r}_{21}$ ➝ Unit Vector Pointing from charge $+q_{2}$ to charge $+q_{1}$

$\overrightarrow{F_{21}}=\frac{1}{4\pi \varepsilon _{0}}\frac{q_{1}q_{2}}{r^2}\:\:\hat{r_{12}}\qquad(2)$

Where$\widehat{r}_{12}$ ➝ Unit Vector Pointing from charge $+q_{1}$ to charge $+q_{2}$

From the above figure, we can conclude that the direction of unit vector $\widehat{r}_{12}$ and $\widehat{r}_{21}$ is opposite. i.e.

$\hat{r_{12}}=-\hat{r_{21}}\qquad(3)$

So from equation $(2)$ and equation $(3)$, we can write as

$\overrightarrow{F_{21}}=-\frac{1}{4\pi \varepsilon _{0}}\frac{q_{1}q_{2}}{r^2}\:\:\hat{r_{21}}\qquad(4)$

Now, Put the value of equation $(1)$ in equation $(4)$. So equation $(4)$, we can write as

$\overrightarrow{F_{21}}=-\overrightarrow{F_{12}}\qquad (5)$

The above equation $(5)$ shows that " The Coulomb's force is Action and Reaction Pair. This force acts on different bodies." If

$\overrightarrow{F_{12}}=\overrightarrow{F_{21}}=\overrightarrow{F}$

And

$ \hat{r_{12}}=\hat{r_{21}}=\hat{r}$

Then generalized vector form of Coulomb's Law$\overrightarrow{F}=\frac{1}{4\pi\varepsilon _{0}}\frac{q_{1}q_{2}}{r^2}\:\hat{r}$

Where $\hat{r}=\frac{\overrightarrow{r}}{r}$

$ \overrightarrow{F}=\frac{1}{4\pi \varepsilon _{0}}\frac{q_{1}q_{2}}{r^3}\:\overrightarrow{r}$

Where $\overrightarrow{r}$ is displacement vector

This is a generalized vector form of Coulomb's law.

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