Solution of electromagnetic wave equations in conducting media

The electromagnetic wave equations in conducting media:

For electric field vector:

$ \nabla^{2}.\overrightarrow{E}-\mu \epsilon\frac{\partial^{2} \overrightarrow{E}}{\partial t^{2}} - \sigma \mu \frac{\partial \overrightarrow{E}}{\partial t}=0 \qquad(1)$

For magnetic field vector:

$\nabla^{2}.\overrightarrow{B} - \mu \epsilon \frac{\partial^{2} B}{\partial t^{2}}-\sigma \mu \frac{\partial \overrightarrow{B}}{\partial t}=0 \qquad(2)$

The wave equation of electric field vector:

$\overrightarrow{E}(\overrightarrow{r},t)=E_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(3)$

The wave equation of magnetic field vector:

$\overrightarrow{B}(\overrightarrow{r},t)=B_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(4)$

Now the solution of electromagnetic wave for electric field vector.

Differentiate with respect to $t$ of equation $(3)$

$\frac{\partial \overrightarrow{E}}{\partial t}=i \omega E_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)}$

Again differentiate with respect to $t$ of the above equation:

$\frac{\partial^{2} \overrightarrow{E}}{\partial t^{2}}=i^{2} \omega^{2} E_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)}$

$\frac{\partial^{2} \overrightarrow{E}}{\partial^{2} t}=- \omega^{2} \overrightarrow{E}(\overrightarrow{r},t)$

Now substitute the value of the above equation in equation$(1)$

$\nabla^{2} \overrightarrow{E}=-\omega^{2} \mu \epsilon \overrightarrow{E} - i \omega \mu \sigma \overrightarrow{E}$

$\nabla^{2} \overrightarrow{E}=- \left( \omega^{2} \mu \epsilon + i \omega \mu \sigma \right) \overrightarrow{E}$

This is the solution of the electromagnetic wave equation in conducting media for the electric field vector.

Now component form of the above equation:

$(\frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}} +\frac{\partial^{2}}{\partial z^{2}})(\hat{i}E_{x}+\hat{j}E_{y}+\hat{k}E_{z}) \\ =- \left( \omega^{2} \mu \epsilon + i \omega \mu \sigma \right)(\hat{i}E_{x}+\hat{j}E_{y}+\hat{k}E_{z}) \qquad(5)$

If the wave is propagating along $z$ direction. Then for uniform-plane electromagnetic waves-

$\frac{\partial}{\partial x}=\frac{\partial}{\partial y}=0$

$\frac{\partial^{2}}{\partial x^{2}}=\frac{\partial^{2}}{\partial y^{2}}=0$

$E_{z}=0$

Now the equation $(5)$ can be written as:

$\frac{\partial^{2}}{\partial x^{2}} (\hat{i}E_{x}+\hat{j}E_{y}) \\ =- \left( \omega^{2} \mu \epsilon + i \omega \mu \sigma \right)(\hat{i}E_{x}+\hat{j}E_{y})$

Now separate the above equation in $x$ and $y$ components so

$\left.\begin{matrix} \frac{\partial^{2} E_{x}}{\partial z^{2}}=- \left( \omega^{2} \mu \epsilon + i \omega \mu \sigma \right) E_{x} \\ \frac{\partial^{2}E_{y}}{\partial z^{2}} =- \left( \omega^{2} \mu \epsilon + i \omega \mu \sigma \right) E_{y} \end{matrix}\right\} \quad(6)$

The solution of electromagnetic wave for magnetic field vector can find out by following the above method.

Therefore $x$ and $y$ components of the solution of the electromagnetic wave equation for magnetic field vector can be written as. i.e.

$\left.\begin{matrix} \frac{\partial^{2} B_{x}}{\partial z^{2}}=- \left( \omega^{2} \mu \epsilon + i \omega \mu \sigma \right) B_{x} \\ \frac{\partial^{2}B_{y}}{\partial z^{2}} =- \left( \omega^{2} \mu \epsilon + i \omega \mu \sigma \right) B_{y} \end{matrix}\right\} \quad(7)$

In the solution of electromagnetic wave equation $(6)$ and equation $(7)$. The term $\left( \omega^{2} \mu \epsilon + i \omega \mu \sigma \right)$ is equal to $k_{z}^{2}$. It is known as propagation constant $k_{z}$. Then

$k_{z}^{2}=\left( \omega^{2} \mu \epsilon + i \omega \mu \sigma \right) \qquad(8)$

The propagation constant is the complex quantity so

$k_{z}=\alpha+i \beta \qquad(9)$

Now from equation $(8)$ and equation $(9)$

$\left(\alpha+i \beta \right)^{2}=\left( \omega^{2} \mu \epsilon + i \omega \mu \sigma \right)$

$\alpha^{2} - \beta^{2} +2 i \alpha \beta =\left( \omega^{2} \mu \epsilon + i \omega \mu \sigma \right)$

Now separate the real and imaginary terms:

$Real \: Term \rightarrow \alpha^{2} - \beta^{2} = \omega^{2} \mu \epsilon \quad (10)$

$Imaginary \: Term \rightarrow 2 \alpha \beta = \omega \mu \sigma \quad (11)$

On solving the equation $(10)$ and equation $(11)$

$\alpha= \omega \sqrt{\frac{\mu \epsilon}{2}} \left[ 1 + \left\{ 1+ \left( \frac{\sigma}{\epsilon \omega} \right)^{2} \right\}^{1/2} \right]^{1/2} \quad(12)$

$\beta= \omega \sqrt{\frac{\mu \epsilon}{2}} \left[ \left\{ 1+ \left( \frac{\sigma}{\epsilon \omega} \right)^{2} \right\}^{1/2} -1 \right]^{1/2} \quad(13)$

The wave equation $(3)$ of the electric field vector also can be written as:

$\overrightarrow{E}(\overrightarrow{r},t)=E_{\circ} e^{-\beta \overrightarrow{r}}e^i{(\alpha \overrightarrow{r} - \omega t)} \qquad(14)$

The above equation has an additional term $e^{-\beta \overrightarrow{r}}$ compared to the purely harmonic solution.

Where
$\alpha \rightarrow$ Attenuation Constant
$\beta \rightarrow$ Absorption Coefficient and Phase Constant

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Electromagnetic Wave Equation in Conducting Media (i.e. Lossy dielectric or Partially Conducting)

Maxwell's Equations: Maxwell's equation of the electromagnetic wave is a collection of four equations i.e. Gauss's law of electrostatic, Gauss's law of magnetism, Faraday's law of electromotive force, and Ampere's Circuital law. Maxwell converted the integral form of these equations into the differential form of the equations. The differential form of these equations is known as Maxwell's equations.

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


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


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


4. $\overrightarrow{\nabla} \times \overrightarrow{B}= \mu \overrightarrow{J}$
   
   
   Modified form:
   
   
   $\overrightarrow{\nabla} \times \overrightarrow{B}= \mu \overrightarrow{J}+\mu \epsilon \frac{\partial \overrightarrow{E}}{\partial t}$

For Conducting Media:

Current density $(\overrightarrow{J}) = \sigma \overrightarrow{E} $
Volume charge distribution $(\rho)=0$
Permittivity of Conducting Media= $\epsilon$
Permeability of Conducting Media=$\mu$

Now, Maxwell's equation for Conducting Media:

$\overrightarrow{\nabla}. \overrightarrow{E}=0 \qquad(1)$
$\overrightarrow{\nabla}. \overrightarrow{B}=0 \qquad(2)$
$\overrightarrow{\nabla} \times \overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t} \qquad(3)$
$\overrightarrow{\nabla} \times \overrightarrow{H}= \overrightarrow{J}$

Modified form for Conducting Media:

$\overrightarrow{\nabla} \times \overrightarrow{B}= \mu \overrightarrow{J}+\mu \epsilon \frac{\partial \overrightarrow{E}}{\partial t}$

$\overrightarrow{\nabla} \times \overrightarrow{B}= \mu \sigma \overrightarrow{E}+ \mu \epsilon \frac{\partial \overrightarrow{E}}{\partial t} \qquad(4)$

Now, On solving Maxwell's equation for conducting media i.e perfect dielectric and lossless media, gives the electromagnetic wave equation for conducting media. The electromagnetic wave equation has both an electric field vector and a magnetic field vector. So Maxwell's equation for conducting medium gives two equations for electromagnetic waves i.e. one is for electric field vector($\overrightarrow{E}$) and the second is for magnetic field vector ($\overrightarrow{H}$).

Electromagnetic wave equation for conducting media in terms of $\overrightarrow{E}$:

Now from equation $(3)$

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

Now take the curl on both sides of the above equation$

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

$(\overrightarrow{\nabla}. \overrightarrow{E}).\overrightarrow{\nabla} - (\overrightarrow{\nabla}. \overrightarrow{\nabla}).\overrightarrow{E}=-\frac{\partial}{\partial t} (\overrightarrow{\nabla} \times \overrightarrow{B}) \qquad(5)$

We know that

$\overrightarrow{\nabla}. \overrightarrow{E}=0 $
$\overrightarrow{\nabla}.\overrightarrow{\nabla}=\nabla^{2}$
$\overrightarrow{\nabla} \times \overrightarrow{B}=\sigma \mu \overrightarrow{E} + \mu \epsilon \frac{\partial \overrightarrow{E}}{\partial t} $

Now substitute these values in equation $(5)$. So

$ -\nabla^{2}.\overrightarrow{E}=-\frac{\partial}{\partial t} \left(\sigma \mu \overrightarrow{E} + \mu \epsilon \frac{\partial \overrightarrow{E}}{\partial t} \right)$

$ -\nabla^{2}.\overrightarrow{E}=-\mu \frac{\partial}{\partial t} \left(\sigma \overrightarrow{E} + \epsilon \frac{\partial \overrightarrow{E}}{\partial t} \right)$

$ \nabla^{2}.\overrightarrow{E}=\mu \epsilon \frac{\partial^{2} \overrightarrow{E}}{\partial t^{2}} + \sigma \mu \frac{\partial \overrightarrow{E}}{\partial t} $

$ \nabla^{2}.\overrightarrow{E}-\mu \epsilon \frac{\partial^{2} \overrightarrow{E}}{\partial t^{2}} - \sigma \mu \frac{\partial \overrightarrow{E}}{\partial t}=0 $

The value of $\frac{1}{\sqrt{\mu \epsilon}}= v$. Where $v$ is the speed of the electromagnetic wave in the conducting medium. So the above equation is often written as

$ \nabla^{2}.\overrightarrow{E}-\frac{1}{v^{2}}\frac{\partial^{2} \overrightarrow{E}}{\partial t^{2}} - \sigma \mu \frac{\partial \overrightarrow{E}}{\partial t}=0 $

This is an electromagnetic wave equation for conducting media in terms of electric field vector ($\overrightarrow{E}$).

Electromagnetic wave equation for conducting media in terms of $\overrightarrow{B}$:

Now from MAxwell's equation $(4)$

$\overrightarrow{\nabla} \times \overrightarrow{B}= \mu \sigma \overrightarrow{E}+ \mu \epsilon \frac{\partial \overrightarrow{E}}{\partial t}$

Now take the curl on both sides of the above equation

$\overrightarrow{\nabla} \times (\overrightarrow{\nabla} \times \overrightarrow{B})=\overrightarrow{\nabla} \times \left( \mu \sigma \overrightarrow{E}+ \mu \epsilon \frac{\partial \overrightarrow{E}}{\partial t} \right) $

$(\overrightarrow{\nabla}. \overrightarrow{B}).\overrightarrow{\nabla} - (\overrightarrow{\nabla}. \overrightarrow{\nabla}).\overrightarrow{B} \\ =\overrightarrow{\nabla} \times \sigma \mu \overrightarrow{E}+ \mu \epsilon \left( \overrightarrow{\nabla} \times \frac{\partial \overrightarrow{E}}{\partial t} \right)$

$(\overrightarrow{\nabla}. \overrightarrow{B}).\overrightarrow{\nabla} - (\overrightarrow{\nabla}. \overrightarrow{\nabla}).\overrightarrow{B} \\ =\sigma \mu \left( \overrightarrow{\nabla} \times \overrightarrow{E} \right)+ \mu \epsilon \frac{\partial }{\partial t}\left( \overrightarrow{\nabla} \times \overrightarrow{E} \right) \qquad(6)$

We know that

$\overrightarrow{\nabla}. \overrightarrow{B}=0$
$\overrightarrow{\nabla}.\overrightarrow{\nabla}=\nabla^{2}$
$\overrightarrow{\nabla} \times \overrightarrow{E}= -\frac{\partial \overrightarrow{B}}{\partial t}$

Now substitute these values in equation $(6)$. So

$-\nabla^{2}.\overrightarrow{B}=- \sigma \mu \frac{\partial \overrightarrow{B}}{\partial t} - \mu \epsilon \frac{\partial^{2} B}{\partial t^{2}}$

$\nabla^{2}.\overrightarrow{B}= \sigma \mu \frac{\partial \overrightarrow{B}}{\partial t} + \mu \epsilon \frac{\partial^{2} B}{\partial t^{2}} $

$\nabla^{2}.\overrightarrow{B}-\sigma \mu \frac{\partial \overrightarrow{B}}{\partial t} - \mu \epsilon \frac{\partial^{2} B}{\partial t^{2}}=0 $

The value of $\frac{1}{\sqrt{\mu \epsilon}}= v$. Where $v$ is the speed of the electromagnetic wave in the conducting medium. So the above equation is often written as

$\nabla^{2}.\overrightarrow{B} - \frac{1}{v^{2}} \frac{\partial^{2} B}{\partial t^{2}}-\sigma \mu \frac{\partial \overrightarrow{B}}{\partial t}=0 $

This is an electromagnetic wave equation for conducting media in terms of electric field vector ($\overrightarrow{B}$).

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Bernoulli's Theorem and Derivation of Bernoulli's Equation

Statement of Bernoulli's Theorem:

When an ideal fluid (i.e incompressible and non-viscous Liquid or Gas) flows in streamlined motion from one place to another, then the total energy per unit volume (i.e Pressure energy + Kinetic Energy + Potential Energy) at each and every of its path is constant.

$P+\frac{1}{2}\rho v^{2} + \rho gh= constant$

Derivation of Bernoulli's Theorem Equation:

Let us consider that an incompressible and non-viscous liquid is flowing in streamlined motion through a tube $XY$ of the non-uniform cross-section.

Now Consider:

The Area of cross-section $X$ = $A_{1}$
The Area of cross-section $Y$ = $A_{2}$

The velocity per second (i.e. equal to distance) of fluid at cross-section $X$ = $v_{1}$
The velocity per second (i.e. equal to distance) of fluid at cross-section $Y$ = $v_{2}$

The Pressure of fluid at cross-section $X$ = $P_{1}$
The Pressure of fluid at cross-section $Y$ = $P_{2}$

The height of cross-section $X$ from surface = $h_{1}$
The height of cross-section $Y$ from surface = $h_{2}$

The work done per second by force on the liquid Entering the tube at $X$:

$W_{1}$ = Force $ \times $ Distance covered in one second

$W_{1}= P_{1} \times A_{1} \times v_{1} \quad \left( Force =Pressure \times Area \right)$

Similarly
The work done per second by force on the liquid leaving the tube at $Y$:

$W_{2}= P_{2} \times A_{2} \times v_{2}$

The net work done on the liquid:

$\Delta W=W_{1}-W_{2}$
$\Delta W= P_{1} \times A_{1} \times v_{1} - P_{2} \times A_{2} \times v_{2} \qquad(1)$

Now according to the principle of continuity:

$A_{1} v_{1} = A_{2} v_{2} = \frac{m}{\rho} \qquad(2)$

Now from equation $(1)$ and equation $(2)$

$\Delta W=\left( P_{1} -P_{2} \right) \frac {m}{\rho} \qquad(3)$

The kinetic energy of the fluid entering at $X$ in 1 second

$K_{1}=\frac{1}{2}mv_{1}^{2}$

The kinetic energy of the fluid leaving at $Y$ in 1 second

$K_{2}=\frac{1}{2}mv_{2}^{2}$

Therefore, The increase in kinetic energy

$\Delta K = K_{2}-K_{1}$

Now substitute the value of $K_{1}$ and $K_{2}$ in above equation then

$\Delta K = \frac{1}{2}mv_{2}^{2} - \frac{1}{2}mv_{1}^{2}$
$\Delta K = \frac{1}{2}m \left( v_{2}^{2} - v_{1}^{2} \right) \qquad(4)$

The potential energy of fluid at $X$

$U_{1}= mgh_{1}$

The potential energy of fluid at $Y$

$U_{2}= mgh_{2}$

Therefore, The decrease in potential energy

$\Delta U = U_{1}-U_{2}$

Now substitute the value of $U_{1}$ and $U_{2}$ in above equation then

$\Delta U = mgh_{1} - mgh_{2}$
$\Delta U = mg \left( h_{1} - h_{2} \right) \qquad(5)$

This increase in energy is due to the net work done on the fluid, i.e.

Net Work done = Net increase in energy

Net Work done $(\Delta W)$ = Net increase in Kinetic Energy $(\Delta K)$ - Net decrease in Potential Energy $(\Delta U)$

$\left( P_{1} -P_{2} \right) \frac {m}{\rho} = \frac{1}{2}m \left( v_{2}^{2} - v_{1}^{2} \right) - mg \left( h_{1} - h_{2} \right) $

$\left( P_{1} -P_{2} \right) = \frac{1}{2}\rho \left( v_{2}^{2} - v_{1}^{2} \right) -\rho g \left( h_{1} - h_{2} \right) $

$P_{1} + \frac{1}{2} \rho v_{1}^{2} + \rho g h_{1} = P_{2} + \frac{1}{2}\rho v_{2}^{2} + \rho g h_{2} $

$P + \frac{1}{2} \rho v^{2} + \rho g h = Constant $

This is Bernoulli's theorem equation.

Pressure Head, Velocity Head, and Gravitational Head of a Flowing Fluid:

According to Bernoulli's Theorem equation

$P + \frac{1}{2} \rho v^{2} + \rho g h = Constant $

Now dividing the above equation by $\rho g$, then we get

$\frac{P}{\rho g}+\frac{v^{2}}{2g}+ h = Constant$

Where

$\frac{P}{\rho g}$ = Pressure Head

$\frac{v^{2}}{2g}$ = Velocity Head

$h$ = Gravitational Head

The dimension of each of these three is the dimension of height. The sum of these heads is called the 'Total Head'

Therefore, Bernoulli's theorem may also state as follows:

In streamlined motion of an ideal fluid, the sum of pressure head, velocity head and gravitational head at any point is always constant.

When the fluid flows in a horizontal plane $(h_{1}=h_{2})$, then Bernoulli's equation

$P_{1} + \frac{1}{2} \rho v_{1}^{2} = P_{2} + \frac{1}{2}\rho v_{2}^{2} $

$\frac{P}{\rho g}+\frac{v^{2}}{2g} = Constant$

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Numerical Problems and Solutions of Power

Formulas and Solutions to Numerical Problems of Power

Important Formula of Power:

1.) $P=\frac{Work(W)}{Time(t)}$

2.) $P=\frac{F.s}{t}$

3.) $P=\frac{F \: s \: cos \theta}{t}$

4.) $P=\frac{F \: s }{t} \qquad \left( \because \theta =0^{\circ} \right)$

5.) $P=F \: v \qquad \left( \because v = \frac{s}{t} \right)$

Numerical Problems and Solutions

Q.1 A woman pulls a bucket of water of mass $5 Kg$ from a well which is $10 m$ deep in 10 sec. Calculate the power used by her $(g=10 \: m/sec^{2})$.

Solution:
Given that:
The mass of bucket of water $(m)=5 \: Kg$ Depth of well $(h)=10 \: m$

The time taken to pull a bucket from the well $t= 10 \: sec $

The value of gravitational acceleration $g=10 \: m/sec^{2}$

The power used $(P)=?$


Now the power used by her:


$P=\frac{W}{T}$

$P=\frac{mgh}{t} \qquad \left( \because W=mgh \right)$

Now substitute the given values in the equation of power:

$P=\frac{5 \times 10 \times 10}{10}$

$P=50 \: Joule/sec=50 W$

Q.2 A man whose mass is $50 \: Kg$ climbs up $30$ steps of the stairs in $30 \: Sec$. If each step is $20 \: cm$ high, Calculate the power used in climbing the stairs $(g=10 \: m/sec^{2})$

Solution:
Given that:
Mass of a man $(m) = 50 \: Kg$

Climbs up the number of Steps $(N) = 30$

The time is taken to climb up the 30 steps $(t)= 30 \: sec$

The length or height of one step $(l)=20 cm \: = \: 0.20 \: m$

The total length or height of 30 steps $h=30 \times 0.20 \: m$

The power used in climbing the stairs=?

Now, from the equation of power:

$P=\frac{W}{t}$

$P=\frac{mgh}{t} \qquad \left( \because W=mgh \right)$

Now substitute the given values in the above equation:

$P=\frac{50 \times 10 \times 30 \times 0.20 }{30}$

$P=100 \: Joule/sec = 100 \: W$

Q.3 A horse exerts a pull of $300 N$ on a cart so that the horse-cart system moves with a uniform speed of $18 Km/h$ on a level road. Calculate the power in watts developed by the horse and also find its equivalent in horsepower.

Solution:
Given that:
The horse exerts the pull, i.e, force $(F)=300 N$

The unifrm speed of the horse-cart $(v)=18 Km/h$

So the distance moved in $(s)=18 Km=18000m$

The time $(t)=1h= 60 \times 60 sec$

The power developed by the horse$(P)=?$

From the equation of the power, the power developed by a horse in one hour:

$P=\frac{W}{t}$ $P=\frac{F.s}{t}$

Now substitute the given values in the above equation:

$P=\frac{3000 \times 18000}{60 \times 60}$

$P=1500 W$

$P=\frac{1500}{746}$

$P=2 \: hp$

Q.4 A man weighing $60Kg$ climbs up a staircase carrying a load of $20 Kg$ on his head. The staircase has 20 steps, each of height $0.2m$. If he takes $10 sec$ to climb, find his power.

Solution:
Given that:
The weight of man $(m_{1})=60Kg$

The weight of load $(m_{2})=20Kg$

The number of steps in the staircase $(N)=20$

The height of each step $(H)=10s$

The total mass of the man and load $(m)=m_{1}+m_{2}= 60+20=80 Kg$

The total heght of stair case $(h)=N \times H = 20 \times .2= 4 m$



The power of man: $P=\frac{W}{t}$

$P=\frac{mgh}{t}$

$P=\frac{80 \times 9.8 \times 4}{10}$

$P=313.6 W$

Q.5 A car of mass $2000 Kg$ and it is lifted up a distance of $30m$ by a crane in $1 \: min$. A second crane does the same job as the first crane in $2 \: min$. Do both cranes consume the same or different amounts of fuel? Find the power supplied by each crane? Neglecting power dissipation against friction.

Solution:
Given that:
The mass of car $(m)=2000 Kg $

The lifted up distance $(h)=30m$

The time taken by first crane $(t_{1}= 1\: min)$

The time taken by second crane $(t_{2}= 2\: min)$

The work done by each crane:

$W=mgh$

$W=2000 \times 9.8 \times 30$

$W=5.88 \times 10^{5} J$

As both the cranes do the same amount of work, both consume the same amount of fuel.

The power supplied by the first crane:

$P_{1}=\frac{mgh}{t_{1}}$

$P_{1}=\frac{2000 \times 9.8 \times 30}{60}$

$P_{1}=9800 W$

The power supplied by the second crane:

$P_{2}=\frac{2000 \times 9.8 \times 30}{1.2}$

$P_{2}=4900 W$

Q.6 The human heart discharges $75 \: mL$ of blood at every beat against a pressure of $0.1 m$ of Hg. Calculate the power of the heart assuming that the pulse frequency is $80$ beats per minute. Density of $Hg=13.6 \times 10^{3} Kg/m^{3}$.

Solution:
Given that:
The volume of blood discharge per beat $(V)=75 \: mL = 75 \times 10^{-6} m^{-3} $

The pressure of blood $(P)=0.1 m \: of \: Hg$ i.e.

$P= 0.1 \times 13.6 \times 10^{3} N-m^{-2} \quad (\because P=\rho g h)$

The work done per beat $=PV$

The workdone in 80 beats$W =80 \times PV$

The power of heart:

$P=\frac{80 \times PV}{60}$

$P=\frac{80 \times 0.1 \times 13.6 \times 10^{3} \times 75 \times 10^{-6} \times 9.8}{60}$

$P=1.33 W$

Q.7 A machine gun fires $60$ bullets per minute with a velocity of $700 m/sec$. If the mass of each bullet is $50 g$, then find the power developed by the gun.

Solution:
Given that:
The mass of one bullet is $M=50 g$

The number of bullets $N=60$

The mass of $60$ bullets $m=M \times N= 50\times 60 = 300g= 3Kg$

The velocity of the bullet $v=700 m/sec$

The time take to fire $60$ bullets $t=1 \: min = 60 sec$

The power developed by gun:

$P=\frac{W}{t}$

According to the work-energy theorem:

$W=\Delta K$

$P=\frac{\Delta K}{t} $

$P=\frac{m v^{2}}{2t} \quad( \because K=\frac{mv^{2}}{2})$

$P=\frac{3 \times (700)^{2}}{2 \times 60} $

$P=12250 W $

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Variation of Mass with Velocity in Relativity

Derivation of variation of mass with velocity: Consider two systems of reference (frame of reference) $S$ and $S’$. The frame $S’$ is moving with constant velocity $v$ relative to frame $S$.

Let two bodies of masses $m_{1}$ and $m_{2}$ be traveling with velocities $u’$ and $-u’$ parallel to the x-axis in the system $S’$. Suppose the two bodies collide and after collision coalesce into one body.

The principles of conservation of mass and of momentum also hold good in relativity same as in classical mechanics. So now apply the principle of conservation of momentum.

$m_{1}u_{1}+m_{2}u_{2}=\left ( m_{1}+m_{2} \right )v\qquad(1)$

Apply the law of addition of velocities, the velocities $u_{1}$ and $u_{1}$ in the system $S$ corresponding to $u’$ and $-u’$ in frame $S’$ are given by $\rightarrow$

$u_{1}= \frac{u'+v}{1+\frac{u'v}{c^{2}}}\quad or \quad u_{2}= \frac{-u'+v}{1-\frac{u'v}{c^{2}}}\qquad(2)$

Now substitute the value of $u_{1}$ and $u_{1}$ in equation $(1)$

$ m_{1}\frac{u'+v}{\left ( 1+\frac{u'v}{c^{2}} \right )}+m_{2}\frac{-u'+v}{\left ( 1-\frac{u'v}{c^{2}} \right )}=\left ( m_{1}+m_{2} \right )v $

$ m_{1}\left [ \frac{u'+v}{1+\frac{u'v}{c^{2}}}-v \right ]=m_{2}\left [ v- \frac{-u'+v}{1-\frac{u'v}{c^{2}}} \right ]$

$ m_{1}\left [ \frac{u'+v-v-u'\frac{v^{2}}{c^{2}}}{1+u'\frac{v}{c^{2}}} \right ]=m_{2}\left [ \frac{v-u'\frac{v^{2}}{c^{2}}+u'-v}{1- u'\frac{v}{c^{2}}} \right ]$

$ \frac{m_{1}}{m_{2}}= \frac{\left ( 1+u'\frac{v}{c^{2}} \right )}{\left ( 1-u'\frac{v}{c^{2}} \right )}\qquad(3)$

From equation$(2)$

$u_{1}^{2} =\left ( \frac{u'+v}{1+\frac{u'v}{c^{2}}} \right )^{2}$

$ 1-\frac{u_{1}^{2}}{c^{2}}=1- \frac{1}{c^{2}} \left ( \frac{u'+v}{1+\frac{u'v}{c^{2}}} \right )^{2}$

$ 1-\frac{u_{1}^{2}}{c^{2}}= \frac{\left ( 1+\frac{u'v}{c^{2}} \right )^{2}-\left ( \frac{u'+v}{c} \right )^{2}}{\left ( 1+\frac{u'v}{c^{2}} \right )^{2}}$

$ 1-\frac{u_{1}^{2}}{c^{2}}= \frac{\left ( 1-\frac{u'^{2}}{c^{2}} \right )\left ( 1-\frac{v^{2}}{c^{2}} \right )}{\left ( 1+\frac{u'v}{c{2}} \right )}$

$ \left ( 1+\frac{u'v}{c^{2}} \right )^{2}=\frac{\left ( 1-\frac{u'^{2}}{c^{2}} \right )\left ( 1-\frac{v^{2}}{c^{2}} \right )}{1-\frac{u_{1}^{2}}{c^{2}}}$

$ \left ( 1+\frac{u'v}{c^{2}} \right ) =\left [ \frac{\left ( 1-\frac{u'^{2}}{c^{2}} \right )\left ( 1-\frac{v^{2}}{c^{2}} \right )}{1-\frac{u_{1}^{2}}{c^{2}}} \right ]^\frac{1}{2}\qquad(4)$

Similarly, square the velocity of $u_{2}$ and solve the as above, so

$\left ( 1-\frac{u'v}{c^{2}} \right ) =\left [ \frac{\left ( 1-\frac{u'^{2}}{c^{2}} \right )\left ( 1-\frac{v^{2}}{c^{2}} \right )}{1-\frac{u_{2}^{2}}{c^{2}}} \right ]\qquad(5)$

Now substituting the values from equation $(4)$ and equation $(5)$ in equation $(3)$

$\frac{m_{1}}{m_{2}}=\frac{\left ( 1-\frac{u_{2}^{2}}{c^{2}} \right )^{\frac{1}{2}}}{\left ( 1-\frac{u_{1}^{2}}{c^{2}} \right )^{\frac{1}{2}}}$

If the body of mass $m_{2}$ is at rest i.e. $m_{2}=m_{0}$ so velocity of the body in frame-S will be zero. i.e. $u_{2}=0$

Where $m_{0}$ is the rest mass.

$m_{1}=\frac{m_{0}}{\left ( 1-\frac{u_{1}^{2}}{c_{2}} \right )^{\frac{1}{2}}}$

Let $ m_{1}=m$ and $u_{1}=v$ so above equation

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

This is the generalized formula of variation of mass with velocity.

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Galilean Transformation Equations and Failure of Galilean Relativity

What is Transformation Equation?

A point or a particle at any instant, in space has different cartesian coordinates in the different reference systems. The equation which provide the relationship between the cartesian coordinates of two reference system are called Transformation equations.

Galilean Transformation Equation:

Let us consider, two frames $S$ and $S'$ in which frame $S'$ is moving with constant velocity $v$ relative to an inertial frame $S$. Let

1. The origin of the two frames coincide at $t=0$


2. The coordinate axes of frame $S'$ are parallel to that of the frame $S$ as shown in the figure below


3. The velocity of the frame $S'$ relative to the frame $S$ is $v$ along x-axis;

The position vector of a particle at any instant $t$ is related by the equation

$ \overrightarrow{r'}=\overrightarrow{r}-\overrightarrow{v}t\qquad (1)$

In component form, the coordinate are related by the equations

$\overrightarrow{x'}=\overrightarrow{x}-\overrightarrow{v}t;\quad y'=y; \quad z'=z\qquad (2)$

The equation $(1)$ and equation $(2)$ express the transformation of coordinates from one inertial frame to another. Hence they are referred to as Galilean transformation.

The equation $(1)$ and equation $(2)$ depending on the relative motion of two frames of reference, but it also depends upon certain assumptions regarding the nature of time and space. It is assumed that the time t is independent of any particular frame of reference. i.e. If $t$ and $t'$ be the times recorded by observers $O$ and $O'$ of an event occurring at $P$ then

$ t=t'\qquad (3)$

Now add the above assumption with transformation equation $(3)$ so the Galilean transformation equations are

$ \begin{Bmatrix} x'=x-vt\\ y'=y\\ z'=z\\ t'=t \end{Bmatrix}\qquad (4)$

The other assumption, regarding the nature of space, is that the distance between two points (or two particles) is independent of any particular frame of reference. For example if a rod has length $L$ in the frame $S$ with the end coordinates $(x_{1}, y_{1}, z_{1})$ and $(x_{2}, y_{2}, z_{2})$ then

$L=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}\quad (5)$

At the same time, the end coordinate of the rod in frame $S'$ are $(x'_{1}, y'_{1}, z'_{1})$ and $(x'_{2}, y'_{2}, z'_{2})$ then

$ L'=\sqrt{(x'_{2}-x'_{1})^{2}+(y'_{2}-y'_{1})^{2}+(z'_{2}-z'_{1})^{2}}\qquad (6)$

But for any time $t$ from equation $(4)$

$ \begin{Bmatrix} x'_{2}-x'_{1}=x_{2}-x_{1}\\ y'_{2}-y'_{1}=y_{2}-y_{1}\\ z'_{2}-z'_{1}=z_{2}-z_{1} \end{Bmatrix}\quad\quad\quad (7)$

So from equation $(5)$, equation $(6)$ and equation $(7)$, we can write as:

$L=L'$

Thus, the length or distance between two points is invariant under Galilean Transformation.

The hypothesis of Galilean Invariance:(Principle of Relativity)

The hypothesis of Galilean invariance is based on experimental observation and is stated as follows:

The basic laws of physics are identical in all reference system which move with uniform velocity with respect to one another.

OR in other words

The basics laws of physics are invariant in inertial frame.

Modify the hypothesis of Galilean Invariance by giving the following statement-

The basic law of physics are invariant in form in two reference system which are connected by Galilean Transformation

Failure of Galilean Relativity OR Galilean Transformation:

There are the following points that could not explain by Galilean transformation:

1. Galilean Transformation failed to explain the actual result of the Michelson-Morley experiment.


2. It violates the postulates of the Special theory of relativity.

According to Maxwell's electromagnetic theory, the speed of light in a vacuum is $c$ $(3\times10^{8} m/sec)$ in all directions. Let us consider a frame of reference relative to which the speed of light is $c$ in all directions, According to Galilean transformation the speed of light in any other inertial system, which is in relative motion with respect to the former, will be different in a different direction. For example- If an observer is moving with speed $v$ opposite or along with the propagation of light, The speed of light $c_{0}$ in the frame of the observer is given by

$ c_{0}=c\:\pm \: v$

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Consequences of Lorentz's Transformation Equations

Consequences:

There are two consequences of Lorentz's Transformation

1. Length Contraction (Lorentz-Fitzgerald Contraction)
2. Time Dilation (Apparent Retardation of Clocks)

Length Contraction (Lorentz-Fitzgerald Contraction): Lorentz- Fitzgerald, first time, proposed that When a body moves comparable to the velocity of light relative to a stationary observer, then the length of the body decreases along the direction of velocity. This decrease in length in the direction of motion is called 'Length Contraction'.

Expression for Length Contraction:

Let us consider two frames $S$ and $S'$ in which frame $S'$ is moving with constant velocity $v$ relative to frame $S$ along the positive x-axis direction. Let a rod is associated with frame $S'$. The rod is at rest in frame $S'$ so the actual length $l_{0}$ is measured by frame $S'$. So

$ l_{0}=x'_{2}-x'_{1}\quad\quad (1)$

Where $x'_{2}$ and $x'_{1}$ are the x-coordinate of the ends of the rod in frame $S'$. 

According to Lorentz's Transformation

$ x'_{1}=\alpha (x_{1}-vt)\quad\quad (2)$

$ x'_{2}=\alpha (x_{2}-vt)\quad\quad (3)$

Now put the value of $x'_{1}$ and $x'_{2}$ in equation $(1)$, then

$ l_{0}=\alpha (x_{2}-vt)-\alpha (x_{1}-vt)$

$ l_{0}=\alpha \left (x_{2}-vt-x_{1}+vt \right )$

$ l_{0}=\alpha \left (x_{2}- x_{1}\right )$

$ l_{0}=\alpha l $ < div>
$ l_{0}= \frac{l}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\quad\quad\left \{ \because \alpha =\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \right \}$

$l=l_{0}\sqrt{1-\frac{v^{2}}{c^{2}}}\quad$

Here $l$ is the length of the rod measured in frame $S$.

Here The factor $\sqrt{1-\frac{v^{2}}{c^{2}}}$ is less than unity. It means that

$ \sqrt{1-\frac{v^{2}}{c^{2}}}< 1 $

so

$l< l_{0}$

So the length of the rod in frame $S$ will be less than the proper length or actual length which is measured in frame $S'$.

Case: If $v=c$, Then $l=0$, i.e. a rod moving with the velocity of light will appear as a point to a stationary observer. So from the above discussion, we can conclude that in relativity there is no absolute length'.

**What is the proper length?

The length of the rod is measured by a stationary observer relative to the length of the rod in the frame.

Time Dilation (Apparent Retardation of Clocks):

Let us consider two frames $S$ and $S'$ in which frame $S'$ is moving with constant velocity $v$ relative to frame $S$ along the positive x-axis direction.

Let two events occur in frame $S$ which is at rest at time $t_{1}$ and $t_{2}$. These two event measured in frame $S'$ at time $t'_{1}$ and $t'_{2}$. So time interval between these two events in frame $S'$ is

$ t=t'_{2}-t'_{1}\qquad (1)$

According to Lorentz's Transformation

$ t'_{1}=\alpha '\left ( t_{1}-\frac{xv}{c^{2}} \right )\qquad(2)$

$ t'_{2}=\alpha '\left ( t_{2}-\frac{xv}{c^{2}} \right )\qquad(3)$

Now put the value of $t'_{1}$ and $t'_{2}$ in equation $(1)$.so

$ t=\alpha '\left ( t_{2}-\frac{xv}{c^{2}} \right )-\alpha '\left (t_{1}-\frac{xv}{c^{2}} \right )$

$ t=\alpha '\left (t_{2}-\frac{xv}{c^{2}}- t_{1}+\frac{xv}{c^{2}} \right )$

$ t=\alpha '\left (t_{2}- t_{1} \right )$

$ t=\alpha 't_{0}$

$t=\frac{t_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

Here the factor $\sqrt{1-\frac{v^{2}}{c^{2}}}$ is less then unity. i.e

$ \sqrt{1-\frac{v^{2}}{c^{2}}}< 1$

Then

$t> t_{0}$

So the time interval between two events in frame $S'$ will be longer than the time interval taken in frame $S$.

The time dilation is a real effect. All clocks will appear running slow for an observer in relative motion. It is incorrect to say that the clock in moving frame $S'$ is slow as compared to the clock in stationary frame $S$. The correct statement would be that All clocks will run slow for an observer in relative motion.

Case:

If $v=c$, then $t=∞ $ i.e. When a clock moving with the speed of light appears to be completely stopped to an observer in a stationary frame of reference.

** Proper and Non-Proper Time:

The time interval between two events that occur at the same position recorded by a clock in the frame in which the events occur (or frame at rest) is called 'proper time'.

The time interval between the same two events recorded by an observer in a frame that is moving with respect to the clock is known as 'Non -proper time or relativistic time'.

Experimental Verification of Time Dilation:

The direct experimental confirmation of time dilation is found in an experiment on cosmic ray particles called mesons. μ-mesons are created at high altitudes in the earth's atmosphere (at the height of about 10 km) by the interaction of fast cosmic-ray photons and are projected towards the earth's surface with a very high speed of about $2.994\times10^{8}$ m/s which is $0.998$ of the speed of light $c$. μ-mesons are unstable and decay into electrons or positrons with an average lifetime of about $2.0\times10^{-6}$sec. Therefore, in its lifetime a μ-mesons can travel a distance.

$ d=vt$

$ d=vt=(2.99\times10^{8})\: (2\times10^{-6})\simeq 600\:m$

$ d=0.6\:km$

Now the question arises how μ-mesons travel a distance of 10 km to reach the earth's surface. This is possible because of the time dilation effect. In fact, μ-mesons have an average lifetime $t_{0}=2.0\times10^{-6}$ sec in their own frame of reference. In the observer's frame of reference on the earth's surface, the lifetime of the μ-mesons is lengthened due to relativity effects to the value $t$ given as,

$t=\frac{t_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=\frac{2.0\times10^{-6}}{\sqrt{1-(0.998)^{2}}}=\frac{2.0\times10^{-6}}{0.063}$

$ t=3.17\times10^{-5} sec$

In this dilated lifetime, μ-mesons can travel a distance

$ d_{0}=(2.994\times10^{8})(3.17\times10^{-5})$

$ d_{0}=9500\: meter=9.5\:km$

This explains the presence of μ-mesons on the earth's surface despite their brief lifetime.

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Derivation of Lorentz Transformation's Equations

Derivation:

Let us consider two inertial frames $S$ and $S'$ in which frame $S'$ is moving with constant velocity $v$ along the positive x-axis direction relative to the frame $S$. Let $t$ and $t'$ be the time recorded in two frames. Let the origin $O$ and $O'$ of the two reference systems coincide at $t=t'=0$.

Now suppose, a source of light is situated at the origin $O$ in the frame $S$, from which a wavefront of light is emitted at time $t=0$. When the light reaches point $P$, the time required by a light signal in travelling the distance OP in the Frame $S$ is

$ t=\frac{OP}{c}$

$ t=\frac{\left (x^{2}+y^{2}+z^{2} \right )}{c}$

$ x^{2}+y^{2}+z^{2}=c^{2}t^{2}\qquad (1)$

The equation $(1)$ represents the equation of wavefront in frame $S$. According to the special theory of relativity, the velocity of light will be $c$ in the second frame $S'$. Hence in frame $S'$ the time required by the light signal in travelling the distance $O'P$ is given by

$ t'=\frac{O'P}{c}$

$ x'^{2}+y'^{2}+z'^{2}=c^{2}t^{2}\qquad (2)$

According to the Galilean transformation equation:

$ \begin{Bmatrix} x'=x-vt\\ y'=y\\ z'=z\\ t'=t \end{Bmatrix}$

Now substitute these values in equation $(2)$ then we get

$ (x-vt)^{2}+y^{2}+z^{2}=c^{2}t^{2}$

$ x^{2}+v^{2}t^{2}-2xvt+y^{2}+z^{2}-c^{2}t^{2}=0$

The above equation is certainly not same as the equation $(1)$ because it contains an extra term $(-2xvt+v^{2}t^{2})$. Thus the Galilean transformation fails.

Further $t=t'$ because $\left( t=\frac{OP}{c} \: and \: t'=\frac{O'P}{c} \right)$ which does not agree with Galilean transformation equations.

The extra term $(-2xvt+v^{2}t^{2})$ indicates that transformations in $x$ and $t$ should be modified so that this extra term is cancelled. So modification in transformation

$ x'=\alpha (x-vt) \quad for \: x'=0,\: x=vt$

$ t'=\alpha (t+fx)$

Where $α$, $α'$ and $f$ are constant to be determined for Galilean Transformations $α= α'=1$ and $f=0$. Now substituting these modified values in equation $(2)$ so

$ \alpha ^{2}(x-vt)^{2}+y^{2}+z^{2}=c^{2}\alpha'^{2}(t+fx)^{2}$

$ \alpha ^{2}(x^{2}+v^{2}t^{2}-2xvt)+y^{2}+ z^{2} =c^{2}\alpha{'}^{2}(t^{2}+f^{2}x^{2}+2fxt)$

$\alpha^{2}x^{2}+\alpha^{2}v^{2}t^{2}-2xvt\alpha^{2}+y^{2}+ z^{2} =c^{2} \alpha'^{2}t^{2}+c^{2} \alpha'^{2}f^{2}x^{2}+c^{2} \alpha'^{2}(2fxt)$

$ x^{2}(\alpha^{2}-c^{2} \alpha'^{2}f^{2})- 2xt(v\alpha^{2}+c^{2}\alpha '^{2}f)+y^{2}+ z^{2} =c^{2}t^{2}\left ( \alpha'^{2}-\frac{\alpha ^{2}v^{2}}{c^{2}} \right )\qquad(3)$

This result obtained from applying transformations from frame $S'$ to $S$ must be identical to equation $(1)$. Therefore

$ \alpha ^{2}-c^{2}\alpha '^{2}f^{2}=1\qquad (i)$

$ v\alpha ^{2}+c^{2}\alpha '^{2}f=0\qquad (ii)$

$ \alpha '^{2}-\frac{\alpha ^{2}v^{2}}{c^{2}}=1\qquad (iii)$

After solving equation $(i)$ equation $(ii)$ and equation $(iii)$, we get

$ \alpha =\alpha '=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$ $ f=-\frac{v}{c^{2}}$

Thus, The new transformation equation is invariant at the velocity of light $c$. These are:

$ \begin{Bmatrix} x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\\ y'=y\\ z'=z\\ t'=\frac{(t-\frac{vx}{c^{2}})}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \end{Bmatrix}$

These equations are called Lorentz Transformations because they were first obtained by Dutch Physicist H. Lorentz.

The above transformation equation shows that frame $S'$ is moving in positive x-direction with velocity $v$ relative to the frame $S$. But if we say that frame $S$ is moving with $v$ velocity relative to frame $S'$ along negative x-direction then the transformation is:

$ \begin{Bmatrix} x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\\ y=y'\\ z=z'\\ t=\frac{(t'+\frac{vx'}{c^{2}})}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \end{Bmatrix}$

These are known as inverse Lorentz Transformations equations.

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Force between multiple charges (Superposition principle of electrostatic forces)

Principle of Superposition for Electric force:

If a system contains a number of interacting charges, then the net force on anyone charge equals the vector sum of all the forces exerted on it by all the other charges. This is the principle of Superposition for electric force.

If a system contains n point charges $ q_{1},q_{2},q_{3}........q_{n}$. Then according to the principle of superposition, the force acting on the charge $q_{1}$ due to all the other charges

$\overrightarrow{F_{1}}=\overrightarrow{F_{12}}+\overrightarrow{F_{13}}+\overrightarrow{F_{14}}+...+\overrightarrow{F_{1n}} \qquad (1)$

Where $\overrightarrow{F_{12}}$ is the force on charge $q_{1}$ due to charge $q_{2}$, $\overrightarrow{F_{13}}$ that is due to $q_{3}$ and $\overrightarrow{F_{1n}}$ that due to $q_{n}$.

If the distance between the charges $q_{1}$ and $q_{2}$ is $\widehat{r}_{12}$ (magnitude only) and $\widehat{r}_{21}$ is unit vector from charge $q_{2}$ to $q_{1}$, then

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

Similarly, the forces on charge $q_{1}$ due to other charges are given by

$ \overrightarrow{F_{13}}=\frac{1}{4\pi \epsilon _{0}}\:\frac{q_{1}q_{3}}{r_{13}^{2}}\:\hat{r_{31}}\qquad (3)$

$.............................$

$.............................$

$ \overrightarrow{F_{1n}}=\frac{1}{4\pi \epsilon _{0}}\:\frac{q_{1}q_{n}}{r_{1n}^{2}}\:\hat{r_{n1}}\qquad (n)$

Hence, putting the value of $\overrightarrow{F_{12}},\overrightarrow{F_{13}},\overrightarrow{F_{14}}......\overrightarrow{F_{1n}}$, in equation $(1)$, the total force on charge $q_{1}$ due to all other charges is given by

$ \overrightarrow{F_{1}}=\frac{1}{4\pi \epsilon _{0}}[\:\frac{q_{1}q_{2}}{r_{12}^{2}}\:\hat{r_{21}}+\frac{q_{1}q_{3}}{r_{13}^{2}}\:\hat{r_{31}}+....\\ \quad\quad\quad ..+\frac{q_{1}q_{n}}{r_{1n}^{2}}\:\hat{r_{n1}}]$

The same procedure can be applied to finding the force on any other charge due to all the remaining charges. For example, the force on $q_{2}$ due to all the other charges is given by

$ \overrightarrow{F_{2}}=\frac{1}{4\pi \epsilon _{0}}[\:\frac{q_{2}q_{1}}{r_{21}^{2}}\:\hat{r_{12}}+\frac{q_{2}q_{3}}{r_{23}^{2}}\:\hat{r_{32}}+... \\ \quad\quad\quad ..+\frac{q_{2}q_{n}}{r_{2n}^{2}}\:\hat{r_{n2}}\:]$

Some Observations Points of Coulomb's Law:

There are the following point has been observed in Coulomb's law, that are

1. Coulomb's force between the two charges is directly proportional to the product of the magnitude of the charge.
   
   
   $ F\propto q_{1}q_{2}$


2. Coulomb's force between the two charges is inversely proportional to the square of the distance between the two charges.
   
   
   $F\propto \frac{1}{r^{2}}$


3. The electrostatic force acts between the line joining the charges. In two charges, one charge is assumed to be at rest for the calculation of the force on the second charge. So It is also known as a central force.


4. The magnitude of the electrostatic force is equal and the direction of force is opposite. So the electrostatic force is also known as the action and reaction pair.


5. The electrostatic force between two charge does not affect by the presence and absence of any other charges but the net force increase on the source charge.

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Definition and derivation of the phase velocity and group velocity of wave

Wave:

A wave is defined as a disturbance in a medium from an equilibrium condition that propagates from one region of the medium to other regions.

When such type of wave propagates in the medium a progressive change in phase takes place from one particle to the next particle.

Propagation of Wave: Wave propagation in the medium occurs with two different kinds of velocity. i.e. phase velocity and group velocity.

1. Phase Velocity:

The velocity with which plane of constant the phase of a wave propagates through the medium at a certain frequency is called the phase velocity or wave velocity.

A plane wave traveling in the positive x-direction is represented by

$y=A \: sin \omega (t-\frac{x}{v})$

Where $ \omega $ – angular frequency

$y=A \: sin(\omega t-\frac{\omega x}{v})$

$y=A \: sin(\omega t-kt) \qquad(1)$

For plane-wave $(\omega t-kx)$ is the phase of wave motion. For the plane of constant phase (wavefront). We have

$( \omega t-kx) = constant (\phi) \qquad(2)$

Differentiate with respect to time $(t)$ of the above equation

$ \omega -k.\frac{dx}{dt}=0$

$\frac{dx}{dt}=\frac{ \omega}{k}\qquad(3)$

$V_{p}=\frac{ \omega}{k} \qquad \left [ \because V_{p}= \frac{dx}{dt} \right ]$

Where $V_{p}$ - Phase Velocity of a wave

Question- Show that the phase velocity of matter-wave always exceeds the velocity of light.

Answer- Method-I

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

$V_{p}=\frac{2\pi \nu }{\frac{2\pi }{\lambda }}$

$V_{p} = \frac{2 \pi h \nu}{\frac{2 \pi h}{\lambda }}$

$V_{P} = \frac{E}{P}$

$V_{P} =\frac{mc^{2}}{mv}$

$V_{P} =\frac{c^{2}}{v}$

Method-II:

$V_{P}=\nu \lambda$

$V_{P}=\frac{mc^{2}}{h} \frac{h}{mv}$

$V_{P}=\frac{c^{2}}{v}$

Where $v$ is the velocity of matter particles.

Wave packet→

A wave packet is an envelope or packet which contains the number of plane waves having different wavelengths or wavenumbers. These numbers of waves superimpose on each other and form constructive and destructive interference over a small region of space and a resultant wave obtain. The spread of amplitude of the resultant wave with distance determines the size of the wave packet. A wave packet is also called a wave group.

Group Velocity→

The velocity of propagation of wave packet through space is known as group velocity.

Group Velocity also represents the velocity of energy flow or transmission of information in a traveling wave or wave packet. It is represented by $V_{g}$. So

$V_{g} = \frac{d\omega }{dk}$

Expression for group velocity $(V_{g})$→

Let two plane simple harmonic waves of the same amplitude and slightly different wavelength traveling simultaneously in the positive x-direction in a dispersive medium be represented by –

$y_{1} = A\: sin{(\omega t-kx)} \qquad (1)$

$y_{2}= A\:sin[(\omega +\delta \omega )t-(k+\delta k)x] \qquad (2)$

From the superposition principle the resultant displacement of waves

$y=y_{1}+y_{2}\qquad (3)$

$y=A \: sin(\omega t-kx)+ A \: sin[(\omega +\delta \omega)t-(k+\delta k)x]$

$y=A[sin(\omega t-kx)+ sin{(\omega +\delta \omega )t-(k+\delta k )x}]$

$y=2Asin(\omega t-kx)cos\frac{1}{2}(t\delta \omega –x\delta k) \qquad (4)$

This is the analytical equation for the group of waves. This equation represents the following points –

1. The sine factors represent a carrier wave that travels with the phase velocity.


2. The amplitude of the resultant wave is

$R=2A cos \frac{1}{2}(t \delta\omega-x \delta k) \qquad (5)$

(A) For maximum amplitude→

$ cos\frac{1}{2}(\delta\omega t-x\delta k)=1 \qquad (6) $

Then the resultant amplitude of the wave packet will be

$R_{m}=2A \qquad $ {from equation $(5)$ }

Where $R_{m}$ - Resultant maximum amplitude

From equation$(6)$-

$cos\frac{1}{2}(t\delta\omega –x\delta k)=1$

$cos\frac{1}{2}(t\delta\omega –x\delta k)=cos0$

$\frac{1}{2}(t\delta\omega –x\delta k)=0$

$\frac{x}{t}=\frac{\delta\omega }{\delta k}$

$V_{g}=\frac{\delta\omega }{\delta k}$

OR

$V_{g}=\frac{d\omega}{dk}$

Thus maximum amplitude i.e. the center of the wave packet moves with velocity $\frac {d\omega}{dk}$ or group velocity.

(B) For minimum amplitude→

Let a wave packet that has minimum (zero) amplitude on two successive points $x_{1}and x_{2}$. So for minimum amplitude at point $x_{1}$ –

$cos\frac{1}{2}(t.\delta \omega –x_{1}.\delta k)=0$

$cos\frac{1}{2}(t.\delta \omega –x_{1}.\delta k)=cos(2n+1)\frac{\pi }{2}$

$\frac{1}{2}(t.\delta \omega –x_{1}.\delta k)=(2n+1)\frac{\pi }{2}$

$t.\delta \omega – x_{1}.\delta k=(2n+1)\pi \qquad(1)$

Similarly minimum amplitude at the second successive point $x_{2}$-

$t\delta \omega –x_{2}.\delta k =(2n-1)\pi \qquad(2)$

Now subtract the equation $(2)$ in equation $(1)$

$(t.\delta \omega – x_{1}.\delta k) – (t.\delta \omega - x_{2}.\delta k)= (2n+1)\pi–(2n-1)\pi $

$(x_{2}-x_{1}).\delta k=2\pi$

$x_{2}-x_{1}=\frac{2\pi}{\delta k}$

$x_{2}-x_{1}=\frac{2\pi}{d(\frac{2\pi}{\lambda})} $

$\delta x=\frac{-\lambda ^{2}}{\delta  \lambda }$

Here $\delta x$ -Length of the wave packet

The above equation also can be written as-

$ dx=\frac{-\lambda^{2}}{d \lambda }$

Prove that→

$V_{g}= -\lambda^{2} \frac{d\nu }{d\lambda}$

Where $\nu $ is the frequency and $\lambda $ is the wavelength.

Proof:

We know that

$V_{g}=\frac{d\omega }{dk}$

$V_{g}=\frac{d(2\pi\nu )}{d(\frac{2\pi }{\lambda })} \qquad \left( \because \omega= 2\pi \nu \: and \: k=\frac{2\pi}{\lambda } \right)$

$V_{g}=\frac{2\pi d\nu }{2\pi d( \frac{1}{ \lambda })}$

$V_{g}=-\lambda ^{2}\frac{d\nu }{d\lambda }$

Note: A moving particle cannot be equal to a single wave train. The speed of a single wave train is called the phase velocity so moving particles are equivalent to a group of waves or wave packets.

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