Electric field intensity due to uniformly charged wire of infinite length

Derivation of electric field intensity due to the uniformly charged wire of infinite length: Let us consider a uniformly-charged (positively charged) wire of infinite length having a constant linear charge density (that is, a charge per unit length) $\lambda$ coulomb/meter. Let P is a point at a distance $r$ from the wire at which electric field $\overrightarrow{E}$ has to find.

Let us draw a coaxial Gaussian cylindrical surface of length $l$ through point $P$. By symmetry, the magnitude $E$ of the electric field will be the same at all points on this surface and directed radially outward. Thus, Now take small area elements $dA_{1}$, $dA_{2}$ and $dA_{3}$ on the Gaussian surface as shown in figure below. Therefore, the total electric flux passing through area elements is:

$ \phi_{E}= \oint \overrightarrow{E} \cdot \overrightarrow{dA_{1}} + \oint \overrightarrow{E} \cdot \overrightarrow{dA_{2}} + \oint\overrightarrow{E} \cdot \overrightarrow{dA_{3}}$

$ \phi_{E}= \oint E\: dA_{1} cos\theta_{1} + \oint E\: dA_{2} cos\theta_{2} + \oint E\: dA_{3} cos\theta_{3}$

The angle between the area element $dA_{1}$, $dA_{2}$ and $dA_{3}$ with electric field are $\theta_{1}= 0^{\circ}$ , $\theta_{2}= 90^{\circ}$, and $\theta_{3}=90^{\circ}$ respectively. So electric flux

$ \phi_{E}= \oint E\: dA_{1} cos0^{\circ} + \oint E\: dA_{2} cos90^{\circ} + \oint E\: dA_{3} cos90^{\circ}$

Here $cos \: 0^{\circ}=1$ and $cos \: 90^{\circ}=0$

Hence the above equation can be written as: $ \phi_{E}= \oint E\:dA_{1} $

The total electric flux passing through the Gaussian surface is

$ \phi_{E}= \oint{E\:dA_{1}} $

$ \phi_{E}= E\:\oint{dA_{1}} $

$ \phi_{E}= E\:\left(2\pi r l \right) \qquad \left\{\because \oint{dA_{1}} =2\pi r l \right\} $

$ \phi_{E}= E\:\left(2\pi r l \right)$

But, By Gaussian's law, The total flux $\phi_{E}$ must be equal to $\frac{q}{\epsilon_{0}}$, where $q$ is the total charge enclosed with the Gaussian surface. so that

$ \frac{q}{\epsilon_{0}}= E\:\left(2\pi r l \right)$

$ E= \frac{q}{2\pi r l \epsilon_{0}}$

For linear charge distribution → $q=\lambda l$. So substitute this value in the above equation which can be written as

$ E= \frac{\lambda l}{2\pi r l \epsilon_{0}}$

$ E= \frac{\lambda}{2\pi\epsilon_{0} r }$

The vector form of the above equation :

$\overrightarrow{E}= \frac{\lambda}{2\pi\epsilon_{0} r }\widehat{r}$

Where $\widehat{r}$ is a unit vector in the direction of $r$. The direction of $\overrightarrow{E}$ is radially outwards(for positively charged wire).

Thus, the electric field ($E$) due to the linear charge is inversely proportional to the distance ($r$) from the linear charge and its direction is outward perpendicular to the linear charge.

Special Note: A charged cylindrical conductor behaves for external points as the whole charge is distributed along its axis.

Read More

Electric field intensity due to point charge by Gauss's Law

Derivation of electric field intensity due to a point charge by Gauss's Law: Let us consider, a source point charge particle of $+q$ coulomb is placed at point $O$ in space. Let's take a point $P$ on the electric field of the source point charge particle. To find the electric field intensity $\overrightarrow{E}$ at point $P$, first put the test charge particle $+q_{0}$ on the point $P$ and draw a gaussian surface which passes through the point $P$. After that take a very small area $\overrightarrow{dA}$ around the point $P$. If the distance between the source charge particle and small area $\overrightarrow {dA}$ is $r$ then electric flux passing through the small area $\overrightarrow{dA}$ →

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

$ d\phi_{E}= E\:dA\: cos\theta$

Electric field due to point charge

from the figure, the direction between $\overrightarrow{E}$ and $\overrightarrow{dA}$ is parallel to each other i.e. the angle will be $0^{\circ}$. So the above equation can be written as →

$ d\phi_{E}= E\:dA\: cos0^{\circ}$

$ d\phi_{E}= E\:dA $

The electric flux passing through the entire Gaussian surface and be found by closed integration of the above equation →

$ d\phi_{E}= \oint {E\:dA} $

$ d\phi_{E}= E\:\oint {dA} $

$ \phi_{E}= E\left(4\pi r^{2} \right) \qquad \left\{ \because \oint {dA}=4\pi r^{2} \right\}$

According to Gauss's Law → $\phi_{E}= \frac{q}{\epsilon_{0}}$ then above equation can be written as →

$ \frac{q}{\epsilon_{0}}= E \left(4\pi r^{2} \right)$

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

The above expression is the electric field intensity due to a point source charged particle.

Read More

Normalization of the wave function of a particle in one dimension box or infinite potential well

Description of Normalization of the wave function of a particle in one dimension box or infinite potential well:

We know that the wave function for the motion of the particle along the x-axis is

$\psi_{n}(x)= A \: sin \left( \frac{n \pi x}{L} \right) \quad \left\{ Region \quad 0 \lt x \lt a \right\}$

$\psi_{n}(x)= 0 \quad \left\{ Region \quad 0 \gt x \gt a \right\}$

The total probability that the particle is somewhere in the box must be unity. Therefore,

$\int_{0}^{L} \left| \psi_{n}(x)\right|^{2}dx =1$

Now substitute the value of the wave function in the above equation. Then

$\int_{0}^{L} \left| A \: sin \left( \frac{n \pi x}{L} \right) \right|^{2}dx =1$

$\int_{0}^{L} A^{2} \: sin^{2} \left( \frac{n \pi x}{L} \right) dx =1$

$ \frac{A^{2}}{2}\int_{0}^{L} \left[ 1- cos \left( \frac{2n \pi x}{L} \right) \right] dx =1$

$ \frac{A^{2}}{2} \left[ x - \left( \frac{L}{2n\pi} \right) sin \left( \frac{2n \pi x}{L} \right) \right]_{0}^{L} =1$

$ \frac{A^{2}}{2} \left[ L - \left( \frac{L}{2n\pi} \right) sin \left( \frac{2n \pi L}{L} \right) \right] =1$

$ \frac{A^{2}}{2} \left[ L - \left( \frac{L}{2n\pi} \right) sin \left( 2n \pi \right) \right] =1$

$ \frac{A^{2}}{2} \left[ L - \left( \frac{L}{2n\pi} \right) sin \left( 2n \pi \right) \right] =1$

$ \frac{A^{2}}{2} \left[ L - 0 \right] =1 \qquad(\because sin2n\pi =0)$

$ \frac{A^{2} L}{2} =1$

$ A= \sqrt{\frac{2}{L}}$

Hence, the normalized wave function

$\psi_{n}(x)=\sqrt{\frac{2}{L}} sin \left( \frac{n \pi x}{L} \right)$

The absolute square $\left| \psi_{n}(x) \right|^{2}$ of the wave function $\psi_{n}(x)$ gives the probability density. Hence

$\left| \psi_{n}(x) \right|^{2} = \frac{2}{L} sin^{2} \left( \frac{n \pi x}{L} \right)$

The wave function for the particle in a box can be viewed in analogy with standing waves on a string. The wave function for a standing wave that has nodes at endpoints is of the form $\psi_{n}(x)= A \: sin \left( \frac{n \pi x}{L} \right)$. The condition for a standing wave can also be expressed in terms of wavelength.

$\lambda_{n}=\frac{2 \pi}{k_{n}}$

$\lambda_{n}=\frac{2 \pi}{\frac{n \pi}{L}} \qquad \left( \because k_{n}=\frac{n \pi}{L} \right)$

$\lambda_{n}=\frac{2 L}{n}$

$L= \frac{n \: \lambda_{n}}{2}$

So,

$L= \frac{\: \lambda_{1}}{2} \qquad \left( for \: n=1 \right)$

$L= \lambda_{2} \qquad \left( for \: n=2 \right)$

$L= \frac{3 \: \lambda_{3}}{2} \qquad \left( for \: n=3 \right)$

$L= 2 \lambda_{4} \qquad \left( for \: n=4 \right)$

Geo structure of wave function $\psi_{n}(x)$ and wave function's density $\left| \psi_{n}(x) \right|^{2}$.

Variation of the wave function and probability of finding the particle in a one-dimensional box:

We know that normalised wave function $\psi_{n}(x)$

$\psi_{n}(x)=\sqrt{\frac{2}{L}} sin \left( \frac{n \pi x}{L} \right)$

The probability density of wave function $\left| \psi_{n}(x) \right|$

$\left| \psi_{n}(x) \right|^{2} = \frac{2}{L} sin^{2} \left( \frac{n \pi x}{L} \right)$

Maximum Condition:

The values of $\psi_{n}(x)$ and $\left| \psi_{n}(x) \right|^{2}$ will be maximum. When

$sin \left( \frac{n \pi x}{L} \right)=1$

$sin \left( \frac{n \pi x}{L} \right )=sin \frac{\left( 2m+1 \right) \pi}{2}$

$ \frac{n \pi x}{L} =\left( 2m+1 \right) \frac{ \pi}{2}$

$ x =\left( 2m+1 \right) \frac{ L}{2n}$

Minima Condition:

The values of $\psi_{n}(x)$ and $\left| \psi_{n}(x) \right|^{2}$ will be minima. When

$sin \left( \frac{n \pi x}{L} \right)=0$

$sin \left( \frac{n \pi x}{L} \right)= \sin \: m\pi$

$ \frac{n \pi x}{L} = \: m\pi$

$x=m\left( \frac{L}{n} \right)$

Read More

The electric potential energy of an electric dipole in the uniform electric field

Derivation of the electric potential energy of an electric dipole in the uniform electric field:

Let us consider an electric dipole $AB$, which is made up of two charges $q_1$ and $q_2$, which are placed at a distance of $2l$ in the electric field $E$. So force acting on each charge due to the electric field will be $qE$. If the dipole gets rotated a small-angle $d\theta$ against the torque acting on it in the uniform electric field $E$ then the small work done is

$dW=\tau. d\theta \qquad (1)$

Force of moment on an electric Dipole

The torque (i.e moment of force) on an electric dipole in a uniform electric field

$ \tau=p.E\:sin\theta$

Now substitute the value of $\tau$ in equation $(1)$. So work done

$ dW=p.E\:sin\theta.d\theta$

If the dipole rotate the angle from $\theta_{1}$ to angle $\theta_{2}$ then workdone

$\int_{W_{1}}^{W_{2}}dW=p.E\int_{\theta_{1}}^{\theta_{2}}sin\theta \: d\theta$

$ W_{2}-W_{1}=p.E\left[-cos\theta \right]_{\theta_{1}}^{\theta_{2}}$

$ \Delta W= p.E \left( cos\theta_{1}-cos\theta_{2} \right)\qquad\qquad (2)$

This work is stored in the form of the electric potential energy of an electric dipole in the electric field. So

$U=\Delta W$

$U=p.E \left( cos\theta_{1}-cos\theta_{2} \right)$

If the electric dipole rotates from $0^{\circ}$ (when the direction of electric dipole moment $p$ is aligned in the direction of the electric field $E$) to an angle $\theta$ in the electric field i.e $\theta_{1}=0^{\circ}$ and $\theta_{2}=\theta$ then the electric potential energy of dipole in a uniform electric field

$U=p.E (1-cos\theta)$

Case-(I) If $\theta=0^{\circ}$ i.e It is stable equilibrium position then

$U_{min}=-pE$

Case-(II) If $\theta=90^{\circ}$ i.e Position of zero energy then

$U=0$

Case-(III) If $\theta=180^{\circ}$ i.e It is unstable equilibrium position then

$U_{max}=pE$

Read More

de-Broglie Concept of Matter wave

Louis de-Broglie thought that similar to the dual nature of light, material particles must also possess the dual character of particle and wave. This means that material particles sometimes behave as particle nature and sometimes behave like a wave nature.

According to de-Broglie –

A moving particle is always associated with a wave, called as de-Broglie matter-wave, whose wavelengths depend upon the mass of the particle and its velocity.

According to Planck’s theory of radiation–

$E=h\nu \qquad(1) $

Where
h – Planck’s constant
$\nu $ - frequency

According to Einstein’s mass-energy relation –

$E=mc^ {2} \qquad (2)$

According to de Broglie's hypothesis equation $ (1)$ and equation $(2)$ can be written as –

$mc^ {2} = h \nu$

$mc^ {2} = \frac{hc}{\lambda }$

$\lambda =\frac{h}{mc}\qquad(3) $

$\lambda =\frac{h}{P}$

Where $P$ –Momentum of Photon

Similarly from equation $(3)$ the expression for matter waves can be written as

$\lambda=\frac{h}{mv}=\frac{h}{P}\qquad(4)$

Here $P$ is the momentum of the moving particle.

1.) de-Broglie Wavelength in terms of Kinetic Energy

$K=\frac{1}{2} mv ^{2}$

$K=\frac{m^{2}v^{2}}{2m}$

$K=\frac{P^{2}}{2m}$

$P=\sqrt{2mK}$

Now substitute the value of $P$ in equation $ (4)$ so

$\lambda =\frac{h}{\sqrt{2mK}} \qquad (5)$

2.) de-Broglie Wavelength for a Charged particle

The kinetic energy of a charged particle is $K = qv$

Now substitute the value of $K$ in equation$(5)$ so

$\lambda =\frac{h}{\sqrt{2mqv}}$

3.) de-Broglie Wavelength for an Electron

The kinetic energy of an electron

$K=ev$

If the relativistic variation of mass with a velocity of the electron is ignored then $m=m_{0}$ wavelength

$\lambda =\frac{h}{\sqrt{2m_{0}ev}}$

So wavelength of de-Broglie wave associated with the electron in non-relativistic cases

4.) de-Broglie wavelength for a particle in Thermal Equilibrium

For a particle of mass $m$ in thermal equilibrium at temperature $T@

$K=\frac{3}{2}kT$

Where $K$ – Boltzmann Constant

$\lambda =\frac{h}{\sqrt{2m.\frac{3}{2}kt}}$

$\lambda =\frac{h}{\sqrt{3mKT}}$

Properties of matter wave →

1. Matter waves are generated only if the material's particles are in motion.


2. Matter-wave is produced whether the particles are charged or uncharged.
3. The velocity of the matter wave is constant; it depends on the velocity of material particles.


4. For the velocity of a given particle, the wavelength of matter waves will be shorter for a particle of large mass and vice-versa.


5. The matter waves are not electromagnetic waves.


6. The speed of matter waves is greater than the speed of light.
   
   
   According to Einstein’s mass-energy relation
   
   
   $E=mc^{2}$
   
   
   $h\nu = mc^{2}$
   
   
   $\nu =\frac{mc^{2}}{h}$
   
   
   Where $\nu$ is the frequency of matter-wave.
   
   
   We know that the velocity of matter-wave
   
   
   $ u =\nu \lambda $
   Substitute the value of $\nu$ in the above equation
   
   
   $u =\frac{mc^{2}}{h}. \lambda $
   $u =\frac{mc^{2}}{h} . \frac{h}{mv}$
   
   

   $u =\frac{c^{2}}{v}$

   
   
   Where $v$ → particle velocity which is less than the velocity of light.


7. The wave and particle nature of moving bodies can never be observed simultaneously.

Read More