### Electric field intensity due to uniformly charged plane sheet and parallel sheet

Electric field intensity due to a uniformly charged infinite plane thin sheet:

Let us consider, A plane charged sheet (It is a thin sheet so it will have surface charge distribution whether it is a conducting or nonconducting sheet) whose surface charge density is $\sigma$. From symmetry, Electric field intensity is perpendicular to the plane everywhere and the field intensity must have the same magnitude on both sides of the sheet. Let point $P_{1}$ and $P_{2}$ be the two-point on the opposite side of the sheet.

To use Gaussian law, we construct a cylindrical Gaussian surface of cross-section area $\overrightarrow{dA}$, which cuts the sheet, with points $P_{1}$ and $P_{2}$. The electric field $\overrightarrow{E}$ is normal to end faces and is away from the plane. Electric field $\overrightarrow{E}$ is parallel to cross-section area $\overrightarrow{dA}$. Therefore the curved cylindrical surface does not contribute to the flux i.e. $\oint \overrightarrow{E} \cdot \overrightarrow{dA}=0$.Hence the total flux is equal to the sum of the contribution from the two end faces. Thus, we get

$\phi_{E}=\int_{A} \overrightarrow{E} \cdot \overrightarrow{dA}+\int_{A} \overrightarrow{E} \cdot \overrightarrow{dA}$

$\phi_{E}= \int_{A} E \: dA \:cos 0^{\circ} +\int_{A} E \: dA \:cos 0^{\circ}$

Here the direction of $\overrightarrow{E}$ and $\overrightarrow{dA}$ is same. So the angle will be $\theta = 0^{\circ}$.
 Infinite plane thin sheet

$\phi_{E}= \int_{A} E \: dA +\int_{A} E \: dA$

$\phi_{E}= \int_{A} 2E \: dA$

$\phi_{E}= 2E \int_{A} \: dA$

$\phi_{E}= 2E\:A$

$\frac{q}{\epsilon_{0}}=2E\:A \qquad \left \{\because \phi_{E}=\frac{q}{\epsilon_{0}} \right \}$

$E=\frac{q}{2\epsilon_{0} A}$

$\because q=\sigma A$, So the above equation can be written as:

$E=\frac{\sigma A}{2\epsilon_{0} A}$

$E=\frac{\sigma}{2\epsilon_{0}}$

Electric field intensity due to the uniformly charged infinite conducting plane thick sheet or Plate:

Let us consider that a large positively charged plane sheet having a finite thickness is placed in the vacuum or air. Since it is a conducting plate so the charge will be distributed uniformly on the surface of the plate. Let $\sigma$ be the surface charge density of the charge

Let's take a point $P$ close to the plate at which electric field intensity has to determine. Since there is no charge inside the conducting plate, this conducting plate can be assumed as equivalent to two plane sheets of charge i.e sheet 1 and sheet 2.
 Plane Charged Plate

The magnitude of the electric field intensity $\overrightarrow {E_{1}}$ at point $P$ due to sheet 1 is →

$E_{1}=\frac{\sigma}{2\epsilon_{0}}$    (away from sheet 1)

The magnitude of the electric field intensity $\overrightarrow {E_{2}}$ at point $P$ due to sheet 2 is →

$E_{2}=\frac{\sigma}{2\epsilon_{0}}$    (away from sheet 2)

Since $\overrightarrow {E_{1}}$ and $\overrightarrow {E_{2}}$ are in the same direction, the magnitude of resultant intensity $\overrightarrow {E}$ at point $P$ due to both the sheet is →

$E=E_{1}+E_{2}$

$\because \quad E=\frac{\sigma}{2\epsilon_{0}}+\frac{\sigma}{2\epsilon_{0}}$

$E=\frac{\sigma}{\epsilon_{0}}$

The resultant electric field will be away from the plate. If the plate is negatively charged, the electric field intensity $\overrightarrow {E}$ would be directed toward the plate.

We have obtained the above formula for a 'plane' charged conductor. In fact, it holds for the electric field intensity 'just' outside a charged conductor of any shape.

Electric field intensity due to two Infinite Parallel Charged Sheets:

When both sheets are positively charged:

Let us consider, Two infinite, plane, sheets of positive charge, 1 and 2 are placed parallel to each other in the vacuum or air. Let $\sigma_{1}$ and $\sigma_{2}$ be the surface charge densities of charge on sheet 1 and 2 respectively.
 Likely positive charged sheet
Let $\overrightarrow {E_{1}}$ and $\overrightarrow {E_{2}}$ be the electric field intensities at any point due to sheet 1 and sheet 2 respectively. Then,

The electric field intensity at points $P'$ →

$E_{1}=\frac{\sigma_{1}}{2\epsilon_{0}}$    (away from sheet 1)

$E_{2}=\frac{\sigma_{2}}{2\epsilon_{0}}$    (away from sheet 2)

Since, Electric field intensities $\overrightarrow {E_{1}}$ and $\overrightarrow {E_{2}}$ are in the same direction, the magnitude of resultant intensity at point $P'$ is given by →

$E=E_{1}+E_{2}$

$E=\frac{\sigma_{1}}{2 \epsilon_{0}}+\frac{\sigma_{2}}{2 \epsilon_{0}}$

$E=\frac{1}{2 \epsilon_{0}} \left (\sigma_{1}+\sigma_{2} \right )$

If both sheets have equal charge densities $\sigma$ i.e. $\sigma_{1}=\sigma_{2}=\sigma$, Then above equation can be written as:

$E=\frac{\sigma}{ \epsilon_{0}}$

This electric field intensity would be away from both sheet 1 and sheet 2.

The electric field intensity at points $P$→

Electric field intensity at point $P$ due to sheet 1 is →

$E_{1}=\frac{\sigma_{1}}{2\epsilon_{0}}$    (away from sheet 1)

$E_{2}=\frac{\sigma_{2}}{2\epsilon_{0}}$    (away from sheet 2)

Now, both electric field intensities $\overrightarrow{E_{1}}$ and $\overrightarrow{E_{2}}$ are in opposite direction. The magnitude of resultant electric field $\overrightarrow{E}$ at point $P$ is given by

$E= E_{1}-E_{2}$

$E=\frac{\sigma_{1}}{2 \epsilon_{0}}-\frac{\sigma_{2}}{2 \epsilon_{0}}$

$E=\frac{1}{2 \epsilon_{0}} \left (\sigma_{1}-\sigma_{2} \right )$

If both sheets have equal charge densities $\sigma$ i.e. $\sigma_{1}=\sigma_{2}=\sigma$, Then above equation can be written as:

$E=0$

The electric field intensity at points $P''$ → $E_{1}=\frac{\sigma_{1}}{2\epsilon_{0}}$    (away from sheet 1)

$E_{2}=\frac{\sigma_{2}}{2\epsilon_{0}}$    (away sheet 2)

Since, Electric field intensities $\overrightarrow {E_{1}}$ and $\overrightarrow {E_{2}}$ are in the same direction, the magnitude of resultant intensity at point $P''$ is given by →

$E=E_{1}+E_{2}$

$E=\frac{\sigma_{1}}{2 \epsilon_{0}}+\frac{\sigma_{2}}{2 \epsilon_{0}}$

$E=\frac{1}{2 \epsilon_{0}} \left (\sigma_{1}+\sigma_{2} \right )$

If both sheets have equal charge densities $\sigma$ i.e. $\sigma_{1}=\sigma_{2}=\sigma$, Then above equation can be written as:

$E=\frac{\sigma}{ \epsilon_{0}}$

This electric field intensity would be away from both sheet 1 and sheet 2.

When one-sheet is positively charged and the other sheet negatively charged:

Let us consider two sheets 1 and 2 of positive and negative charge densities $\sigma_{1}$ and $\sigma_{2}$ ($\sigma_{1} > \sigma_{2}$)
 Unlike charged parallel Sheet
The electric field intensities at point $P'$ →

$E_{1}=\frac{\sigma_{1}}{2\epsilon_{0}}$    (away from sheet 1)

$E_{2}=\frac{\sigma_{2}}{2\epsilon_{0}}$    (toward from sheet 2)

The magnitude of the resultant electric field $E$

$E=E_{1}-E_{2}$

$E= \frac{1}{2\epsilon_{0}} \left ( \sigma_{1}- \sigma_{2}\right )$

If both sheets have equal charge densities $\sigma$ i.e. $\sigma_{1}=\sigma_{2}=\sigma$, Then above equation can be written as:

$E=0$

The electric field intensities at point $P$ →

$E_{1}=\frac{\sigma_{1}}{2\epsilon_{0}}$    (away from sheet 1)

$E_{2}=\frac{\sigma_{2}}{2\epsilon_{0}}$    (towards sheet 2)

Since, Electric field intensities $\overrightarrow {E_{1}}$ and $\overrightarrow {E_{2}}$ are in the same direction, the magnitude of resultant intensity at point $P$ is given by →

$E=E_{1}+E_{2}$

$E=\frac{\sigma_{1}}{2 \epsilon_{0}}+\frac{\sigma_{2}}{2 \epsilon_{0}}$

$E=\frac{1}{2 \epsilon_{0}} \left (\sigma_{1}+\sigma_{2} \right )$

If both sheets have equal charge densities $\sigma$ i.e. $\sigma_{1}=\sigma_{2}=\sigma$, Then above equation can be written as:

$E=\frac{\sigma}{ \epsilon_{0}}$

The electric field intensities at point $P''$ →

$E_{1}=\frac{\sigma_{1}}{2\epsilon_{0}}$    (away from sheet 1)

$E_{2}=\frac{\sigma_{2}}{2\epsilon_{0}}$    (toward from sheet 2)

The magnitude of the resultant electric field $E$ →

$E=E_{1}-E_{2}$

$E= \frac{1}{2\epsilon_{0}} \left ( \sigma_{1}- \sigma_{2}\right )$

If both sheets have equal charge densities $\sigma$ i.e. $\sigma_{1}=\sigma_{2}=\sigma$, Then above equation can be written as:

$E=0$

From the above expression, we can conclude that the magnitude of $E$ is free from the 'position' of the point taken in the electric field between the sheet and outside the sheet. It is also shown that the electric field between the sheet is uniform everywhere and independent of separation between the sheets.

### Numerical Aperture and Acceptance Angle of the Optical Fibre

Angle of Acceptance → If incident angle of light on the core for which the incident angle on the core-cladding interface equals the critical angle then incident angle of light on the core is called the "Angle of Acceptance. Transmission of light when the incident angle is equal to the acceptance angle If the incident angle is greater then the acceptance angle i.e. $\theta_{i}>\theta_{0}$ then the angle of incidence on the core-cladding interface will be less than the critical angle due to which part of incident light is transmitted into cladding as shown in the figure below Transmission of light when the incident angle is greater than the acceptance angle If the incident angle is less then the acceptance angle i.e. $\theta_{i}<\theta_{0}$ then the angle of incidence on the core-cladding interface will be greater than the critical angle for which total internal reflection takes place inside the core. As shown in the figure below Transmission of light w

### Fraunhofer diffraction due to a single slit

Let $S$ be a point monochromatic source of light of wavelength $\lambda$ placed at the focus of collimating lens $L_{1}$. The light beam is incident normally from $S$ on a narrow slit $AB$ of width $e$ and is diffracted from it. The diffracted beam is focused at the screen $XY$ by another converging lens $L_{2}$. The diffraction pattern having a central bright band followed by an alternative dark and bright band of decreasing intensity on both sides is obtained. Analytical Explanation: The light from the source $S$ is incident as a plane wavefront on the slit $AB$. According to Huygens's wave theory, every point in $AB$ sends out secondary waves in all directions. The undeviated ray from $AB$ is focused at $C$ on the screen by the lens $L_{2}$ while the rays diffracted through an angle $\theta$ are focussed at point $p$ on the screen. The rays from the ends $A$ and $B$ reach $C$ in the same phase and hence the intensity is maximum. Fraunhofer diffraction due to

### Particle in one dimensional box (Infinite Potential Well)

Let us consider a particle of mass $m$ that is confined to one-dimensional region $0 \leq x \leq L$ or the particle is restricted to move along the $x$-axis between $x=0$ and $x=L$. Let the particle can move freely in either direction, between $x=0$ and $x=L$. The endpoints of the region behave as ideally reflecting barriers so that the particle can not leave the region. A potential energy function $V(x)$ for this situation is shown in the figure below. Particle in One-Dimensional Box(Infinite Potential Well) The potential energy inside the one -dimensional box can be represented as $\begin{Bmatrix} V(x)=0 &for \: 0\leq x \leq L \\ V(x)=\infty & for \: 0> x > L \\ \end{Bmatrix}$ $\frac{d^{2} \psi(x)}{d x^{2}}+\frac{2m}{\hbar^{2}}(E-V)\psi(x)=0 \qquad(1)$ If the particle is free in a one-dimensional box, Schrodinger's wave equation can be written as: $\frac{d^{2} \psi(x)}{d x^{2}}+\frac{2mE}{\hbar^{2}}\psi(x)=0$ \$\frac{d^{2} \psi(x)}{d x