### 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.

### 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