Coulomb′s Law and Applications

Coulomb’s Law: This law was first published by French physicist Charles-Augustin de Coulomb in the year 1785. According to Coulomb’s Law-

The electric force acting between the two point charges is directly proportional to the product of magnitude of the two charges and inversely proportional to square of the distance between these two charges. The electric force always acts along the line joining the charges.

Coulomb's Force between the two positive charges

Coulomb's force between the two charges

Let us consider two positive charges whose magnitude $q_{1}$ and $q_{2}$ are placed at a distance $‘r’$. According to Coulomb’s Law (magnitude only):

$F\propto q_{1}q_{2} \qquad(1)$

$F\propto \frac{1}{r^{2}} \qquad(2)$

From equation$(1)$ and equation$(2)$, we can write as:

$F\propto \frac{q_{1}q_{2}}{r^{2}} \qquad(3)$

$F=\frac{1}{4\pi \varepsilon K} \frac{q_{1}q_{2}}{r^{2}} \qquad(4)$

Where
$\epsilon$= Permittivity of any medium,
$K$ = Dielectric constant

For air and vacuum:
$\epsilon= \epsilon_{0}$ and $K=1$

So Coulomb’s Law for Air and Vacuum:

$F=\frac{1}{4\pi \varepsilon_{0}}\frac{q_{1}q_{2}}{r^{2}} \qquad(5)$

Where $\varepsilon _{0}=8.0854 \times 10^{-12} \:\: C^{2}N^{-1}m^{-2}$

Then, the value of $\frac{1}{4\pi \varepsilon_{0}}=9\times10^{9} N-m^{2}/C^{2}$ So, From Equation$(5)$

$F=9\times 10^{9} \frac{q_{1}q_{2}}{r^{2}}$

This is the equation of Coulomb's Law that applies to the medium of air or vacuum. The above equation of Coulomb Law shows only the magnitude value of electrostatic force.

Properties of Coulomb's law:-

There are the following properties of Coulomb's law:-

Coulomb force is an action and reaction pair and follows Newton's third law.

Coulomb force is a conservative force.

Coulomb force is central force i.e. it is always acting along the line joining between two charges.

If the net force is zero then momentum will be conserved.

If the center of mass is at rest and momentum is conserved then it follows the mass conservation law.

The force between two charges is independent of the presence or absence of other charges but the net force increases on that particular charge.

Limitation of Coulomb's Law:

This law does not apply to moving charges i.e. it applies to static charges (charge at rest) and charges must be stationary relative to each other.

It applies to charges of regular and smooth shape. It is very difficult to apply to irregular shapes.

The charges must not overlap for example they must be distinct point charges.

This law can not be directly applicable to calculate the charge on big planets.

Application of Coulomb's Law:

Coulomb's law is used to calculate the distance between the charges.

Coulomb's law is used to calculate the electrostatic force between the charges.

Coulomb's law is used to calculate the electrostatic force on a point charge due to the presence of several point charges. It is also known as the Superposition theorem of electrostatic force.

**What is $1$ Coulomb's?

Answer: When two unit charges are placed in a vacuum at one meter apart then the force acting between charges is $9\times10^{9}$ $N$.This force is known as 1 Coulomb.

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 than 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 the 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 than 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 lig

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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

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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

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