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Showing posts from April, 2022

Energy Stored in a Charged Capacitor

Description→ When a capacitor is charged, the work is done by charged battery (i.e the chemical energy of the battery is used to charge the capacitor). As the capacitor gets charged, the potential difference between its plates increases. Due to this increase in the potential difference between the plates, the battery has to give the same amount of charge to the capacitor. Because of that, the battery has to do more and more work. The total amount of work done in charging the capacitor is stored in the form of electric potential energy in between the capacitor plates. This energy is retrieved as heat when the capacitor is discharged through a resistance. Derivation→ Let us consider, a capacitor of capacitance $C$with a potential difference of $V$ between the plates. In the process of charging, electrons are transferred from the positive to negative, unit each plate acquires an amount of charge $q$. Suppose during the process of charging, we increase the charge fr

Force between the plates of a Charged Parallel Plate Capacitor

Derivation and Description→ Consider, A parallel plate capacitor with a charge $+q$ on one of its plates and $-q$ on the other plate. Let initially the plates of the capacitor are almost, but not quite touching. Due to opposite polarity, there is an attractive force $F$ between the plates. Now If these plates are gradually pulling and apart to a distance $d$, in such a way that $d$ is still small compared to the linear dimension of the plates, then the approximation of a uniform field between the plates is maintained and thus the force remains constant. The force between the charged parallel plates of the capacitor Now the work done in separating the plates from near $0$ to $d$, $W=F.d \qquad(1)$ This work done $(W)$ is stored as electrostatic potential energy $(U)$ between the plates, i.e. $U=\frac{1}{2}q V $ But $V=E.d$, then the above equation can be written as $U=\frac{1}{2}q \: E \: d \qquad(2)$ But the equation $(1)$ and equation $(2)$ both

Capacitance of a Parallel Plate Capacitor Partly Filled with Dielectric Slab between Plates

Derivation→ Let us consider, The charge on a parallel-plate capacitor = $q$ The area of parallel-plate = $A$ The distance between the parallel-plate = $d$ The dielectric constant of the slab of a material =$K$ The thickness of the material =$t$ The vacuum (or air) between the plates =$(d-t)$ The surface charge density on the plates= $\sigma$ The capacitance of Parallel Plate Capacitor with Dielectric Slab between Plates The electric field in the air between the plates is $E_{\circ}=\frac{\sigma}{\epsilon_{\circ}}$ $E_{\circ}=\frac{q}{\epsilon_{\circ}\: A} \qquad(1)$ The electric field in the dielectric material $E=\frac{q}{\epsilon_{\circ}\: K\: A} \qquad(2)$ The potential difference between the plates $V=E_{\circ}\left( d-t \right)+E\:t \qquad(3)$ Now substitute the value of $E_{\circ}$ and $E$ in the equation $(3)$, Then we get $V=\frac{q}{\epsilon_{\circ}\: A} \left( d-t \right)+ \frac{q}{\epsilon_{\c

Parallel Plate Air Capacitor and Its Capacitance

Parallel Plate Air Capacitor→ A parallel-plate capacitor consists of two long, plane, metallic plates mounted on two insulating stands and placed at a small distance apart in a vacuum (or air). The plates are exactly parallel to each other. Parallel Plate Air Capacitor Derivation of the capacitance of parallel plate capacitor in the air→ Let us consider, Two plates $X$ and $Y$ are separated at a small distance $d$ in the vacuum (or air). If the area of plates is $A$ and the plates $X$ and $Y$ have charge $+q$ and $-q$ respectively. If the surface charge density on each plate is $\sigma$ then electric field intensity at a point between two parallel plates is $E=\frac{\sigma}{\epsilon_{\circ}}$ $E=\frac{q}{\epsilon_{\circ}A}\qquad(1) \qquad (\because \sigma=\frac{q}{A}) $ The potential difference between the parallel plates capacitor in air (or vacuum) is $V=E.d$ Now substitute

Potential Energy of a Charged Conductor

Definition : The work done in charging the conductor is stored as potential energy in the electric field in the vicinity of the conductor is called the potential energy of a charged conductor. Derivation → Let us consider, a conductor of capacitance $C$. If charge $+Q$ is given into small amount of $dq$ to the surface of the conductor. Then the work done will be $dw=V\:dq \qquad(1) $ $dw=\frac{q}{C} \:dq \qquad \left (\because V=\frac{q}{C} \right)$ Therefore, as the amount of charge on the conductor will increase from $0$ to $Q$ that causes also increase in work done. So integrate the above equation for total work done $ \int_{0}^{W} dw=\frac{1}{C} \int_{0}^{Q} q dq$ $W=\frac{1}{C} \int_{0}^{Q} q dq$ $W=\frac{1}{C} \left[ \frac{q^{2}}{2} \right]^{Q}_{0}$ $W=\frac{1}{2} \frac{Q^{2}}{C}$ This work is stored in the form of electric potential energy $U$. Then $U=\frac{1}{2} \frac{Q^{2}}{C}$ $U=\frac{1}{2} C V^{2}$ $U=\fr

Capacitance of an Isolated Spherical Conductor

Derivation → Capacitance of an Isolated Spherical Conductor Let us consider an isolated spherical conductor of radius $a$ is placed in a vacuum or air. Let a charge $+q$ be given to the sphere and this charge is distributed uniformly on the surface of conducting sphere. Then the electric potential at the surface of the conducting sphere is $V=\frac{1}{4\pi \epsilon_{\circ}} \frac{q}{a} \qquad(1)$ So the capacitance of the sphere $C=\frac{q}{V} \qquad(2)$ Now substitute the value of $V$ in equation $(2)$ Then we get $C=\frac{q}{\frac{q}{4\pi \epsilon_{\circ}a}}$ $C=4\pi \epsilon_{\circ}a$ Thus, The capacitance of a spherical conductor is directly proportional to its radius. i.e If the radius of conducting sphere is large then the sphere will hold a large amount of the given charge without running up too high a voltage. Unit: The unit of capacitance is "Farad" i.e. $F$

Magnetic Field at the axis of a current-carrying Circular Loop

Derivation→ Let us consider, A circular loop which have radius $a$, carrying a current $i$. Let take a point $P$ on the axis of the loop at a distance $x$ from the center $O$ of the loop. Now to find the magnetic field at point $P$, take a small current element of length $dl$ at the top of the loop. Let $r$ is the distance between the current element and point $P$. Now apply the Biot-Savart's Law to find the magnitude of the magnetic field due to small current-element $dl$ at point $P$ i.e. $dB=\frac{\mu_{\circ}}{4 \pi} \frac{i \: dl\: sin\theta}{r^{2 }}$ Where $\theta$ is the angle between the length of the element $dl$ and the line joining the element to the point $P$. Here this angle is $90^{\circ}$ $dB=\frac{\mu_{\circ}}{4\pi} \frac{i \: dl}{r^{2}}\qquad(1)$ The direction of the magnetic field $d\overrightarrow{B}$ is in the plane of the paper and at right angles to the line $r$ as shown in the figure below (By applying the right-hand screw rule). It can be

Magnetic Field due to a Straight Current-Carrying Conductor of Finite Length

Derivation→ Let us consider, a current-carrying conductor $XY$ having length $l$ in which current $i$ is flowing from $X$ to $Y$. Now, To find the magnetic field due to the conductor, take a point $P$ at a distance $d$ from point $O$ of the conductor. Now consider a small length element $dl$ at the conductor which is making an angle $\theta$ from point $P$. The length element $dl$ is also making the angle $d\theta$ from point $O$. If $\theta_{1}$ and $\theta_{2}$ is the angle from point $O$ to point $X$ and $Y$ respectively. Than magnetic field at point $P$ due to small length element $dl$ which is at a distance $r$ $dB=\frac{\mu_{0}}{4\pi} \frac{i\: dl \: sin (90+\theta)}{r^{2}}$ $dB=\frac{\mu_{0}}{4\pi} \frac{i \: dl \: cos\theta}{r^{2}} \qquad(1)$ Magnetic Field due to a Straight Current-Carrying Conductor of Finite Length But from the figure, In $\Delta NOP$ $cos\theta=\frac{d}{r}$ $r=\frac{d}{cos\theta} \qquad(2)$ $tan\theta=\frac{l}{d}$

Motion of Charged Particles in Uniform Magnetic Field

Description: The motion of any particle is depends upon the force applying on it. In the magnetic field, the motion of charge is perpendicular to the magnetic force applied on the charge particle because of that the path of the charge particle becomes circular. If the direction of force and the direction of motion are not perpendicular to each other that is they are at any angle then the path of the charge particle becomes becomes helical. A.) When $\overrightarrow{v}$ is perpendicular to $\overrightarrow{B}$→ Let us consider a charge $q$ enters in the magnetic field $\overrightarrow{B}$ from a point $O$ with velocity $\overrightarrow{v}$ directed perpendicular to the magnetic field $\overrightarrow{B}$.The general Expression for the force acting on the particle is $\overrightarrow{F}=q(\overrightarrow{v} \times \overrightarrow{B})$ Here the magnetic field $\overrightarrow{B}$ is perpendicular to the plane of the page directed downwards which is s

Normalized and Orthogonal wave function

Description: We know that $\psi^{*}\psi$ or $\left|\psi \right|^{2} d\tau $ represent the probability of finding the particle in volume element $d\tau$. The total probability of finding the particle in the entire space is 1 so $ \int \left|\psi(r,t) \right|^{2} d\tau=1 $ Where integral extends overall space. $\int \psi^{*}(r,t) \psi(r,t) d\tau=1$ A wave function satisfies the above equation so it is called normalized to unity. For any wave function that is a solution of the time-dependent Schrodinger equation $\int \psi^{*} \psi d\tau=N$ $\frac{1}{N} \int \psi^{*} \psi d\tau=1$ $\int \frac{\psi{*}}{\sqrt{N}} \frac{\psi}{\sqrt{N}} d\tau = 1$ Where $\sqrt{N}$ → Normalized Factor $\frac{\psi}{N} $ → Normalised wave function If independent coordinate $x$,$y$,$z$, and $\psi$ satisfy the Schrodinger wave equation. Then it is evident that $\frac{\psi}{\sqrt {N}}$ also satisfies the Schrodinger wave equation. If $\psi_{i}$ and $\psi_{j}$ are tw

Quantum Mechanical Operators

Operator → An operator is defined as a mathematical term that is used in the operation of a function so that this function may or may not be transformed into another function. Operators of Quantum Mechanics → There are the following quantum mechanical operators which are used in the wave function of particles:- Momentum Operator Kinetic Energy Operator Total Energy Operator (Hamiltonian Operator) Total Energy Operator in terms of the differential with respect to time Momentum Operator → The wave function for a free particle moving along the position $x$-direction is $\psi(x,t)=A e^{\frac{i}{\hbar}(P_{x}x-Et)}$ Differentiate the above equation with respect to $x$ then we get $\frac{\partial \psi}{\partial x}= A e^{\frac{i}{\hbar}(P_{x}x-Et)} \frac{i}{\hbar} P_{x} $ $\frac{\partial \psi}{\partial x}= \psi \frac{i}{\hbar} P_{x} $ $ P_{x} \psi = \frac{\hbar}{i}\frac{\partial \psi}{\partial x}$

Postulate of wave mechanics or Quantum Mechanics

Postulate of Wave or Quantum Mechanics (or Operator formalism in Quantum mechanics) → The formulation of mathematical equations of quantum mechanics is based on the linear operator. This operator formulation of quantum mechanics is known as postulates of quantum mechanics. These postulates are given below:- For a system consisting of particles moving in a field of a conservative force, there is an associated complex wave function $\psi(x, y, z, t)$ where $x$,$y$,$z$ space coordinates, and $t$ is the time. This wave function enables us to obtain a description of the behavior of the system, consistent with the principle of uncertainty. There is an operator with every observable dynamical quantity. The operator corresponding to the pertinent dynamical quantities is:- Dynamical Variable Symbol Quantum Mechanical Operator

Eigenfunction, Eigenvalues and Eigenvectors

Eigenfunction and Eigenvalues → If $\psi$ is a well-behaved function, then an operator $\hat{P}$ may operate on $\psi$ in two different ways depending upon the nature of function $\psi$ ` When an operator $\hat{P}$ operates on any function $\psi$ then this function $\psi$ changes into another function $\phi$. i.e. $\hat{P} \psi =\phi$ Where $\phi$ is a new function linearly depending upon the initial function $\psi$. Example: Let us consider a function $f(x)=x^{2}$ and an operator ie. differential operator $\frac{d}{dx}$ is operate on the function. Then we get $\frac{d}{dx}f(x)= \frac{d}{dx} (x^{2})$ $\frac{d}{dx}f(x)= 2x$ Now the given function $f(x)=x^{2}$ change into another function $f(x)=x$. When an operator $\hat{P}$ opera

Kepler's Laws of Planetary Motion

Kepler's Laws → Kepler's found important uniformity in the motion of the planets. This uniformity is known as "Kepler's Laws of planetary motion". There are basically three laws → First law (Law of orbits) Second Law (Law of Areal Speed) Third Law (Law of Periods) First law (Law of orbits) → All the planets move around the sun in elliptical path and the sun is at the one foci of the ellipse. Second Law (Law of Areal Speed) → A line joining the any planet to sun sweeps out equal areas in equal interval of of times. i.e The areal speed of the planets remains constant. According to the second law, When the planet is farthest from the sun, then its speed is maximum, and when it is nearest the sun, then its speed is maximum. Planetary Motion From the above figure, A planet is moving around the sun from $A$ to $B$ in a given time interval, and from $C$ to $D$ in the same time inter

Relation between gravitational acceleration and gravitational force

Relation between $g$ and $G$ → Let us consider: The mass of earth = $M_{e}$ The radius of earth = $R_{e}$ The mass of the object = $m$ If the object is placed on the surface of the earth then the gravitational force on the object is → $F=G \frac{M_{e} m}{R_{e}^{2}} \qquad(1)$ The force on the object due to gravitational acceleration is → $F=mg \qquad(2)$ From equation $(1)$ and equation $(2)$ $mg=G \frac{M_{e} m}{R_{e}^{2}}$ $g=G \frac{M_{e}}{R_{e}^{2}}$ $G M_{e}=g R_{e}^{2}$

Newton's law for Gravitational Force

Gravitational Force → Newton's Gravitational Law statement is a combination of three individual statements. These are The force between the two-particle is directly proportional to the product of their masses i.e. $F \propto m_{1} \: m_{2} \qquad(1)$ Where $m_{1}$ & $m_{2}$ are the masses of the particles. The force between the two-particle is inversely proportional to the square of the distance between them i.e. $F \propto \frac{1}{r^{2}} \qquad(2)$ Where $r$ is the distance between the particles. This force always acts between the line joining the masses. From the above the equation $(1)$ and equation $(2)$ $F\propto \frac{m_{1} \: m_{2}}{r^{2}}$ $F=G \frac{m_{1} \: m_{2}}{r^{2}}$ Where $G$ is Newton's gravitaional constant and its experimental value $6.67\times 10^{-11} \frac{N-