ABSTRACT OF WORK UNDERTAKEN:

As per as possible I have taken every care to eliminate omission and errors. It is too much to expect that no inaccuracy/error has crept in.

I would like to thank my teachers and my classmates for their co-operation.

Mostly I collected the information’s from internet with various websites like wikipedia, yahoo.mail and also from google. I also got various datas from many text books.

TABLE OF CONTENTS

Topics page no.

01.energy expression for

a capacitor………………….. 6-8

02. energy expression for

a inductor…………………… 9-10

03. energy expression for

a simple harmonic oscillator.. 11-

04. energy expression for

electric and magnetic field

energy densities……………… 12-13

05. energy expression for

a rotating body……………….

06. energy expression for

a uniformly moving body……

07.

INTRODUCTION TO THE PROBLEM

We must be done by external agente to charge a capacitor . We visualize the work required to charge a capacitor as being stored in the form of electric potential energy U in electric field between the plates. We can recover the energy by discharging the capacitor in a circuit, just we can recover the potential energy stored in a stretched bow by realeasinng the bow string to transfer the energy to the kinetic energy of an arrow.

The circuit parameter of capacitance is represented by the letter C, is measured in farads

(F), and is symbolized graphically by two short parallel conductive plates, as shown in

Figure 1. Because the farad is an extremely large quantity of capacitance, practical capacitors

are based on submultiples of the farad. The most frequently encountered values

in microelectronics lie in the femtofarad (fF), picofarad (pF) and nanofarad range

Inductors can be used to produced a desired magnetic field . If we stablish a current I in the windings of solenoid we are taking as our inductor, the current produces a magnetic flux through the central region of the inductor.

The simple harmonic oscillator can also be called as torsion pendulum, the element of springiness or elasticity associated with the twisting of a suspension wire rather than extension and compression of a spring.

Storing Energy in a Capacitor

The energy stored on a capacitor can be expressed in terms of the work done by the battery. Voltage represents energy per unit charge, so the work to move a charge element dq from the negative plate to the positive plate is equal to V dq, where V is the voltage on the capacitor. The voltage V is proportional to the amount of charge which is already on the capacitor.

Element of energy stored:

If Q is the amount of charge stored when the whole battery voltage appears across the capacitor, then the stored energy is obtained from the integral:

This energy expression can be put in three equivalent forms by just permutations based on the definition of capacitance C=Q/V.

Storing Energy in a Capacitor

When the switch is closed to connect the battery to the capacitor, there is zero voltage across the capacitor since it has no charge buildup. The voltage on the capacitor is proportional to the charge

Storing energy on the capacitor involves doing work to transport charge from one plate of the capacitor to the other against the electrical forces. As the charge builds up in the charging process, each successive element of charge dq requires more work to force it onto the positive plate. Summing these continuously changing quantities requires an integral.

Note that the total energy stored QV/2 is exactly half of the energy QV which is supplied by the battery, independent of R!

Capacitor Energy Integral

Transporting differential charge dq to the plate of the capacitor requires work

But as the voltage rises toward the battery voltage in the process of storing energy, each successive dq requires more work. Summing all these amounts of work until the total charge is reached is an infinite sum, the type of task an integral is essential for. The form of the integral shown above is a polynomial integral and is a good example of the power of integration.

Energy Stored in an Inductor

Suppose that an inductor of inductance is connected to a variable DC voltage supply. The supply is adjusted so as to increase the current flowing through the inductor from zero to some final value . As the current through the inductor is ramped up, an emf ${\cal E} = - L\,d i/dt$ is generated, which acts to oppose the increase in the current. Clearly, work must be done against this emf by the voltage source in order to establish the current in the inductor. The work done by the voltage source during a time interval is

$\begin{displaymath} d W = P\,dt = -{\cal E}\,i\,dt = i\,L\,\frac{d i}{d t} \,d t = L\,i\,di. \end{displaymath}$

(247)

Here, $P=-{\cal E}\,i$ is the instantaneous rate at which the voltage source performs work. To find the total work done in establishing the final current in the inductor, we must integrate the above expression. Thus,

$\begin{displaymath} W = L\int_0^I i\,di, \end{displaymath}$

(248)

giving

$\begin{displaymath} W = \frac{1}{2}\,L\,I^2. \end{displaymath}$

(249)

This energy is actually stored in the magnetic field generated by the current flowing through the inductor. In a pure inductor, the energy is stored without loss, and is returned to the rest of the circuit when the current through the inductor is ramped down, and its associated magnetic field collapses.

Consider a simple solenoid. Equations (244), (246), and (249) can be combined to give

$\begin{displaymath} W = \frac{1}{2} \,L\,I^2 =\frac{\mu_0\,N^2\,A}{2\,l} \,\left(\frac{B\,l}{\mu_0\,N} \right)^2, \end{displaymath}$

(250)

which reduces to

$\begin{displaymath} W = \frac{B^2}{2\,\mu_0} \,\,l\,A. \end{displaymath}$

(251)

This represents the energy stored in the magnetic field of the solenoid. However, the volume of the field-filled core of the solenoid is $l\,A$ , so the magnetic energy density (i.e., the energy per unit volume) inside the solenoid is $w = W/(l\,A)$ , or

$\begin{displaymath} w = \frac{B^2}{2\,\mu_0}. \end{displaymath}$

(252)

It turns out that this result is quite general. Thus, we can calculate the energy content of any magnetic field by dividing space into little cubes (in each of which the magnetic field is approximately uniform), applying the above formula to find the energy content of each cube, and summing the energies thus obtained to find the total energy.

When electric and magnetic fields exist together in space, Eqs. (122) and (252) can be combined to give an expression for the total energy stored in the combined fields per unit volume:

$\begin{displaymath} w = \frac{\epsilon_0\,E^2}{2} + \frac{B^2}{2\,\mu_0}. \end{displaymath}$

(253)

Simple Harmonic Oscillator

We have already discussed the solution of the quantum mechanical simple harmonic oscillator (s.h.o.) in class by direct substitution of the potential energy

(3.1)

into the one-dimensional, time-independent Schroedinger equation. Recall that C is the spring constant of the spring attached to a mass m . The spring is assumed to obey Hooke’s Law so that the force on the mass is

(3.2)

The resulting differential equation is solved by a series solution to find the quantized energies and the energy eigenfunctions. (The details of the solution are discussed in Appendix I of Eisberg & Resnick.) Recall that the allowed energies are given by

(3.3)

where

. (3.4)

The series solution is quite involved and a bit “messy”. We are going to solve the problem again using an operator theory approach. There is one interesting difference in the two approaches that we will observe. Using the operator theory approach, we will find the energies and will be able to evaluate the averages of quantities like position and momentum without knowing the specific forms of the eigenfunctions! This is remarkable since we have said before that you must know the wavefunction of the particle in order to solve for physical quantities of its motion. In the differential equation approach that we originally used, we had to make some guesses about the nature of the wavefunctions in order to find the energies, and to find the averages of position or momentum, we had to know the wavefunctions exactly.

Energy in Electric and Magnetic Fields:

Both electric fields and magnetic fields store energy. For the electric field the energy density is

Show

This energy density can be used to calculate the energy stored in a capacitor.

For the magnetic field the energy density is

Show

which is used to calculate the energy stored in an inductor.

For electromagnetic waves, both the electric and magnetic fields play a role in the transport of energy. This power is expressed in terms of the Poynting vector

Energy density of electric and magnetic fields

Electric and magnetic fields store energy. In a vacuum, the (volumetric) energy density (in SI units) is given by

$U = \frac{\varepsilon_0}{2} \mathbf{E}^2 + \frac{1}{2\mu_0} \mathbf{B}^2$ ,

where E is the electric field and B is the magnetic field. In the context of magnetohydrodynamics, the physics of conductive fluids, the magnetic energy density behaves like an additional pressure that adds to the gas pressure of a plasma.

In normal (linear) substances, the energy density (in SI units) is

$U = \frac{1}{2} ( \mathbf{E} \cdot \mathbf{D} + \mathbf{H} \cdot \mathbf{B} )$ ,