Full-wave center-tap rectifier - eee.iub.edu.bd

IUB Dr. Mustafa Chowdhury 1

HF Electromagnetic Waves

Lecture – 6

Spring 2016

Faculty

Dr. Mustafa Chowdhury

Spring 2016

EEE304/ETE309_L6

Q

E

dLaL

Energy&

Potential


Energy expended in moving a point charge in an electric field

Suppose we wish to move a charge Q at a distance dL in an

electric field E.

The force on Q due to the electric field is

The component of this force in the direction dL

where aL = a unit vector in the direction of dL.

The force which we must apply, to move a charge Q at a distance

dL, is equal and opposite to the force due to the field,

Q

E

dLaL

EEE304/ETE309_L6

Spring 2016

EF QE

LLEEL Q aEaFF

Lappl Q aEF


Energy expended in moving a point charge in an electric field

Expenditure of energy is the product of the force and distance,

therefore, the differential work done by external source is

where we have replaced aLdL by the simpler expression dL.

Q. Under which condition the required work done is zero?

A: One condition is that E and dL are perpendicular to each other.

The work required to move the charge a finite distance must be

determined by integrating,

EEE304/ETE309_L6

Spring 2016

LEaE dQdLQdW L

LELE dQdQWfinal

init

final

init


Example 1

An electric field is given as E = 6y2zax + 12xyzay + 6xy2az V/m.

An incremental path is represented by L = -3ax + 5ay - 2az µm.

Find the work done in moving a 2-µC charge along this path if

the location of the path is at: (a) PA(0,2,5) (b) PB(l,l,l) (c) PC(-

0.7,-2,-0.3).

Solution:

pJ720J1072010360102 1266 W

226 126018102 xyxyzzy

QQWfinal

init

LELE(a)

Therefore, W at point PA(0,2,5)

EEE304/ETE309_L6

Spring 2016


Example 2

Consider a nonuniform electric field E = yax+ xay + 2az.

Determine the work expended in carrying 2C charge from point

B(1,0.1) to A(0.8,0.6,1) along the shorter arc of the circle

x2 + y2 = 1 and z = 1.

Solution:

EEE304/ETE309_L6

Spring 2016


Line integral

Let us first find the work done in carrying the positive charge Q

about a circular path of radius 1, centered at the line charge, as

illustrated in figure below.

Without any calculation,

we can say that the work

must be nil, since the

path is always

perpendicular to the

electric field intensity, or

the force on the charge is

always exerted at right

angles to the direction in

which we are moving it.

EEE304/ETE309_L6

Spring 2016


Line integral (contd..)

The electric field has only the component

And the differential length has only component

The work is then

EEE304/ETE309_L6

Spring 2016

aaE

102

LE

aL dd 1

02

2

0

2

0

1

10

aa

aaLE

dQ

dQdQW

L

Lfinal

init

final

init


Line integral (contd..)

Let us now find the work done in carrying the positive charge Q

from = a to = b along a radial path as shown in figure below.

Here the differential length has only component

Therefore, the work done is

EEE304/ETE309_L6

Spring 2016

aL dd

a

bQ

dQ

dQW

L

b

a

L

Lfinal

init

ln2

1

2

2

0

0

0

aa


The potential difference V is defined as the work done by an

external source in moving a unit positive charge from one point

to another in an electric field.

We know that, the work done by an external source in moving a

charge Q from one point to another in an electric field E is

From the above equation,

Potential and potential difference

final

initdQW LE

differencePotential VdQ

W final

initLE

EEE304/ETE309_L6

Spring 2016


Potential difference

The potential difference between points A and B is the work done

in moving the unit charge from B to A.

Potential difference is measured in joules per coulomb, for which

the volt is defined as a more common unit, abbreviated as V.

If the potential at point A is VA and that at B is VB, then

where we necessarily agree that VA and VB shall have the same

zero reference point.

LELE ddVB

A

A

BAB

BAAB VVV

EEE304/ETE309_L6

Spring 2016


Let us now find the work done in carrying the positive charge Q

from = b to = a along a radial path as shown in figure below.

Thus the potential difference

between points = b to = a

Potential difference (contd..) EEE304/ETE309_L6

Spring 2016

a

bQW L ln

2 0

a

b

Q

WV L

ab ln2 0


Potential difference (contd..)

Let us now find the potential difference between points A and B

at radial distances rA and rB from a point charge Q.

Choosing an origin at Q,

We have

If rB > rA, the potential difference VAB is positive, indicating that

energy is expended by the external source in bringing the

positive charge from rB to rA.

EEE304/ETE309_L6

Spring 2016

r

rrr

drd

r

QE

aL

aaE

2

04

BA

r

r

A

BAB

rr

Qdr

r

QdV

A

B

11

44 0

2

0 LE


Potential gradient

We know that

Now consider the elemental length L is very short along which

E is essentially constant, leading to an incremental potential

difference V

If L = LaL, E = EaL

and is the angle between

L and E, then

LE dV

LE V

cosLEV

EEE304/ETE309_L6

Spring 2016


Potential gradient (contd..)

By taking the limit

or

Now V/L will be maximum when cos = -1, and this will

happen when L is opposite to E. The maximum value is

The magnitude of the electric field intensity is given by the

maximum value of the rate of change of potential with

distance.

coslim0

EL

V

L

cosE

dL

dV

EdL

dV

max

Do you remember the unit of

electric field intensity?

EEE304/ETE309_L6

Spring 2016



In vector form

Since dV/dL|max occurs when L is in the direction of aN, we may

remind ourselves of this fact by letting

which gives

By definition,

Therefore,

NdL

dVaE

max

dN

dV

dL

dV

max

NdN

dVaE

VV gradE

EEE304/ETE309_L6

Spring 2016

NdN

dTTT a gradofGradient



Since we have shown that V is a unique function of x, y, and z,

we may take its total differential

But we also have

Since both expressions are true for any dx, dy, and dz, then

The result yield

EEE304/ETE309_L6

Spring 2016

dzz

Vdy

y

Vdx

x

VdV

dzEdyEdxEddV zyx LE

zyx

z

V

y

V

x

VaaaE

z

VE

y

VE

x

VE zyx

,,



Therefore,

Hence, the vector operator is written as

The gradient of V in different coordinate systems

EEE304/ETE309_L6

Spring 2016

zyxz

V

y

V

x

VV aaa

grad

zyxzyx

aaa

)(sin

11

)(1

)(

SphericalV

r

V

rr

VV

lCylindricaz

VVVV

Cartesianz

V

y

V

x

VV

zr

z

zyx

aaaa

aaa

aaa


Example 3

Find the electric flux density at point P (-4,3,6) for an electric

field E = -4xyax –2x2ay + 5azV/m. Find also the volume charge

density.

Solution:

Value of E at point P is

Therefore the flux density D is

The volume charge density v is

EEE304/ETE309_L6

Spring 2016


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Conducting

Materials

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Spring 2016

Full-wave center-tap rectifier - eee.iub.edu.bd

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