Electric field Lecture

Topic:ELECTRIC FIELD97.315 Basic E&M and Power Engineering

Lecture 2 TITLETITLE

• Theory• Field and force

• One point charges, n = 1

• Units

• Examples of calculation• Point charges

• Theory

• Multiple point charges, n > 1

• Examples of calculation

• Two point charges

• Assignment

• References

• Summary

•Theory

• Charge distributions

• Linear

• Surface

• Volume

•Examples of calculation

• Line

• …..

97.315 Basic E&M and Power Engineering Topic: ELECTRIC FIELD

Lecture 1 OUTLINEOUTLINE


Action at a distance!!!!!

Qoq oq

How do the charges qoknow of the presence of the large charge Q?

sun

earth

Similar to the action at a distance force of gravitation

Lecture 2 THEORYTHEORY



Gravity: Another action at a distance effect!!!!

sun

earth

Introduce the idea of a gravitational field. A mass placed in the gravitational field will experience a gravitational force.

Field

Test mass

Force


We define an electric field similar to that of a gravitational field.

A charge produces an electric field such that when another “test” charge is placed in the field it will experience an electrical force.

Electric field

Test charge

Electric force



qFE�

�

�

q

F�

The electric field at any point is the force per unit charge experienced by a charge at that point.

Since the electric force is a vector then the electric field is also a vector.



Electric field and electric force are vectors which point in the same direction when the test charge q is positive.

q F�

qFE�

�

�


q

F�

The electric field lines for a positive source charge point away from the source charge.

The electric field lines for a negative source charge point towards the source charge.


Consider two point charges (+Q and +q) again and the force that exists between them.

q

r

F�

rrQqkF ˆ2�

�

Electric force on q

Electric field at qQ

rrQk

qFE ˆ

2��

�

�



Lecture 1 UNITSUNITS

[k]-Coulomb constant; meter/Farad {m/F}

[q]-charge; Coulomb {C}

[r]-distance; meters {m}

[E]-Electric field; Newton/Coulomb {N/C}

[E]-Electric field; Volt/meter {V/m}

Q

r�P Observation point

rrQkE ˆ2�

�

ELECTRIC FIELD


Lecture TEXTTEXT

(1) Electric field is a vector quantity. Thus at all points where the electric field exists it has magnitude and direction.

Some properties of the electric field for a POINT CHARGE QSome properties of the electric field for a POINT CHARGE Q

(2) The charge q must be small and positive such that it does not disturb the source charge Q.

(3) For a positive source charge Q the electric field vector and the electric force on the test charge qare in the same direction.

(4) For a positive source charge Q, the electric field lines are directed away from the charge.

(5) For a point charge Q located at the origin the electric field vector is:

rrQkE ˆ2�

�Y r�

� � � �

� � � � ��

��

��

��

�

��

�

��

�

yxrˆˆ

cossinsincos

ˆˆ

��

��

�

x

y

+Q X

97.315 Basic E&M and Power Engineering Topic: COULOMB FORCE

PROBLEM SOLVING STRATEGYPROBLEM SOLVING STRATEGY

1. Consistent units are essential. Distances must be in meters,charge in coulombs. Don’t forget to convert.

2. Remember that the electric force is a vector quantity. You may want to go back and review vector algebra. It’s often useful to use components in an (x, y, z) coordinate system. Be sure to use the correct vector notation. Indicate your coordinate axes clearly on your diagram, and be certain that the components are consistent with your choice of axes.

3. In working out directions of electric field vectors, be careful to distinguish between source point and the field point P. The field produced by a positive point charge always points in the direction from the source point to field point; the opposite is true for a negative point source.

Lecture 1 ASSIGNMENTASSIGNMENT


Example (Question)What is the electric field 30 cm from a charge q = 4.0 nC?

Lecture 2 EXAMPLEEXAMPLE


Example (Solution)What is the electric field 30 cm from a charge q = 4.0 nC?

CnCq 9100.40.4 �

��

mcmr 110330 �

��

2

29100.9

41

CNm

o

�

��

��

rrqE

o

ˆ4

12

��

�

�

rCNE ˆ400�

�

q

r

P

X

Y

Z r

�

�




Example (Alternate Solution)r

rqqF

o

ˆ4

12

��

��

�Obtain the force on a test charge

CnCq 9100.40.4 �

��

q�

r

X

r

�

�

Zmcmr 110330 �

��

2

29100.9

41

CNm

o

�

��

��

rCNqF ˆ400 ��

�

q Y

rCN

qFE ˆ400��

�

�

�

Then obtain the electric field

END


Example (Question)A charged raindrop carrying 10 �C experiences an

electric force of 0.30 N in the +x direction. What is the electric field at this location? What would be the force on a -5.0 �C drop at the same location?



Example (Solution)A charged raindrop carrying +10 �C experiences an



Cq �10��

xNF ˆ30.0�

�

CxNE 61010ˆ30.0�

��

�

�

xCNE ˆ100.3 4

��

�

Electric field at location of charge q�

qFE�

�

�

�

Y

X

Z


Example (Solution)A charged raindrop carrying +10 �C experiences an


EqF��

��


Cq �0.5��

xCNE ˆ100.3 4

��

�

xCNCF ˆ100.3105 46

��

�

xNF ˆ15.0��

�

Electric force on negative charge q�

Y

X

Z


Example (Summary)A charged raindrop carrying +10 �C experiences an


You might wonder if the field should point in the -x direction when we talk about putting a negative charge in the field. It doesn’t because the whole point of the field concept is to provide a description that’s independent of the particular charge experiencing that force. The electric field in this example points in the +x direction no matter what charge we may choose to put in the field. For a positive charge the force q’E points in the same direction as the field; for a negative charge q’ < 0, and the force is opposite the field direction. As always the algebra takes care of the sign.

Cq �0.5��

F�

E�

x

Lecture 2

Cq �10��

F�

E�

ENDEXAMPLEEXAMPLE


Electric field produced by multiple point charges (n = 2)

q1


X

Y

q2

r�1r�

2r�

ZP

q1 and q2 are the source point charges.

P is the field point

The charges (q1 and q2) produce the electric field observed at the point P


Consider each charge in turn, independently of all other charges present.

q1


X

Y

1E�

2E�

Charge q1 produces an electric field at point P.

Charge q2 produces an electric field at point P2E

�

1E�

ZP

q2

The total electric field at P is the vector sum of the electric field produced by each individual charge. 21 EEEP

��

��


Consider charge q1 only


X

Y

Zq1

q2

r�

1rr ��

�

1r�

Electric field produced by charge q1 at P

��

��

��

��

�

�

�

1

12

1

11 rr

rrrr

kqE��

��

��

�

P

Distance separating q1 and P

Unit vector along line joining q1 and P


Consider charge q2 only

q1


X

Y

q2

r�

2rr ��

�

2r�

Electric field produced by charge q2 at P

��

��

��

��

�

�

�

2

22

2

22 rr

rrrr

kqE��

��

��

�

ZP

Distance separating q2 and P

Unit vector along line joining q2 and P


The total electric field at P is the vector sum of the electric field produced by each individual charge.

2rr ��

�

X

Y

Zq1

q2P

r�

1rr ��

�

1r�

2r�

��

��

��

��

�

�

�

��

��

��

��

�

�

�

2

22

2

2

1

12

1

1

rrrr

rrkq

rrrr

rrkqE

��

��

��

��

��

��

�

21 EEEP

��

��

� � � �3

2

223

1

11

rrrrkq

rrrrkqE

��

��

��

��

�

�

��

�

��



Electric field produced by multiple point charges (n > 1)

The total electric field at P is the vector sum of the electric field produced by each individual charge.

��

�

�

n

i i

ii

rrrrqkE

13��

��

�

....4321 �� EEEEE��

X

Y

Zq1

q2P

q3

q4

q5

qn

qi

ir�

irr ��

�

r�

��

�

n

iiEE

1

��



PRINCIPLE OF SUPERPOSITIONPRINCIPLE OF SUPERPOSITION

Given a group of charges we find the net electric field at any point in space by using the principle of superposition. This is a general principle that says a net effect is the sum of the individual effects. Here, the principle means that we first compute the electric field at the point in space due to each of the charges, in turn. We then find the net electric field by adding these electric fields vectorially, as usual.

....4321 �� EEEEE��

��

�

n

iiEE

1

��

Lecture 2 TEXTTEXT


Example (Question)The figure shows two point charges each of +10 nC

separated in air by 8.0 m. Compute the electric field at the points A, B, and C

+ +A

B

C

x

y4

4�

4 )(m

)(m

)(m

nC10�

nC10�



Example (Solution)The figure shows two point charges each of +10 nC


Point A: Make a sketch of the layout and then draw in vectors for the fields E1 produced by q1 and E2 produced by q2. To do that imagine a positive test charge at A. The force on it due to the charge q1 acts along the center-to-center line, is repulsive, and so points to the right. That means the E1 at A is to the right along the axis. Similarly, the force due to q2 on our imaginary test charge is to the left as is E2. Next calculate E1 and E2 and add them vectorially. We are spared this effort since E1 = E2, the two cancel and the field at A is zero

+ +A

2E�

1E�





Point B: At point B the fields act as drawn in the figure, and we must find their components. First we will calculate E1 and E2.


+ +

B2E�

1E�

� o45

X


Example (Solution)Point B: Since the charges and distances happen to be the same, the magnitudes of the two contributing fields are equal:

� �

2

9

2

29

221

45sin0.4

)100.10()100.9(��

��

�

��

�

om

CC

NmrqkEE

CNE 81.21 �

Now for the vector components

+ +

B2E�

1E�

� o45

XLecture 2 EXAMPLEEXAMPLE


Example (Solution)Point B: Since the charges and distances happen to be the same, the magnitudes of the two contributing fields are equal:

� �� yxCNE ˆ)45sin(ˆ)45cos(81.21 ��

�

� �� yxCNE ˆ)45sin(ˆ)45cos(81.22 ��

�

+ +

B2E�

1E�

� o45

XLecture 2 EXAMPLEEXAMPLE


Example (Solution)Point B: The horizontal field components are equal and act in opposite direction. They will cancel. Only the vertical field components contribute, and in the same direction.

� � � � )707.0)(81.2(245sin45sin 21 CNEEE oo

B ��

CNEB 0.4�

The direction is straight up in the positive y-direction

Y

X+ +

B2E�

1E�

� o45



Example (Solution)Point C: The point C is similarly located with respect to the charges as point B is.The field magnitude at C is the same as at B except the direction is straight down in the negative y-direction.at

� � � � )707.0)(81.2(245sin45sin 21 CNEEE oo

B ��

+ +A

B

C

x

y4

4�

4 )(m

)(m

)(m

nC10�

nC10�

CNEC 0.4�

The direction is straight down in the negative y-direction





Additional figures related to this example question

+ +

ENDLecture 2 EXAMPLEEXAMPLE


Example (Question)A molecule consist of separate regions of positive and

negative charge, modeled approximately as a positive charge q at x = a and a negative charge -q at x = -a. Find the general expression for the electric field at any point on the y axis and an approximate expression valid at large distances (y >> a)

- +ax �� ax ��

x

y r r�

E�

�E�

�E�

�

+ +�

+

-



Example (Solution)A molecule consist of separate regions of positive and

negative charge, modeled approximately as a positive charge q atx = a and a negative charge -q at x = -a. Find the general expression for the electric field at any point on the y axis and an approximate expression valid at large distances (y >> a)

The figure shows the individual field vectors E+ and E- along with their sum. The y components cancel to give a net field parallel to the x axis.

- +ax �� ax ��

x

y r r�

E�

�E�

�E�

�



Example (Solution)

The x components of the two fields are clearly the same so we have:

� ��

��

��

��sin2 2r

qkEEE xxx

The minus sign occurs because the net field points in the negative x direction.

- +ax �� ax ��

x

y r r�

E�

�E�

�E�

�Also22 ayr ��

� �22

sinay

ara

�

��

Then

� �x

ay

kqaxEE x ˆ2ˆ2

322�

��

�



Example (Solution)

Does this result make sense: Try the point y = 0.

� �x

ay

kqaxEE x ˆ2ˆ2

322�

��

�

xakqxEE x ˆ2ˆ 2��

�

At y = 0 gives

- +ax �� ax ��

x

y r r�

E�

�E�

�E�

�

Which is indeed twice the field of either charge at a distance a.



Example (Solution)

+ +�

+-

We are frequently interested in the field far from a dipole which is the reason why this example asks for an approximate expression for the electric field at y >> a.

In this case we can thus neglect a2 compared to y2 in our expression for the electric field giving:

xykqaE

ayˆ2lim 3��

��

�

xakqxEE x ˆ2ˆ 2��

�

- +ax �� ax ��

x

y r r�

E�

�E�

�E�

�



Electric field produced by a continuous charge distributionElectric field produced by a continuous charge distribution


Charged volume

Charged lineIn is often useful to imagine that there is a continuous distribution of charge.

The principle of superposition applies provided the charge distribution is divided into small elements of charge �q and the total electric field at an observation point is obtained by summing all electric field contributions from each element �q.

Charged surface


Electric field produced by a continuous charge distributionElectric field produced by a continuous charge distribution


Charged volume

Charged lineIn is often useful to imagine that there is a continuous distribution of charge.

Charged surface Recall in calculus

��

dqqq 0


rdq

dq

dq

r

r r

rr

P

Ed�

Ed�

Ed�

E�

The electric field at the point P is the sum of the vectors arising from the individual charge elements dq in the entire distribution, each calculated using the appropriate distance r and unit vector .

Ed�

r�

Lecture 2 SUMMARYSUMMARY


Electric field produced by a continuous line of chargeElectric field produced by a continuous line of charge

r�

dq�d

�� Linear charge density on the line

� �� Units; {C/m}

��ddq �� Charge on length segment �d

��

LL rdrkEdE 2

ˆ�

��

�

rrdqkEd ˆ2�

�

�

Electric field produced by all segments along line of length L

L

Electric field produced at P by one segment dq

P



Electric field produced by a continuous line of chargeElectric field produced by a continuous line of charge

P

r�

dq�d

L

��

LL rdrkEdE 2

ˆ�

��

�

,.....dxd ��

may be a function of the coordinatesusually a constant

��

��41

�k

,....222 zyxr ��

Integration over length of line charge unit vector

function of (x,y,z),….Lecture 2 THEORYTHEORY


Electric field produced by a continuous surface of chargeElectric field produced by a continuous surface of charge

P Electric field produced at P by one segment dq

r�

dqdA

Ss�

rrdqkEd ˆ2�

�

�

Electric field produced by all segments of surface SSurface charge density

Units; {C/m2}� �s�

dAdq s��

S

s

S rdArkEdE 2

ˆ�

��

Charge on surface element dA



Electric field produced by a continuous surface of chargeElectric field produced by a continuous surface of charge

r�

dqdA

S��

S

s

S rdArkEdE 2

ˆ�

��

,.....dxdydA �

�s may be a function of the coordinatesusually a constant

,....222 zyxr ��

��41

�k

Integration over surface of charge

unit vectorfunction of (x,y,z),….

P



Electric field produced by a continuous volume of chargeElectric field produced by a continuous volume of charge

Electric field produced at P by one segment dq

P

r�

dV

dq

V� Volume charge density

rrdqkEd ˆ2�

�

�

Electric field produced by all segments in volume V

V

��

V

V

V rdVrkEdE 2

ˆ�

�� V� Units; {C/m3}

dVdq V�� Charge in volume element dV



Electric field produced by a continuous volume of chargeElectric field produced by a continuous volume of charge

P

r�

��

V

V

V rdVrkEdE 2

ˆ�

��

dV

dq

,....222 zyxr ��

,.....dxdydzdV ��41

�k

�V may be a function of the coordinatesusually a constant

Integration over volume of charge

unit vectorfunction of (x,y,z),….

V



Example (Question)Wires, antennas, and similar

elongated structures can often be considered as thin rods carrying electric charge. Suppose a rod of length carries a positive charge Q distributed uniformly over its length. Find the electric field at the point P a distance a from the end of the rod.

� dq

dy

y

P0�y

a

�



Example (Solution)Wires, antennas, and similar elongated

structures can often be considered as thin rods carrying electric charge. Suppose a rod of length carries a positive charge Q distributed uniformly over its length. Find the electric field at the point P a distance a from the end of the rod.

� dq

dy

y

P0�y

a

�

Let the y axis lie along the rod, with origin at P. Consider a small length dy of the rod, containing charge dq, and located a distance y from P. A unit vector from dq to P is: y�

The field at P due to dq is:

yydqkEd ˆ

2��

�



Example (Solution)Wires, antennas, and similar elongated

structures can often be considered as thin rods carrying electric charge. Suppose a rod of length carries a positive charge Q distributed uniformly over its length. Find the electric field at the point P a distance a from the end of the rod.

� dq

dy

y

P0�y

a

�

The net field at P is the sum- that is the integral- of all the fields arising from all the dq’s along the rod:Ed

�

��

��

��

��

�� ay

ayy

ydqkEdE ˆ2



Example (Solution)

dq

dy

y

P0�y

a

�

In order to evaluate the integral we must relate dq to y. The rod carries a uniform charge Q distributed over the length . The line charge density is therefore: �

��

Q��

This is the charge per unit length. Thus a length dycarries charge dq given by:

��

��

��

��

�� ay

ayy

ydqkEdE ˆ2



Example (Solution)

dq

dy

y

P0�y

a

�

��

��

��

��

�� ay

ay ydqykE 2ˆ

��

��

��

��

�

�

� ay

ay ydyyQkE 2ˆ

�

�

��

��

��

� ��

a

ayyQkE 1ˆ

��

��

�

��

��

�

aayQkE 11ˆ

� �y

aakQE ˆ�

�

�

��

Does this answer make sense?Lecture 2 EXAMPLEEXAMPLE


END

dq

dy

y

P0�y

a

�

Example (Solution)� �

yaakQE ˆ�

�

�

��

Consider ��a

yakQE

aˆlim 2

�

�

��

�

�

The rod appears as a small concentration of total charge Q at a distance a away from the point P

Does this answer make sense?Lecture 2 EXAMPLEEXAMPLE


Example (Question)A long straight power line coincides with the x axis and

carries a linear charge density C/m. What is the electric field at point P on the y axis. Use the approximation that the line is infinitely long.

��



Example (Solution)A long straight power line coincides with the x axis and carries a linear

charge density C/m. What is the electric field at point P on the y axis. Use the approximation that the line is infinitely long.

��

Here both the direction and magnitude of the electric field element dE arise from charge elements on the line vary with the position x of the charge element.


The figure shows that charge elements on opposite sides of the y axis give rise to electric fields whose x components cancel. Thus the net field points in the y direction, that is away from the positively charged rod.

dq

22 yxr ��

x

y

P�

dq

Ed�

Ed�

yEd�

�




��

Each element of charge dq contributes an amount dEy to the net electric field at P.

� ��cos2rdqkdEy �

dxdq�

��

With:

22 yxr ��

� �22

cosyx

y�

��

dq

22 yxr ��

x

y

P�

dq

Ed�

Ed�

yEd�

�





��

Each element of charge dq contributes an amount dEy to the net electric field at P.

� � 2322 yx

ydxkdEy�

��

�

� ��

��

��

�

��

2322 yx

ydxkdEEE yy�

�

dq

22 yxr ��

x

y

P�

dq

Ed�

Ed�

yEd�

�

We integrate from x = -� to x = +�



Example (Solution)

� ��

��

��

�

��

2322 yx

ydxkdEEE yy�

�

��

��

��

�

�

��

�

�

��

222 yxyxykE

��

��

��

� �� 22

11yy

ykE�

�

ykE ��2

�

dq

22 yxr ��

x

y

P�

dq

Ed�

Ed�

yEd�

�

x limits



Example (Solution)

Since the line is infinite in both directions and has cylindrical symmetry, the expression for the electric field holds for any point a distance y from the line. The result thus shows that the electric field from a positively charged infinite line points radially away from the line.

ykE ��2

�

dq

22 yxr ��

x

y

P�

dq

Ed�

Ed�

yEd�

�

The magnitude of the E field drops as

1/y



Example (Question)A thin ring of radius a is centered on the origin and

carries a total charge Q distributed uniformly around the ring. Find the electric field at a point P located a distance x along the axis of the ring, and show that the result makes sense at x >> a.

dq

a

O

�

22 ax �

x

PEd�

xEd�

�

Ed�



Example (Solution)A thin ring of radius a is centered on the origin and carries a total

charge Q distributed uniformly around the ring. Find the electric field at a point P located a distance x along the axis of the ring, and show that the result makes sense at x >> a.

A point on the ring axis is equidistant from all points on the ring, so the field magnitudes dE are the same but their directions vary.

dq

a

O

�

22 ax �

x

PEd�

xEd�

�

Ed�

The figure shows that any components perpendicular to the x axis cancel for any pair of charge elements on opposite sides of the ring, leaving a net field in the x direction.



Example (Solution)A thin ring of radius a is centered on the origin and carries a

total charge Q distributed uniformly around the ring. Find the electric field at a point P located a distance x along the axis of the ring, and show that the result makes sense at x >> a.

Each charge element dq contributes an amount dEx to the total field.

� ��cos2rdqkdEx �

with22 axr ��

� �22

cosax

x�

��

� � 2322 ax

xdqkdEx�

�

Gives

dq

a

O

�

22 ax �

x

PEd�

xEd�

�

Ed�



Example (Solution)

In this example k, x, and a are constants so we have:

��

ringxx dEEE

� ��

�

�

ring ax

xdqkE2

322

� ��

�

�

ring

dqax

xkE2

322

� � 2322 ax

xQkE�

�

dq

a

O

�

22 ax �

x

PEd�

xEd�

�

Ed�

Does this result make sense?Lecture 2 EXAMPLEEXAMPLE


Example (Solution)A thin ring of radius a is centered on the origin and carries a

total charge Q distributed uniformly around the ring. Find the electric field at a point P located a distance x along the axis of the ring, and show that the result makes sense at x >> a.

At a large distance from the ring x >> a.dq

a

O

�

22 ax �

x

PEd�

xEd�

�

Ed�

2limxQkE

ax�

��

The ring appears as a small concentration of total charge Q at a distance x from the point P.



Example (Question)Find the electric field caused by a uniform surface charge

density �s on a disk of radius R, at a point along the axis of the disk a distance x from its center. Assume that x is positive.

P xdExx

Qdr

dQ

R

rO



Example (Solution)Find the electric field caused by a uniform surface charge density �s

on a disk of radius R, at a point along the axis of the disk a distance x from its center. Assume that x is positive.

We can represent the charge distribution as a collection of concentric rings of charge. We already have the expression for the field produced from one ring, so all we have to do is add the contributions from all the rings.

P xdExx

Qdr

dQ

R

rO

� � 2322 ax

xQkE�

�

Result for one ring of charge Q and radius aLecture 2 EXAMPLEEXAMPLE




Execute a change of variables:

rdrQ s��2�

ra �

xdEE �

P xdExx

Qdr

dQ

R

rO

� � � � 23222

322

2

rx

rdrxkdEax

xQkE sx

�

��

�

��Then






To find the total field due to all the rings, we must integrate over r from 0 to R.

� ��

�

��

Rs

R

xrx

rdrxkdEE0 2

3220

2��

P xdExx

Qdr

dQ

R

rO

Then

� ��

�

��

R

s

R

xrx

rdrkxdEE0 2

3220

2��




� ��

�

�

R

srx

rdrxkE0 2

3222��

� �

R

srx

xkE0

2122

12��

�

�

��

�

�

�

� ��

P xdExx

Qdr

dQ

R

rO

� � ��

�

�

��

�

��

��

xRxxkE s

1122

122��



Example (Question)A spherical shell of radius R is uniformly charged

with surface charge density �s. Find the electric field at an exterior point.

R

P

Uniform surface charge on shell.



Example (Solution)A spherical shell of radius R is uniformly charged with surface

charge density �s. Find the electric field at an exterior point.

R

r�

� � � � � �2122

122

12 zzyyxxr ��

here � �222 ,, zyx

dA (x1,y1,z1)P (x2,y2,z2)

constants and

variables that depends on location of dA. � �111 ,, zyx





P (x2,y2,z2)

R

(x1,y1,z1)r�

� � � � � �� r

zzyyxxr �

121212 ,,ˆ ��

�

rdA

variable that depends on location of dA. Lecture 2 EXAMPLEEXAMPLE




R

(x1,y1,z1)r�

variable that depends on location of dA.

dA

dAdq s��

P (x2,y2,z2)





R

variable that depends on location of dA.

dAdq s��

� � �� ddRdq s sin2�

�

�



Example (Solution)Field at P produced by charge element dq:

R

dq

�

�

rrdqkEd ˆ2�

�

�

r� r

� �

� � � � � �� 121212

232

122

122

12

2

,,sin zzyyxxzzyyxx

ddRkEd s��

��

��

P



Example (Solution)Field at P produced by all charge elements dq is obtained by vector summation of all the field contributions from each surface charge element.

� �

� � � � � ��

��

��

Surface

s

Surface

zzyyxxzzyyxx

ddRkEdE 1212122

3212

212

212

2

,,sin ��

This is not an easy task. It can be done but requires a lot of work. It can best be solved by converting from (x,y,z) coordinates to (r,�,�) coordinates.

We will see an easier way to obtain the electric field from surface and volume charge distributions: Gauss’s law.Gauss’s law.



Example (Question)A cylindrical volume of radius R is uniformly

charged with volume charge density �V. Find the electric field at an exterior point.


P

R

Uniform volume charge in cylinder.


Example (Solution)A cylindrical volume of radius R is uniformly charged with

volume charge density �V. Find the electric field at an exterior point.


R

r�

� � � � � �2122

122

12 zzyyxxr ��

here � �222 ,, zyx

(x1,y1,z1)dVP (x2,y2,z2)

constants and

variables that depends on location of dV. � �111 ,, zyx





P (x2,y2,z2)

R

r�

� � � � � �� r

zzyyxxr �

121212 ,,ˆ ��

�

rdV(x1,y1,z1)

variable that depends on location of dV.





R

(x1,y1,z1)

r�


dV

dVdq V��

P (x2,y2,z2)




z



dVdq V��

dzrdrddq V ��

r

21

21

21 zyxr ��


Example (Solution)Field at P produced by charge element dq:

�

r

dq rrdqkEd ˆ2�

�

�

r� r

� � � � � �� 121212

232

122

122

12

,, zzyyxxzzyyxx

dzrdrdkEd V��

��

��

P



Example (Solution)Field at P produced by all charge elements dq is obtained by vector summation of all the field contributions from each volume charge element.

� � � � � ��

��

��

Volume

V

Volume

zzyyxxzzyyxx

dzrdrdkEdE 1212122

3212

212

212

,,��

This is not an easy task. It can be done but requires a lot of work. It can best be solved by converting from (x,y,z) coordinates to (r,�,z) coordinates.

We will see an easier way to obtain the electric field from surface and volume charge distributions: Gauss’s law.Gauss’s law.




ELECTRIC FIELDELECTRIC FIELDvector

single chargecharge distribution

ELECTRIC FORCEELECTRIC FORCEvector

on chargeon charge distribution

TESTTESTCHARGECHARGE



These questions are straight forward. Plug in the numbers and get your answer. Being able to solve this type of question ensures you of at least a grade of 25% on a quiz or final exam containing questions related to this lecture.

25

These questions require a few manipulations of equations or numbers before the answer can be obtained. Being able to solve this type of question ensures you of at least a grade of 50% on a quiz or final exam containing questions related to this lecture.

50

These question are quite involved and requires a thorough understanding of the topic material. Being able to solve this type of question ensures you of at least a grade of 75% on a quiz or final exam containing questions related to this lecture.

75

These questions are the most difficult and require a thorough understanding of the topic material and also pull in topics from other lectures and disciplines. Being able to solve this type of question ensures you an A grade on a quiz or final exam containing questions related to this lecture.

100

These form excellent review questions when preparing for the quiz and final exam.10075

25 10050 75

SELL EVALUATION SCALESELL EVALUATION SCALE



25 A 65 �C point charge is a the origin. Find the electric field at the points (a) x = 50 cm, y = 0 cm; (b) x = 50 cm, y = 50 cm; (c) x = -25 cm, y = 75 cm.



25 The electron of a hydrogen atom is 0.0529 nm from the proton. What is the proton’s electric field strength at this distance?



25 A 1.0 �C charge and a 2.0 �C charge are 10 cm apart. Find a point where the electric field is zero.



50 A 30 cm long rod carries a charge of 80 �C spread uniformly over its length. Find the electric field strength on the rod axis, 45 cm from the end of the rod.

mVans /101312.2: 6�



50 (a) Find an expression for the electric field on the y axis due to the two charge q. (b) At what point does the field on the y axis have its maximum strength?

aq

22 yar ��

YQ

y

q Xa

� �y

ay

yqaanso

ˆ2

:)(2

322�

��



75 Three identical charges q form an equilateral triangle of sides a, with two charges on the x axis and one on the positive y axis. (a) Find an expression for the electric field at a point on the y axis above the uppermost charge. (b) Show that your result reduces to the field of a point charge 3q for y >> a.

yay

y

ay

qaanso

ˆ

4

2

23

14

:)(2

32

2

2

2

��

�

�

��

�

�

��

��

��

�

��

��

��

��



75 The figure shows four point charges fixed at the corners of a rectangle, in vacuum. (a) Compute The magnitude and direction of the electric field at the center of the rectangle. (b) Compute the electrostatic force acting on the +100 �C charge.

1q

2q3q

4q m0.4

C�125�

C�100�C�32�

C�36�

� � mVyxaans /ˆ1050.2ˆ1042.2:)( 44��



75 A thin rod of length carries a charge Q distributed evenly over its length. A point charge with the same charge Q lies a distance b from the end of the rod. Find a point where the electric field is zero.

�

Q Q

b



75 The rods shown in the figure are both 15 cm long and both carry 1.2 �C of charge. (a) Find the magnitude and direction of the electric field at the point P. (b) repeat the problem where the right hand rod carries -1.2 �C of charge.

P

o45 o45

cm15

cm15 C�2.1 C�2.1

mVxbans /ˆ10339:)( 3�



75 Two identical rods of length lie on the x axis and carry uniform charges � Q. (a) Find an expression for the electric field strength as a function of the position x for points to the right of the right-hand rod. (b) Show that your result has the 1/x3 dependence of a dipole field for x >> . �

�

Q� Q�

��x0�x��x

xxx

Qaanso

ˆ12

:)( 23 ��

��

�

� �

�

��



75 A semicircular loop of radius a carries positive charge Q distributed uniformly over its length. Find the electric field at the center of the loop.

P

dq

a�

�d

xa

anso

ˆ24

:��

��



100 A semicircular loop of radius a carries positive charge Q non-uniformly distributed over its length. Find the electric field at the center of the loop. The charge density varies linearly with the angle � with b constant.

�� b��

P

dq

a�

�d



100 Determine the magnitude and direction of an electric field if the electron placed in it, in vacuum, is to experience a force that will cancel its weight at the earth’s surface.



100 A section of an advertising sigh consists of a long tube filled with neon gas having electrodes inside at both ends. An electric field of 20 kN/C is set up between the electrodes, and neon ions accelerate along the length of the tube. Given that the ions have a mass of 3.35 X 10-26 kg and are slightly ionized, determine (a) their acceleration and (b) terminal velocity when they reach the end of a 1 m long tube.


A spherical shell of radius R is uniformly charged with surface charge density �s. Find the electric field at an exterior point.

100

R

P

Uniform surface charge on shell.



A cylindrical volume of radius R is uniformly charged with volume charge density �V. Find the electric field at an exterior point. Assume cylinder is infinite in length.

100


P

R

Uniform volume charge in cylinder.


(29) A shown in the figure a positive charge +Q is located at the origin and an array of equally spaced negative charges (-q) are placed along the x axis. Obtain a compact expression for the electric field at the point P due to all the other charges when: (a) N = 1, (b) N = 2, (c) N = 10, (d) N = 100, (e) N = 1000, (f) N = infinity. You may find it instructive to plot electric field versus charge number N.

100

P -q-q -q-q -q -q -q

Xa a a a a a a



(30) A shown in the figure a positive charge +Q is located at the origin and an array of equally spaced alternating sigh charges (�q) are placed along the x axis. Obtain a compact expression for the electric field at the point P due to all the other charges when: (a) N = 1, (b) N = 2, (c) N = 10, (d) N = 100, (e) N = 1000, (f) N = infinity. You may find it instructive to plot electric field versus charge number N.

100

P -q+q -q-q +q -q +q

Xa a a a a a a



Lecture 2 REFERENCESREFERENCES

(0) Inan p. 246 - 255(1) Kraus p. 12 - 15(2) Reitz p. 27 - 31(3) Plonus p. 2 - 4(4) Winch p. 258 - 266(5) Lorrain p. 40 - 42(6) Duckworth p. 5 - 8(7) Jackson p. 27 - 28(8) Ulaby p. 7, 143 - 144

(0) Textbook: U. S. Inan, A. S. Inan “Engineering Electromagnetics”

(1) J.D. Kraus, K. R. Carver “Electromagnetics” 2nd

(2) Reitz, Milford, Christy “Foundations of Electromagnetic theory” 4th

(3) M. Plonus “Applied Electromagnetics”(4) R. P. Winch “Electricity and Magnetism”(5) P. Lorrain, D. Corson “Electromagnetic fields and Waves”

2nd

(6) Duckworth “Electricity and Magnetism”(7) J.D. Jackson “Classical Electrodynamics” 2nd

(8) F. Ulaby, “Fundamentals of applied Electromagnetics”


Lecture 2 REFERENCESREFERENCES

(0) Textbook: U. S. Inan, A. S. Inan “Engineering Electromagnetics”

(1) J.D. Kraus, K. R. Carver “Electromagnetics” 2nd

(2) Reitz, Milford, Christy “Foundations of Electromagnetic theory” 4th

(3) M. Plonus “Applied Electromagnetics”(4) R. P. Winch “Electricity and Magnetism”(5) P. Lorrain, D. Corson “Electromagnetic fields and Waves”

2nd

(6) Duckworth “Electricity and Magnetism”(7) J.D. Jackson “Classical Electrodynamics” 2nd

(8) F. Ulaby, “Fundamentals of applied Electromagnetics”

(0) Inan p. 255 - 269(1) Kraus p. 15 - 20(2) Reitz p. 31 - 33(3) Plonus p. 4 - 6(4) Winch p. 266 - 271(6) Lorrain p. 42 - 43(6) Duckworth p. 8 - 16(7) Jackson p. 28 - 29(8) Ulaby p.


X

Y

Zq1

q2

q3

q4

q5

qn

qi

ir�

irr ��

�

r�

��

�

n

iiEE

1

��

The electric field at the point P is the sum of the vectors arising from the individual charge elements dq in the entire distribution, each calculated using the appropriate distance r and unit vector .

Ed�

r�

rdq

dq

dq

r

r r

rr

P

Ed�

Ed�

Ed�

E�

PSUPERPOSITION PRINCIPLESUPERPOSITION PRINCIPLE



Lecture 2 ENDEND

Electric field Lecture

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