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Electric Field Lines
See Sec. 23.6

Electric Field Line= s, cont.

Electric Field Line Patterns

Electric Field Line Patterns

Electric Field Line Patterns

Electric Field Line Patterns

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Electric Field Lines
See Sec. 23.6

A convenient aid for visualizing electric field patterns is to draw lines pointing in the direction of the field vector at any point

These are called el= ectric field lines and were introduced by Michael Faraday

Electric Field Line= s, cont.

The field lines are= related to the field by

The electric field = vector, E, is tangent to the electric field lines at each point

The number of lines= per unit area through a surface perpendicular to the lines is proportional to the strength of the electric field in a given region

Electric Field Line Patterns

Point charge=

The lines radiate e= qually in all directions

For a positive sour= ce charge, the lines will radiate outward

Electric Field Line Patterns

For a negative sour= ce charge, the lines will point inward

Electric Field Line Patterns

An electric dipole = consists of two equal and opposite charges

The high density of= lines between the charges indicates the strong electric field in this region

Electric Field Line Patterns

Two equal but like = point charges

At a great distance= from the charges, the field would be approximately that of a single charge of 2q

The bulging out of = the field lines between the charges indicates the repulsion between the charges

The low field lines= between the charges indicates a weak field in this region

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Second level

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=	<= /td>	<= /td>
�� PHYSICS� 2212K

TEXT: PHYSICS FOR SCIENTISTS

�� AND ENGINEERS, 7^TH = EDITION

AUTHORS: Serway, R.;� Jewett, J.<= br>

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= CHAPTER 23� -� ELECTRIC FIELDS

	=	<= /td>	<= /td>	<= /td>	<= /td>		<= /td>	<= /td>	<= /td>
SECTION 23.1� -� ELECTRIC CHARGE

q� and q₀�are symbols used for charge

These are not two different kinds of charge, they represent two differ= ent
roles for charge to play:
�� q� -� source charge�� q₀� -� test charge

�� SI Unit= For Charge : Coulomb, C

�� e�� -= �� charge on the proton

�� e� =3D� 1.6x10^-19 C

�� -e� is the charge on the electron

�� e� is� a� positive� number

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F=

F=

-= F

-= F

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=	=
�� Benjamin Franklin introduced� the� designations �positive� and
and �negative�� although he did = not think that there were two kinds
of charge as we= do today.

=	<= /td>
�� Franklin thought that objects tha= t had an excess of charge were
�po= sitive� and objects with a deficit of charge were �negative�. His
model does not = explain some phenomena that are observed.

Today we accept= two kinds of charge, positive and negative

+<= /font>

-<= /font>

F�� -F

SEE<= span style=3D'mso-spacerun:yes'>� SEC.� 23.2

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<= /td>	<= /td>
CHAR= GING BY
COND= UCTION
(CONTACT)=

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=
= SECTION S 23.4 � THE ELECTRIC FIELD
SECTION 23.7� - MOTION OF CHARGED PARTI= CLES

=	<= /td>
= E� - the electric field

�� charges� produce E fields

�� charges respond t= o E fields

(1)<= span style=3D'mso-spacerun:yes'>� F� =3D� q₀ E<= /b>

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=	<= /td>
NOTE: Eq.23.8, p 652� F =3D q = E is wrong
compare with eq 23.7 on p 651. q₀ is supposed
to = be used for �test� charges.

Ex

F=

q0

<= /td>
Fx =3D q₀ E_x > 0
�� F_{x is in the direction of E_x}

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(ii)= q₀ < 0

Ex

F=

q0

=
F_x� =3D� q₀ E_x;�� F_x < 0
Fx� is� opposite� E_x=

<= /td>
(iii= ) q₀ =3D 0

F_x� =3D� q₀ E_x� =3D� 0

=
i� and ii� show the First Major

Con= vention of Electricity:

		=	<= /td>	=
+ charges experience a force in the direction

of an applied E

-	charges experience a force opposite the

direction of an applied E

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See= Eq 23.7, p 651

=
EXAMPLE: a CRT emits electrons that are subjected to a horizontal deflecting
Ele= ctric field, E_x�=3D 0.5x10⁴⁼� N/C. Find F_x, the force = on the electrons

scre= en

y

Fx

Ex

-e

x

A

C

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=	=	<= /td>	<= /td>	<= /td>
ACCELERATION� DUE� TO� ELECTROSTATIC� FORCE:
�� A particle of mass, m has a net = test charge, q₀ and is subjected to electric
Fie= ld, E. the force, F on the charge is:
��= �� NOTE: on p 661 above Eq 23.12
�� F� =3D� q₀E�� q₀ should be used for the �test� charge

From Newton�s 2^nd Law, the charge may undergo an acceleration, a:

Whe= re F_net is the net force acting on the charge

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=
EXAMPLE:� Find the acceleration, a_{x of the electron in the CRT if the}
Hor= izontal deflecting field, E_x =3D 0.5x10⁴ N/C

d

Fx

l

defl= ection plates

Ex

v= ₀

-e

<= /td>
A

C

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ax

-e

Fx

Ex

Δ= ;

-e

=	<= /td>	<= /td>
HOMEWORK: If the electron�s speed is v₀ =3D 5x10⁶ m/s
As it enters the deflection plate region which is 1 cm
In length, find Δ, the deflection distance of the
Ele= ctron upon leaving the plate region.

a= _x

v0

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=	=
= II PRODUCTION �
�� A source charge, q is located at = the origin, O. It produces an electric field, E
Everywhere in space. E at point, P a distance r from q is:

E=

P

r

q O<= /font>

<= /td>	<= /td>
The� r� shows� that� E is in the
r dimension. +r is away from q
-r i= s toward q

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<= /td>		=
Removing the vector notation:
(6)	�� E_{r� =3D� k q / r²}

= 49;� -� electric permittivity; ε₀ � value of ε� in� a� �free space� environm= ent

<= /td>
A vacuum is a free space environment and air is treated as �approximately=
free= space�.

=
In a dielectric medium, ε > ε₀.� If a charge distribution produces an electric
Fie= ld of magnitude, E₀ in air, the field in a dielectric would be <= ; E₀.

Diel= ectric, ε > ε₀

<= /td>
air gap
^= 9; =3D ε₀

-

-

-

-

-

+

+

+

+

+

+

+

+

+

+

-�� +

-�� +

E= ₀

E=

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Er

<= /td>
Direction of E_r �
(i) = q > 0

P

r

<= /td>
Er =3D (k q) / r² > 0

E points away from q

+q

(ii)= � q < 0

Er

P

r

Er =3D (k q) / r²�< 0

-q

= E points toward q

=	=
(i) and (ii) show : the second major convention of electricity
�� E �POINTS AWAY FROM + SOURCE CHARGE = AND
�� TOWARD � SOURCE CHARGE<= /font>

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=
It�s often said that �Electric Field Lines start on + charge and end on nega= tive
cha= rge or at infinity.

= 49;₀� =3D� 8.85x10^-12 C² / N m²�� and

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<= /td>	=
Electric Field Lines

	See Sec. 23.6

<= /td>		=	<= /td>		<= /td>
�	A convenient aid for visualizing electric

	field patterns is to draw lines pointing in the

	direction of the field vector at any point

�	These are called electric field lines and were

	introduced by Michael Faraday

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Electric Field Lines, cont.

<= /td>			<= /td>		=	=	<= /td>	<= /td>
�	The field lines are related to the field by

	�	The electric field vector, E, is tangent to the

	electric field lines at each point

	�	The number of lines per unit area through a

	surface perpendicular to the lines is

	proportional to the strength of the electric field

	in a given region

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Electric Field Line Patterns

<= /td>		<= /td>	<= /td>		<= /td>		<= /td>
�	Point charge

�	The lines radiate

	equally in all

	directions

�	For a positive source

	charge, the lines will

	radiate outward

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Electric Field Line Patterns

<= /td>		<= /td>	<= /td>
�	For a negative source

	charge, the lines will

	point inward

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Electric Field Line Patterns

<= /td>		<= /td>	<= /td>		<= /td>	<= /td>
�	An electric dipole

	consists of two equal

	and opposite charges

�	The high density of

	lines between the

	charges indicates the

	strong electric field in

	this region

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Electric Field Line Patterns

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Two equal but like point

charges

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At a great distance from the

charges, the field would be

approximately that of a single

charge of 2q

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The bulging out of the field

lines between the charges

indicates the repulsion between

the charges

�

The low field lines between the

charges indicates a weak field

in this region

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=	=
EXAMPLE: Find the electric field, E_r a distance, a₀ =3D� 5x10^-11 m from
an ¹H nucleus. (Note: a₀ is the �Bohr Radius� the radius of the pa= th of the
ele= ctron in the ground state of the �Bohr Atom�)

Er

r = =3D a₀

-e

+e

WHY = SO BIG ?

Thi= s is not the force on the electron it�s just the electric field

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=
= E Due to a Discreet Charge Distribution �
For multiple source point charges E is found as a vector sum = of the E_i

E= ₂

y

E= ₁

E= ₃

P

q3

q2

x

q1

=
For instance, for three source charges, q₁, q₂, q3, E at point P is:

�(7)�� E� =3D� E₁ + E₂ + E₃

=
The sum in (7) must be done vectorially:
�� E_x =3D E_1x= + E_2x + E_3x
_{��}�E_y =3D E_1y + E= _2y + E_3y

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E= ₂

y

E= ₁

E= ₃

P

q3

q1

q2

x

Ex� =3D� E_1x + E_2x + E_3x

Ey� =3D� E_1y + E_2y + E_3y

F<= /b>

q4

E=

P

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+y

E= ₁

E= XAMPLE -

θ= ;

P

x

E= ₂

r

b

θ= ;

q1

a

a

q2

+x

q1� �=3D� +10 μC;�� q_=
2� =3D� -10 μC

a� =3D� 3 m;� b� =3D� 4 m;�� θ� =3D� 53.13�

1st Find Magnitudes:

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+y

E= ₁

θ= ;

P

x

E= ₂

b

θ= ;

a

a

q1

q2

2nd, Find x and y components:

=3D<= span style=3D'mso-spacerun:yes'>� 2.16 x 10³ N/C

E2x� =3D� E₂ (cos θ)� =3D� 2.16 x 10^{3 N/C}

E1y� =3D� E₁ (sin θ);��= �� E_2y� =3D� -E₂ (sin θ)� =3D� -E_1y

3rd� Find Σ E_x�s�� and Σ E_y�s:

Ex� =3D� E_1x + E_2x� =3D� 2.16 x 10^{3
N/C� +� 2.16 x 10^3�N/C�
=3D� 4.32 x 10^{3 N/C}}

Ey� =3D� E_1y + E_2y� =3D� 0

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E= ₁

θ= ;

P

x