chapter 17 electric charge and electric field …people.physics.tamu.edu/adair/phys202/chapter...
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CHAPTER 17
ELECTRIC CHARGE AND
ELECTRIC FIELD
TWO BASIC CONCEPTS:
COULOMB’S LAW
� = � ������
ELECTRIC FIELD
� = �
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ATOMS – are neutral
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IONS – are charged
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SOME MATERIALS –
e’s move easily – conductors
metals
SOME MATERIALS –
e’s don’t move easily –
insulators
glass
wood
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FORCES ON CHARGED OBJECTS
LIKE CHARGES REPEL
UNLIKE CHARGES ATTRACT
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USE COULOMB’S LAW
� = � � ����
Example
Two point charges q1 = 25 nC
and q2 = -75 nC are separated
by a distance of 3.0 cm. Find
the magnitude and direction of
the electric force that q1 exerts
on q2.
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� = � � ����
= �9�10�� �+25�10����−75�10���
�0.030���= 0.01875!
Attractive Force
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Also
Newton’s third law – if body A
exerts a force on body B, then
body B exerts an equal and
opposite force on body A.
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ELECTRIC FIELD
Electric force per unit charge.
� = �
Or
� = �"#$"
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In direction that positive charge
would move.
For more than one charge the
total electric field equals the
vector sum of all electric fields
due to each charge.
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For two charges q1 and qtest
� = � % %&'(&��
So
� = �"#$" =
)*�*"#$"+��"#$"
Therefore
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� = � ����
Or
� = � ���
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ALSO
� = ,-./
Where
01 = 8.854�10� � 3�!��4
Therefore can write
� = ,-./
������
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And
� = ,-./
���
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ELECTRIC FIELD LINES
1. IN DIRECTION OF FIELD
(POSITIVE TEST CHARGE
MOVES ALONG LINE.)
2. NUMBER OF LINES
PROPORTIONAL TO
ELECTRIC FIELD.
3. ELECTRIC FIELD LINES START
ON POSITIVE CHARGE AND
END ON NEGATIVE
CHARGE.
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Example (continued)
Find the electric field 3.0 cm
from an electric charge
q1 = +25 nC.
r = 3 cm
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What is E at distance r due to q1
� = ,-./
��� =
,-./�56 1781.19�
� = 2.5�105! 34
DIRECTION ?
Away from positive q1
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If we place a charge -75nC at a
point 3 cm away from q1 what
is the force on this charge?
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The electric field at q2 is
2.5�105! 34
Therefore
� = %�
= �−75�10��3��2.5�105! 34 )
= 0.01875!
WE OBTAINED FORCE IN TWO
WAYS
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Example- Field from two
charges
Q1 = -50µC at x=0.52m, y=0
Q2 = 50µC at x=0, y=0 and
find E at point A (x=0, y=0.3m).
y
. A
300
. . X
Q2 Q1
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E; =k�−50x10�>C�
�0.6m��
= −1.25x10> N C4
(towards Q1)
�C� = DE51F 17GHI�1.9J��
= 5�10>! 34
(away from Q2)
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EA2
EA
∅
A EA1
300
�C6 = �C LMN301= 1.1�10>! 34
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�CO = �C� − �C NPQ301= 4.4�10>! 34
�C = R�C6� + �CO�= S�1.1�10>�� + �4.4�10>��
�C = 4.5�10>! 34
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TUQ∅ = �CO�C6 =
4.4�10>! 341.1�10>! 34
= 4
∅ = 761
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NOW CALCULATE THE FORCE ON A
CHARGE Q3 = 50µC PLACED AT POINT A.
� = %� = �50V10�>3�4.5�10>! 34 = 225!
WHAT DIRECTION?
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ANYWHERE NEAR AN INFINITE SHEET OF
CHARGE THE ELECTRIC FIELD WILL BE
W2X1
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GAUSS’S LAW
TWO BASIC CONCEPTS
ELECTRIC FLUX
AND
GAUSS’S LAW
ELECTRIC FLUX
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Consider a number of guns at the left
shooting to the right.
Bullets
Sheet
Flux of bullets is number of bullets times
area of sheet perpendicular to bullets.
�YZV[\]]^_` = ![\]]^_`�a`b^^_
Turn sheet on edge.
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Bullets
Sheet
Flux now near zero since area of sheet is
parallel to bullets.
Turn sheet to angle.
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Bullets
Sheet
Flux somewhere between previous two
values.
Assign area vector to sheet.
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Bullets
φ
Sheet A
Flux will be
Φ[dee'&( = 
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FOR ELECTRIC FLUX
ELECTRIC
FIELD φ
Sheet A
f = �aLMNf
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GAUSS’S LAW
We will state the law then work some
examples. After that we will do an example
to justify the law.
Statement of Gauss’s law:
The flux through a closed surface is equal to
the net charge enclosed by the surface
divided by ε0.
Equation:
g�ah = 4i��'jke
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g�aLMNf = 4i��'jke
� = 14iX1
Therefore
g�aLMNf = 4i 14iX1 �'jke = �'jke
X1
Consider a conducting sphere of radius R
with charge q on the surface.
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Draw a Gaussian surface around this sphere
with radius r.
Apply Gauss’s Law
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g�aLMNf =�'jkeX1
�'j = %
��4i��� = %01
� = 14iX1
%��
The same equation we had for a point
charge.
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Example
Positive electric charge Q is distributed
uniformly throughout the volume of an
insulating sphere with radius R.
a. Find the electric field at point p where
r < R.
b. Find the electric field at point p where
r > R.
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Part a. Radius of Gaussian Surface = r
Gauss’s Law
g�aLMNf =�'jkeX1
Need charge enclosed by surface.
Charge density l = _m&neopn�q'_m&nermedJ' = �
st-ut
Charge enclosed by Gaussian Surface
�'j = l�vMwx�yyQLwMNyz
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�'j = l {43i�9| =�
43i}9{43i�9| =
�9}9 �
g�aLMNf =�9}9 �01
��4i��� =�9}9 �01
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� =�9}9 �01
4i��
� = 14i01
��}9
For r < R
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Part b.
Draw Gaussian Surface outside sphere.
g�aLMNf =�'jkeX1
�'j = �
Same as before
��4i��� = �01
� = 14i01
���
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