chapter 18 electric potential - texas a&m...
TRANSCRIPT
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CHAPTER 18
ELECTRIC POTENTIAL
BASIC CONCEPTS:
ELECTRIC POTENTIAL ENERGY
ELECTRIC POTENTIAL
ELECTRIC POTENTIAL GRADIENT –
POTENTIAL DIFFERENCE
POTENTIAL ENERGY
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h
PE = U = mgh
PE KE
Or
U K
And U + K = total energy = constant
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BOOK EXAMPLE
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Charged Particle in Electric Field is similar
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Consider a point charge q that sets up an
electric field in space
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Now a test charge q0 is placed at position a
a distance ra from q0. Then q0 moves to
position b a distance rb from q0.
What is the change in potential energy?
The change in potential energy is the
negative of the work done to move the test
charge from a to b.
The force on the test charge is
� = 14���
���
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The work done is force times distance. But
the force changes as q0 moves away from q
See Figure 18.6
Must integrate (Not necessary to
understand but the procedure is in the box
on the next page.)
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����� = � ��� =��
� 14���
�����
= ���4��� �1� −
1��
Δ� = �� − �� = − ���4��� �1� −
1��
Use �� = ���
Then
The change in potential energy,Δ�, is the
negative of this work.
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So difference in potential energy between
two points is
Δ� = �� − �� = − ���4��� �1� −
1��
Now define the potential energy to be zero
when the two charged particles are
separated by an infinite distance (b = ∞; Ub
= 0).
0 − �� = − ���4��� �1� −
1∞�
�� = ���4���1�
Or
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�� = ���4���1�
Use �
� !" = # and replace the symbol a
with r the distance from the charge
providing the field to the charge.
� = # $$%�
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ELECTRIC POTENTIAL
Start with:
Difference in potential energy is
Δ� = �� − �� = − ���4��� �1� −
1��
DEFINITION:
ELECTRICAL POTENTIAL IS POTENTIAL
ENERGY PER UNIT CHARGE
&'()*+)�',-*(.*+�'= ,-*(.*+�'&.(/0
�.+*1ℎ�/(
Therefore divide all terms by ��
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���� −���� =
− ���4��� 31� −
1�4��
Thus
5� − 5� = − �4��� �
1� −
1��
5� − 5� = �4��� �
1� −
1��
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Then for potential at a point
The potential energy at a point is
� = # ��6
&'()*+)�',-*(.*+�'= ,-*(.*+�'&.(/0
�.+*1ℎ�/(
U= # $$%�
5 = ��6 = # ��6�6 = # �
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POTENTIALS ADD (SCALERS)
Just as we did with the electric field we can
add the potentials for many charges in an
area.
EXAMPLE
A ∙ 60cm
30cm
8� ∙ ∙ 8
50µC -50 µC
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What is the potential at A?
59 = 59: + 59<
59 = 14�=�
4�=�−50?10@A1
0.6D
59 = 1.5?10A5 − 7.5?10G5 = 7.5?10G5
ELECTRON VOLT
An electron volt is a unit for energy. It is
the work necessary to move an electron
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(charge ( = 1.6?10@�H1) a potential
difference of 1 volt.
1 Volt Batt
The work to move a charge � across a
potential difference is � = �5
� = �5 = I1.6?10@�H1JI15J= 1.6?10@�HK
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Therefore
1(5 = 1.6?10@�HK
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18
CAPACITANCE AND
DIELECTRICS
MAJOR TOPICS
Calculating Capacitance
Capacitors in Circuits
Energy Storage in Capacitors
Dielectrics
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Capacitors store:
Charge
Energy
+Q + - -Q
+ -
+ -
+ -
+ V _
BATTERY
+ v -
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The battery supplies charge to the plates.
The charge Q is proportional to V.
8 ∝ 5
Choose the proportionality constant C.
8 = 15
C is the CAPACITANCE of the capacitor.
1 = 85
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If there is charge on a plate
there is an electric field E.
+Q + - -Q
+ P -
+ -
+ -
+ V _
BATTERY
E at P due to + plate &M = NO"
+ v -
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E at P due to - plate &@ = NO"
Both fields point to the right
d
+Q + - -Q
+ P -
+ E -
+ -
+ V _
& = NO" + N
O" = NO"
+ v -
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For all capacitors 1 = PQ
And for parallel plate capacitors
1 = ��R�
Parallel plate capacitors are easy:
Area and distance between plates gives C.
If know C then:
Know Q can get V
Know V can get Q
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CAPACITORS IN CIRCUITS
Series
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5 = 5� + 5
1 = 85STUVWXYZZZ[ 5 = 8
1
81U$ =
81� +
81
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11U$ =
11� +
11 \-](+(]
Parallel
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8 = 8� + 8
1 = 85STUVWXYZZZ[ 8 = 15
1U$5 = 1�5 + 15
1U$ = 1� + 1\-^��''('
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ENERGY IN CAPACITOR AND ELECTRIC FIELD
The potential difference across the
capacitor is
5 = ��
Where U is the energy stored in the electric
field of the capacitor and q is the charge on
the plates.
Thus the change in potential difference is
Δ5 = Δ��
Now charge a capacitor
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dq
ΔW=VΔq
5 = $_
Δ� = �1 Δ�
Add up all of the work done to charge the
capacitor to charge q
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� = 11� ��� = 1
1 `�2 b�
PP�
� = 11 `82 − 02b
� = 1281
Once again we use calculus but only to show you how something is done. To
“add up” the work.
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� = 1281
This is the amount of work to charge the
capacitor from 0 charge to a charge of Q.
This is the energy stored in the capacitor.
� = 1281
Use definition of capacitance 1 = PQ
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� = �15
Or � = �85
Then for parallel plate capacitors
1 = !"9W and 5 = &�
� = 1215 =
12��R� &�
� = 12 ��&R�
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R� = c-'dD(-\)�^�)+*-
Energy density d = eQ
Therefore d = eQ =
:<!"f<9W9W = �
��&
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DIELECTRICS
Add material between plates and C
increases.
Increases by K the dielectric constant
Parallel Plate Capacitor
1 = =� R� �Ug�hUXYZZZZZ[ i =� R�
Define � = i=�
Then can use 1 = � 9W
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Other quantities
Energy density
d = 12 =�&
�Ug�hUXYZZZZZ[12i=�&=12 �&
Charge on capacitor connected to V.
V 8� = 1�5
Introduce dielectric
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V dielectric
1 = i1�
1 = i P"Q
15 = i8�
15 = 8 = i8�
Insert dielectric and Q increases if V
remains constant.
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Voltage across capacitor without battery.
1� = PQ"
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1 = i1�
1 = P"Q
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i1� = 8�5
5 = P"_"
�j
But P"_" = 5�
So 5 = Q"j
When C not connected to battery inserting
dielectric decreases V.
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Electric Field in dielectric.
&k = QW =
Q" jlW = Q" Wl
j = f"j
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CAPACITOR SUMMARY
11U$ =
11� +
11 \-](+(]
1U$ = 1� + 1\-^��''('
) = 85
ENERGY
� = 1281
� = 1215
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� = 1285
ENERGY DENSITY
d = �5
d = 12 ��&
DIELECTRICS
� = i=�
1 = i1�
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1 = i8�5
5 = 5�i
&k = &�i