mit8_02sc_lectureslides08
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Module 08: Electric Potentialand Gausss Law; Configuration
Energy
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Module 08: Outline
Usin Gausss Law to find V from EConfiguration Energy
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E Field and Potential: Creatin
A point charge qcreates a field and potential around it:r
2;E r
e ek V k
r r
= = se superpos on orsystems of charges
They are related:
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;E E sB AA
V V V V d= = r
rr
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= =
If you put a charged particle, (charge q), in a field:
=
r r, ,
and the particle does not change its kinetic energyen:
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ext
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Derivin E from V
E sB
V d = r
r
A
r
A = x,y,z , B= x+x,y,z
s x=( , , )
E s
x x y z
V d
+
=
r
r
E s
r
r ( )
E ixx E x= =
r
V V E = Rate of chan e in V
( , , )x y z
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x x x with y and z held constant
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Derivin E from VIf we do all coordinates:
E i + j k
V V V = +
r
x y z
E V= r
+x y z= +
Gradient (del) operator: i j + k
x z +
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E from V
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Demonstration:Making & Measuring
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Two Concept QuestionQuestions:
o en a e
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Concept Question: E from V
The above shows potential V(x). Which is true?
1. Ex > 0 is > 0 and Ex < 0 is > 0
. x > 0 x < 03. Ex > 0 is < 0 and Ex < 0 is < 0
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4. Ex > 0 is < 0 and Ex < 0 is > 05. I dont know
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Potential from E
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E=
r r2
0
,4
r a r br
< b No field160
V r V=
123 0 dr= 0= No change in V!
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Potential for Nested Shells
Region 2: a < r< b
= drQr
( ) ( )V r V r b = 0r
0=14243
r
= 40r
r=b
=1
Q
1
1
0
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Electric potential is DIFFERENT surroundings matter
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Potential for Nested Shells
Region 3: r< a
= dr0
a
r
=0( ) ( )V r V r a =
14 431 1kQ
a b
=
V r( )= V a( )= Q 1 1 r0
=
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Confi uration Ener
How much energy to put two charges as pictured?
1 First char e is free2) Second charge sees first:
1 q q 12
=2
= q2 1
=4 r
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= =
Confi uration Ener
How much energy to put three charges as pictured?
1) Know how to do first two
r ng n r :
q3
q1
q2
3 3 1 2
40 r13 r23
Total configuration energy:
U = W2
+ W3
=1 q
1q
2 +q
1q
3 +q
2q
3 = U12 + U13 + U23
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0 12 13 23
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Problem: Build It
1) How much energy
did it take to assemblethe charges at left?
2) How much energy
would it take to add ath +
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8.02SC Physics II: Electricity and Magnetism
Fall 2010
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