Prof. D. Wilton ECE Dept. Notes 12 ECE 2317 Applied Electricity and Magnetism Notes prepared by the...
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Prof. D. Wilton ECE Dept. Notes 12 ECE 2317 ECE 2317 Applied Electricity and Applied Electricity and Magnetism Magnetism Notes prepared by the EM Group, University of Houston
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Transcript of Prof. D. Wilton ECE Dept. Notes 12 ECE 2317 Applied Electricity and Magnetism Notes prepared by the...
Prof. D. Wilton ECE Dept.
Notes 12
ECE 2317 ECE 2317 Applied Electricity and MagnetismApplied Electricity and Magnetism
Notes prepared by the EM Group,University of Houston
ExampleExample
y
x
0B
s
0A
sx = - d/2
x = d/2
(a) x > d/2,0
0
0
02 2
A Bs
xsE
(c) x < -d/2,
(b) -d/2 < x <d/2, 0
0
0
02 2
A Bs
xsE
0 0
002 2
A
x
BssE
By superposition,
Example (cont.)Example (cont.)
y
x
x = - d/2
x = d/2
(a) x > d/2, 0xE
(c) x < - d/2,
(b) - d/2 < x < d/2,0
0
sxE
0xE
Choose: 0 0 0 0A B
s s s s
0s
0s
Example (cont.)Example (cont.)
0
0
sxE
-d/2 < x < d/2
ideal parallel-plate capacitor
0s
0s
y
x
h
Example (cont.)Example (cont.)
y
x
x = - d/2
x = d/2
(a) x > d/2,
0xE
(c) x < - d/2,
(b) -d/2 < x < d/2,
0
0
sxE
This time, choose: 0 0 0
A Bs s s
0s
0s
0
0
sxE
Example (cont.)Example (cont.)
0
0
,2
sx
hE x
0s
0s
yxh
ideal conductor of thickness h
0,2 2x
h hE x
Perfect Electric Conductors (PEC)Perfect Electric Conductors (PEC)
[S/m]J E
AB
0B
AB
A
V E dr
PEC:
Also, constant
0E
since
(Ohm’s law)
Perfect Electric Conductors (PEC)Perfect Electric Conductors (PEC)
Inside PEC:
Proof: assume point where v > 0
Only s on the surface is allowed
S
assume v > 0 inside small volume
0 0
encl
S
encl
D n dS Q
Q
Contradiction !
+ +
+++
+ +
0v
Perfect Electric Conductors (PEC)Perfect Electric Conductors (PEC)
• Inside an empty, hollow PEC shell there is no electric field
• Only s on the outer surface is allowed
PEC shell
+
+
+
+
+
+
+
+
+
+
+
+
E = 0
+
“Faraday-cage effect”:
Perfect Electric Conductors (PEC)Perfect Electric Conductors (PEC)Proof:
Assume an electric field exists inside the shell at some point
PEC shell
+
+
+
+
+
+
+
+
+
+
+
+
Assume E 0
A
B
A flux line must exist through this point.
The flux line must end on the shell (otherwise it contradicts Gauss’s law)
VAB 0 Contradiction !
+
There is also no charge density on the inner surface (otherwise, there would be a flux line coming from the inner surface).
Hence, there is no electric field (and no flux lines) inside the cavity.
Perfect Electric Conductors (cont.)Perfect Electric Conductors (cont.)Faraday-cage effect
The men in the Faraday cages are protected from the high voltage arcs of the Tesla coil!
Perfect Electric Conductors (cont.)Perfect Electric Conductors (cont.)Faraday-cage effect
Entrance to a Faraday roomFaraday shield at Art Nouveau power plant in Heimbach, Germany
Shielding and GroundingShielding and Grounding
q
Spherical PEC shell Neutral (no charge)a b
Drill hole and insert point charge, then solder hole.
Neutral shell (no net charge)
Find E:q
a b
(a) r < a
(c) r > b
(b) a < r < b
2
20
4
4
encl
S
r
D n dS Q
D r q
qE r
r
0E
2
20
04
4
rD r q
qE r
r
q
a b
Neutral shell
(PEC)
Shielding and GroundingShielding and Grounding
Shielding and GroundingShielding and Grounding
204
qE r
r
q
(outside metal)
The neutral metal shell does not block the static electric field
Shielding and GroundingShielding and Grounding
Find QA, QB:
q-
---
-
--
-
+
+
+
+
+
+
+
++
+
+
+
+
+
+
+
QA QB
q
QA
QB
Neutral shell
Shielding and GroundingShielding and Grounding
q-
---
-
--
-
+
+
+
+
+
+
+
++
+
+
+
+
+
+
+
QA QB
24
0
encl
S
r encl A
encl
A
B
D n dS Q
D r Q q Q
Q
Q q
Q q
(since neutral shell)
S so
Hence
and
A Gaussian surface is chosen inside the metal shell.
Example (cont.)Example (cont.)Alternative solution (Gauss’s law):
0
0
2
2
encl
S
x s
s
sx
D n dS Q
D A A
A
D
0
0
[V/m], / 22
sE x x h
y
x
h PEC
S
+ ++
+
+ ++
+
A
(The electric field outside the slab is the same as from a single sheet of surface charge.)
Next, “ground” the shell:
q-
---
-
--
--q
E = 0
PEC wire
Earth
r > b : E = 0
0B
AB
A
V E dr A
B
This must hold for every path.
proof:
The earth is modeled as a very large conductor.
If the electric field were non-zero at a point, there would be a flux line through the point ending on conductors and implying a voltage drop between them
flux line
charge on outer surface:
q-
---
-
--
-
QBG
-q
0
0
encl
S
encl
Gencl B
GB
D n dS Q
Q
Q q q Q
Q
PEC wire
Hence
r > b: E = 0
Earth
charge on outer surface (cont.):
The charge q on the outer surface flows down to ground.
q-
---
-
--
-
QBG = 0
-q
+
+
++PEC wire
Earth
q
- q
before grounding after grounding
ExampleExample
Given:2
0 [C/m ]s
y
x
h
r
PEC
Find E everywhere
(This is the total surface charge density per square meter on the metal slab, which is the sum of the surface charge densities on the upper and lower surfaces.)
ExampleExample
Charge Model:
y
x
20 [C/m ]s
h
r
PEC
y
x
0 / 2s
0 / 2sx = - h/2
x = h/2+ ++
+
+ ++
+
+ ++
+
+ ++
+
ExampleExample
y
x
0 / 2s
0 / 2sx = - h/2
x = h/2+ ++
+
+ ++
+
+ ++
+
+ ++
+0 0
0 0
0
0
/ 2ˆ 2 [V/m], / 2
2 2
0, / 2 / 2
ˆ [V/m], / 22
s s
s
x x x h
E h x h
x x h
By superposition,
ExampleExample
y
x
h PEC
0 0 / 2s s
0 0 / 2s s
The surfaces are initially charged as shown above.
The bottom of the plate is then grounded.
Find the charge and field everywhere
Example (cont.)Example (cont.)(a) charge on lower surface
0
0 0
Gencl s
S
Gs
D n dS Q A
h PEC
S0G
s
A
E = 0
+ + + + + + + ++ + + + + + + +
Hence
Earth
Example (cont.)Example (cont.)(b) charge on earth
0
earthencl s
S
earths
D n dS Q A
h PEC
Searths
+ + + + + + + + + + + + + + + +
A
Hence
Earth
Example (cont.)Example (cont.)(c) charge on top surface
0 0 0
0 0 0
02 2 2
G G earths s sE x
0 0Gs HenceSo
(After grounding)
E = 0 (inside earth)
0G
s
0 0Gs
0 0earths
x
0 0 0 0G G Gs s s
+ + + + + + + + + + + + + + + + + + + +
Example (cont.)Example (cont.)
(before grounding)PEC
0 0 / 2s s
(after grounding)PEC
zero
zero
0 0 / 2s s
Earth
