Download Potential Difference: Path Independence

Potential Difference: Path Independence
f
f
i
DV = - ò E · dl
i
DV = Vf - Vi
1 q
V=
4pe 0 r
Path independence principle:
V between two points does not
depend on integration path
Potential Difference in Metal
In static equilibrium
What is E inside metal? E = 0
In static equilibrium the electric field is zero at
all locations along any path through a metal.
f
What is the potential difference (Vf – Vi)?
f
DV = Vf - Vi = - ò E · dl
(
i
DV = - Ex Dx + Ey Dy + Ez Dz
i
)
DV = 0
The potential difference is zero between any two locations
inside the metal, and the potential at any location must be
the same as the potential at any other location.
Is V zero everywhere inside a metal?
No! But it is constant
Potential in Metal
In static equilibrium A Capacitor with large plates and a small gap
of 3 mm has a potential difference of 6 Volts
-Q
+Q
from one plate to the other.
E»
E
Q/A
e0
f
DV = - ò E · dl
i
DV = Ed = 6 V
E = (6 Volts)/(0.003 m) = 2000 Volts/m
d =3 mm
V = 6 Volt
Potential in Metal
In static equilibrium
-Q1 1 mm +Q1
X
Insert a 1 mm thick metal slab into the
center of the capacitor.
Metal slab polarizes and has charges +Q2
and -Q2 on its surfaces.
What are the charges Q1 and Q2?
At X
d =3 mm
V = 4 V
Charges +Q2 and –Q2
E1 »
Q1 / A
E2 »
Q2 / A
e0
e0
E=0 inside metal
Q2=Q1
𝐸1 = 𝐸2
Now we have 2 capacitors instead of one
Ignoring the fringe fields, E = 2000V/m
in each capacitor (from previous slide).
DVleft = DVright = ( 2000 V/m ) ( 0.001 m ) = 2 V
V inside metal slab is zero!
There is no “conservation of potential”!
Potential in Metal
There can be a potential in metal if is NOT in static equilibrium
Metal is not in static equilibrium:
• When it is in the process of being polarized
• When there is an external source of mobile charges (battery)
f
DV = - ò E · dl
i
For each step 𝐸 ∙ ∆𝑙 , the potential
difference is: V = -EL
If a metal is not in static equilibrium,
the potential isn’t constant in the metal.
Nonzero electric field of uniform magnitude E throughout
the interior of a wire of length L.
Direction of the field follows the direction of the wire.
Question
300 V/m
300 V/m
0 V/m
A
B
0.02m
What is VB-VA?
0.03m
A)
B)
C)
D)
E)
270 V
-270 V
-18 V
6V
-6 V
0.04m
Question
EA = 300 x̂ V / m
EB = -300 x̂ V / m
0 V/m
A
B
0.02m
x
0
0.04m
0.03m
B
.02
DV = VB - VA = - ò E × dl = - ò E × dl A
0
.09
ò E × dl
dl = x̂dx
.05
V
V
−
−300
𝑉𝐵 − 𝑉𝐴 = − 300
0.02m − 0
m
m
𝑉𝐵 − 𝑉𝐴 = −6V − −12V = 6V
0.09m − 0.05m
The electric field is uniform in the region below left. The point B
is at <0, 0, 0> m. The point C is at <0,-2, 0>m. The potential
difference along the path from B to C is ∆𝑉 = +500V.
Round Trip Potential Difference
+
Potential difference due to a stationary point charge is
independent of the path
Potential difference along a closed loop is zero
Predicting Possible Field Configuration
Is the following “curly” pattern of electric field possible?
f
DV = - ò E · dl
dl
i
E is always parallel to dl
dl
dl
DVAA = - ò E · dl = -El = -E2p R ¹ 0
This “curly” pattern of electric field is impossible to produce by
arranging any number of stationary point charges!
Potential of a Uniformly Charged Ring
Q
z 2 + R2
Method 1: Divide into point charges and
add up contributions due to each charge
1 DQ
DV =
4pe 0 r
V = å DV = å
V=
V=
1
DQ
4pe 0
z 2 + R2
1
1
4pe 0
z +R
1
Q
4pe 0
2
2
z2 + R2
å DQ
Potential of a Uniformly Charged Ring
Q
z 2 + R2
Method 2: Integrate electric field
along a path
z
V = - ò E · dl
¥
z
V = -ò
¥
Ez =
1
(
Qz
4pe 0 R + z
2
V =-
)
2 3/2
1
Qz
(
4pe 0 R 2 + z
1
4pe 0
z
Qò
¥
(R
)
2 3/2
z
2
+z
)
dz
2 3/2
z
é -1 ù
V =Qê
4pe 0 ë z 2 + R 2 úû ¥
1
Q
V=
4pe 0 z 2 + R 2
1
dz
Potential of a Uniformly Charged Ring
Q
z 2 + R2
V=
1
4pe 0
Q
z2 + R2
What is V for z>>R ?
1 Q
V»
4pe 0 z
The same as for a point charge!
Potential of a Uniformly Charged Disk
one ring: DV =
DQ
1
4pe 0
z2 + r2
1 Q rDr
DV =
2e 0 A z 2 + r 2
R
integrate:
1 Q
rdr
V=
2e 0 A ò0 z 2 + r 2
1 Q é 2 2 ùR
V=
z +r
ë
û0
2e 0 A
2prDr
DQ =
Q
A
1 Qé 2
V=
z + R2 - z ù
û
2e 0 A ë
Potential of a Uniformly Charged Disk
1 Qé 2
V=
z + R2 - z ù
û
2e 0 A ë
Can find E:
¶V
Ez = ¶z
ù
1 Qé
1
Ez = - 1ú
ê
2e 0 A ë z 2 + R 2
û
ù
1 Qé
1
Ez =
1ê
2
2 ú
2e 0 A ë
z +R û
Potential Difference in an Insulator
Electric field in capacitor filled with
insulator: Enet=Eplates+Edipoles
Eplates=const (in capacitor)
Edipoles,A
Edipoles,B
A
B
Edipoles complicated f(x,y,z)
𝐵
∆𝑉 = −
𝐸 ∙ 𝑑𝑙
𝐴
Travel from B to A:
Edipoles is sometimes parallel to dl, and
sometimes antiparallel to dl
1
2
3
4
5
Potential Difference in an Insulator
Instead of traveling through inside – travel outside from B to A:
A
Edipoles, average
B
E · dl ³ 0
𝐴
𝐸𝑑𝑖𝑝𝑜𝑙𝑒𝑠 ∙ 𝑑𝑙 < 0
∆𝑉𝐴𝐵 = −
𝐵
Enet=Eplates+Edipoles,average
Enet< Eplates
Effect of dielectric is to reduce the potential difference.
Dielectric Constant
Electric field in capacitor filled with
insulator: Enet=Eplates-Edipoles
K – dielectric constant
E plates
Enet =
K
E plates
Q / A)
(
=
Enet
Q / A)
(
=
e0
Ke0
Dielectric Constant
Inside an insulator:
Enet =
Eapplied
K
Dielectric constant for various insulators:
vacuum
air
typical plastic
NaCl
water
strontium titanate
1 (by definition)
1.0006
5
6.1
80
310
Potential Difference in a Capacitor with Insulator
B
DV = - ò E · dl
E plates
A
DV = Es =
s
E plates
s
K
Q / A)
(
DV = Es =
s
Ke0
DVvacuum
DVinsulator =
K
Q / A)
(
=
e0
Potential Difference in Partially Filled Capacitor
B
d +Q
-Q
DV = - ò E · dl
E plates = -
A
K
DVinsulator
A
e0
DV = DVvacuum + DVinsulator
DVvacuum =
s
(Q / A)
DV =
B
x
(Q / A)
e0
(s - d)
(Q / A)
=
d
Ke0
(Q / A)
e0
[s - d(1 - 1 / K )]
x̂
Energy Density of Electric Field
Energy can be stored in electric fields
Eone _ plate
Q / A)
(
=
2e 0
Fby _ you = QE = Q
(for small s)
(Q / A)
2e 0
DU electric = Fby _ you Ds = Q
(Q / A)
Ds
2e 0
2
1 æQ / Aö
÷÷ ADs
DU el = e 0 çç
2 è e0 ø
volume
E
DU el
1
2
=
e
E
Field energy density:
(J/m3)
0
D ( volume) 2
Energy expended by us was converted into energy stored in the electric field
Energy Density of Electric Field
In the previous slide, the “system” is the set of two plates. Work,
Wexternal > 0, is done on the system by you – part of the
“surroundings.”
DEsystem = DKE + DUelectric = Wexternal
If the force exerted by you just offsets the attractive force,
Fby-plates, so that the plate moves with no gain in KE,
DUelectric = Wexternal = Fby _ you Ds
Electric Field and Potential
90V
100V
x
Ex
1 mm
¶V
10 V
Ex = »= -10,000 N/C
¶x
0.001 m
Exercise
Suppose in some area of space V(x,y,z)=x2+yz. What is E(x,y,z)?
¶V E = - ¶V
¶V
Ex = Ez = y
¶y
¶x
¶z
(
¶V
Ex = =¶x
¶ x 2 + yz
¶V
Ey = =¶y
¶ x 2 + yz
¶V
Ez = =¶z
¶ x 2 + yz
(
(
¶x
¶y
¶z
) = - ¶ ( x ) - ¶ ( yz ) = -2x
2
¶x
¶x
) = - ( ) - ¶ ( yz ) = -z
¶ x2
¶y
¶y
) = - ( ) - ¶ ( yz ) = -y
E ( x, y, z ) = -2x,-z,-y
¶ x2
¶z
¶z
Potential Inside a Uniformly Charged Hollow
Sphere
¥
VA = VA - V¥ = ò E · dl
1 Q
VA =
4pe 0 R
A
=0
A
¥
B
A
VB = VB - V¥ = ò E · dl + ò E · dl
1 Q
VB =
4pe 0 R
In general, integration
path may be complex
Example
0V
¶V
Ex = ¶x
35 V
E
If an electron moves from rest through a potential
difference of 35 V, what would be its kinetic energy?
K = 35 eV
DU el = -eDV = -35 eV
What if we had a proton?
K = 35 eV
DU el = eDV = -35 eV
0V
35 V
E
Which particle will
move faster?
Shifting the Zero Potential
In most cases we are interested in V, not the absolute values of V
(V
f
)
+ V0 - (Vi + V0 ) = Vf - Vi
¶DV100
V V
Ex = 0¶D.x0002
x
m
100 V
Ex = 0.0002 m
100 V
Ex = 0.0002 m
E = 5´ 105 N/C
E = 5´ 105 N/C
E = 5´ 105 N/C