Download +Q - Purdue Physics

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
0.04m
0.03m
x
0
B
.02
DV = VB - VA = - ò E × dl = - ò E × dl A
0
.09
ò E × dl
dl = x̂dx
.05
VB-VA = -300*(0.02-0) - (-300)*(0.09-0.05)=-6+12 V = +6 V
Potential in Metal
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, the potential
difference is: V = -EL
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.
Electric Field and Potential
If we know electric field ( ) everywhere we can compute potential
(V) in every point in space.
Can we compute ( ) if we know V?
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
Wire in a Circuit and Electric Field
f
DV = - ò E · dl
i
In a current-carrying wire in a
circuit, there can be a nonzero
electric field, so there is a
difference in potential between two
locations in the wire.
DV = EL
Electric field is not confined to
the wire in a circuit – there must
be an electric field E around the
wire in the air!
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
dl = ẑdz
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!
The difference between metals and
insulators
-Q metal +Q
E inside metal = 0
-Q
insulator +Q
E inside insulator is non zero
Dielectric Constant
+Q -q
-
+q -Q
+
+
+
+
+
+
+
+
+
+
Electric field in capacitor filled with
insulator: Enet=Eplates-Edipoles
qdipoles = b*Q
Enet
Q / A)
(
=
Ke0
Enet =
K – dielectric constant
E plates
K
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
(Q / A)
e0
K
DV = DVvacuum + DVinsulator
DVvacuum =
DVinsulator
A
s
B
x
DV =
(Q / A)
e0
(s - d)
(Q / A)
=
d
Ke0
(Q / A)
e0
[s - d(1 - 1 / K )]
x̂
Finding Potential Difference
1. Subtract the potential at the initial location A from the
potential at final location B
DV = VB - VA
A
B
2. Travel along a path from A to B adding up - E · dl at
each step:
B
DV = - ò E · dl
A
A
dl
B
E
Common Pitfall
Assume that the potential V at a location is defined by the
electric field E at this location.
A
Example: E = 0 inside a charged metal sphere, but V is not!
A negative test charge Q = -0.6C was moved from point A to
point B In a uniform electric field E=5N/C. The test charge is at
rest before and after the move. The distance between A and
B is 0.5m and the line connecting A and B is perpendicular to
the electric field. How much work was done by the net external
force while moving the test charge from A to B?
A
0.5m
B
A.
B.
C.
D.
E.
1.5J
0J
–1.5J
3.0J
–3.0J
E = 5 N/C
After moving the -0.6C test charge from A to B, it was then
moved from B to C along the electric field line. The test charge
is at rest before and after the move. The distance between B
and C also is 0.5m. How much work was done by the net
external force while moving the test charge from A to C?
A
E = 5 N/C
0.5m
A.
B.
C.
D.
E.
1.5J
0J
–1.5J
3.0J
-3.0J
B
C
Instead of moving the test charge from A to B then to C, it is
moved from A to D and then back to C. The test charge is at
rest before and after the move. How much work was done by
the net external force while moving the test charge this time?
A
D
0.5m
B
A.
B.
C.
D.
E.
0.5m
C
1.5J
0J
–1.5J
Infinitely big
Do not know at this time.
E = 5 N/C
Conducting charged sphere and concentric charged
conducting shell
Q2
Q1
rA
rB
rC
A solid conducting sphere of radius rA has charge Q1 uniformly
distributed over its surface. Concentric with the solid sphere is a
conducting shell of inner radius rB and outer radius rC with charge Q2
on its outer surface. What is the electric potential at the center of the
solid sphere? Take the potential at infinity to be equal to zero and the
origin at the center of the solid sphere.
Energy Density of Electric Field
How much work do I need to do to
increase separation of plates?
Wby_you=Fby_you*s
Note that by doing so you increase
area in space where electric field
(E = Q/A0) is present.
Fby _ you
(Q / A)
= QE = Q
2e 0
Eone _ plate
Q / A)
(
=
2e 0
(for small s)
Energy Density of Electric Field
Energy can be stored in electric fields
DU electric = Fby _ you Ds = Q
(Q / A)
Ds
2e 0
Multiply by A/A and 0/0
2
1 æQ / Aö
÷÷ ADs
DU el = e 0 çç
2 è e0 ø
E
DU el
1
2
=
e
E
Field energy density:
(J/m3)
0
D ( volume) 2
Volume
in which
we created
electric
field
Formula holds for any charge configuration.
Energy expended by us was converted into energy stored in the electric field
Potential Energy and Field Energy
A different way to express
all space
instead of calculating a work it takes to assemble charges.
The idea of energy stored in fields is a general one:
Magnetic and gravitational fields can also carry energy.
An Electron and a Positron
System
Surroundings
e+
e-
Release electron and positron – the electron (system) will gain
kinetic energy
Conservation of energy  surrounding energy must decrease
Does the energy of the positron decrease? - No, it increases
Where is the decrease of the energy in the surroundings?
- Energy stored in the fields must decrease
An Electron and a Positron
System
Surroundings
e+
Single charge:
1
E~ 2
r
eEnergy:
1
2
e
E
ò 2 0 dV
Dipole:
1 (far)
E~ 3
r
Energy stored in the E fields decreases as e+ and e- get closer!