Transcript Your turn…

For a non-uniform electric field,
we integrate along the path:
b
DVab = - ò E · dl
a
which we can write as the sum
of three separate integrals:
xf
yf
zf
xi
yi
zi
Vab    E x dx   E y dy   E z dz
Your turn…
Your turn…
Uniformly charged rod.
To calculate VC – VA we should use:
1) 1 path segment, uniform field
2) 2 path segments, uniform fields
3) 3 path segments, uniform fields
4) 1 path segment, integral
5) 2 path segments, integrals
6) 2 path segments: 1 integral, 1 unif. field
Example: A negatively charged metal sphere
Potential difference around a
closed loop is zero.
 
E

d
l

0

Potential at one location
Let rA go to infinity…
VAB
1 1

q1   
40  rB  
+
q1
1
B
Potential at one location
The potential near a point
charge, relative to infinity:
1
q1
V (r ) 
40 r
+
q1
r
q2
Potential energy of two charges
The potential energy of two point
charges, relative to infinity:
1
q1q2
U (r ) 
40 r
+
q1
r
+
q2
Summary
b
• The potential
difference between two
points:
DVab = - ò E · dl
• The potential energy
difference for a charge
q, moved between two
points:
U electric  qV
• The potential near a
point charge, with
respect to infinity:
a
1
q1
V (r ) 
40 r
Potential in a
conductor
B
At equilibrium,
the field inside
the conductor
must be zero.
A
+
VAB  0
Capacitor
Does ΔV change when we
insert a slab of metal?
Your turn…
Now what is the magnitude of the
electric field in the air gaps?
1)
2)
3)
4)
In a capacitor E = (Q/A)/0
Same as originally
Less than originally
Greater than originally
Not enough information to tell
Your turn…
Without the metal slab, VB – VA was
–1000 volts. Then a metal slab was
inserted into the capacitor.
Now V = VB – VA =
1) + 1000 volts
2) +500 volts
3) 0 volts
4) –500 volts
5) –1000 volts
6) Not enough information to tell
In a capacitor E = (Q/A)/0
Your turn…
Originally ΔV was –1000 volts. A plastic slab
is inserted into the capacitor.
Now ΔV = VB – VA=
1) –1000 volts
2) +1000 volts
3) between 0 and –1000 volts
4) between 0 and +1000 volts
5) 0 volts
6) not enough information to tell
In a capacitor E = (Q/A)/0
Enet =
Eapplied
K
dielectric
constant
介电常数
（相对电容率）
Material
K
Vacuum
1
Air
1.0006
Plastic
5
Water
80
Your turn…
Originally V was –1000 volts. A plastic
slab is inserted into the capacitor. The
dielectric constant of the plastic is 2.
Now V = VB – VA=
1) –1000 volts
2) –750 volts
3) –250 volts
4)
0 volts
5) +250 volts
In a capacitor E = (Q/A)/0
Try to pull the plates apart…

Fby you
+
+
+
+
+
+
+

Eother plate

Fby plate
-
Try to pull the plates apart…
Electrical force on
plate:

Fby you
+
+
+
+
+
+
+
d

Eother plate

Fby plate
-


Q A
Fby plate  Q
2 0
Work done by you:
Q A
d
W  Q
 2 0 
This equals the
change in potential
energy of the
system, ΔU.
Q A
d
U  Q
 2 0 
I will now re-write this in a different way…
2
1 Q A
 Ad
U   0 
2  0 
This is the electric
field E between the
plates.
This is the change
in volume (体积)
of space between
the plates.
1
2
U   0 E Volume 
2
U
1
2
 0E
Volume  2
Energy density in an electric
field
1
2
0E
2
This is just another way of thinking
about the potential energy U of the
system of charges.
It is not a ‘new’ energy.
Energy density in an electric
field
1
2
0E
2
It is true for any electric field, not
just in a capacitor!
Electric fields contain energy.
(Also momentum!)
Can you think of a distribution of static charges to

make this field?
E
dl
Electrostatic forces are
conservative.
ò E ·dl = E2p r
The change in potential
around a loop must be zero.
ò E · dl = 0
for fields made by charges at rest.
ò E · dl = 0 means:
No curly electric fields.
BUT: This is only true for “Coulomb” fields
(fields caused by stationary charges).