Jan 19, 2016
A
B
+
Potential near a point charge
What is the potential difference between A and B?
1q
0,0,
0,0,
dxsd
EE x
+1q
A
Bsd
E
x
Now integrate along the path:
B
A
x
x
B
A
x
B
A
AB dxx
qdxEdVV
21
04
1
dxEdV
dxEsdEdV
x
x
0,0,0,0,
Change in potential along a short section of the path:
0,0,
0,0,
dxsd
EE x
+1q
A
Bsd
E
x
dxEdV
dxEsdEdV
x
x
0,0,0,0,
Change in potential along a short section of the path:
ABAB xx
qV11
4
11
0
Potential difference near a point charge
ABAB rr
qV11
4
11
0
+1q
AB
Ar
Br
Potential at one location
11
4
11
0 BAB r
qV
+1q
B
Let rA go to infinity…
Potential at one location
r
qrV 1
04
1)(
1q+
r
The potential at a distance r from a point charge, relative to infinity:
r
qrV 1
04
1)(
Potential at one location
r
qrV 1
04
1)(
1q
The potential at a distance r from a point charge, relative to infinity:
+r
2q
Potential energy of two charges
r
qqrU 21
04
1)(
1q
The potential energy of two point charges, relative to infinity:
+r
+2q
Potential energy of a system of charges
01 U
12
21
02 4
1
r
qqU
23
32
013
31
03 4
1
4
1
r
r
qqU
23
32
13
31
12
21
0tot 4
1
r
r
r
qqU
Finding the field from the potential
dzEdyEdxE
sdEdV
zyx
The change in potential along a very small path:
sd
E
E
E
Finding the field from the potential
dzEdyEdxEdV zyx
Choose a path that only goes in the x-direction (dy = dz = 0):
sd
E
E
E
Finding the field from the potential
dxEdV x
Choose a path that only goes in the x-direction (dy = dz = 0):
sd
E
E
E
dx
dVEx (holding y and z
fixed)
Finding the field from the potential
dxEdV x
Choose a path that only goes in the x-direction (dy = dz = 0):
sd
E
E
E
x
VEx
(partial derivative)
Finding the field from the potential
dxEdV y
Choose a path that only goes in the y-direction (dx = dz = 0):
sd
E
E
E
y
VEy
Finding the field from the potential
dzEdV z
Choose a path that only goes in the z-direction (dx = dy = 0):
sd
E
E
E
z
VEz
Electric field is the negative gradient (梯度 ) of the
potential
z
VE
y
VE
x
VE zyx
,,
V
xxE xE
Electric field is the negative gradient (梯度 ) of the
potential
zyxVE
,, where,
V
xxE xE
The potential is like the height of the hill.
The field is like the slope of the hill.
Just remember:- positive charges go down the hill- negative charges go up!
Field around a point charge
r
qrV 1
04
1)(
1q
The potential near a point charge, relative to infinity:
+r
r
q
dr
d
dr
dVrE 1
04
1)(
The field strength is the gradient of the potential:
Field around a point chargeThe potential near a point charge, relative to infinity:
21
04
1)(
r
qrE
r
qrV 1
04
1)(
1q+
r
The field strength is the gradient of the potential:
Potential along the axis of a ring
Potential obeys the superposition principle, just like the field.
R
x
22 Rx
2204
1
Rx
dQdV
Potential due to one small piece:
dQ
Potential along the axis of a ring
Potential obeys the superposition principle, just like the field.
R
x
22 Rx
dQ
RxdVV
220
1
4
1
Integrate:
dQ
Potential along the axis of a ring
Potential obeys the superposition principle, just like the field.
R
x
22 Rx
2204
1
Rx
QV
Integrate:
dQ
Field along the axis of a ring
The strength of the field is the negative of the potential gradient:
220
1
4 Rxx
Q
x
VEx
E
Field along the axis of a ring
The strength of the field is the negative of the potential gradient:
2/32204
1
Rx
QxEx
E
Field along the axis of a ringWe already calculated this field the hard way.
It is often easier to first calculate the potential, then use its gradient to get the field.
2/32204
1
Rx
QxEx
E
Potential due to a uniformly charged sphere
+ + +
++
+
++
++
++
Q
Remember: The field outside a charged sphere is the same as the field of a point charge.
The same is true for the potential.
Potential due to a uniformly charged sphere
+ + +
++
+
++
++
++
Q r
r
QrV
04
1)(
V(∞) = 0
Potential at the surface
+ + +
++
+
++
++
++
Q
R
R
QV
0surface 4
1
V(∞) = 0
Potential at the surface
+ + +
++
+
++
++
++
Q
R
0surface
RV
Define the surface charge density:
24 R
Q
A
Q
V(∞) = 0
Surface potential of a protein
Positive (+)Negative (-)
0surface
RV
Atserpin 1
Arabidopsis thaliana
Potential in a conductor
+
At equilibrium, the field inside the conductor must be zero.
A
B
0AB V
Potential in a conductor
+
So the potential inside a conductor (and at the surface) must be constant.
constantV
Example: A negatively charged metal sphere