Download potential

Electric Potential
Electric Energy
•Electric fields produce forces; forces do work
•Since the electric fields are doing work, they must have potential energy
•The amount of work done is the change in the potential energy
•The force can be calculated from the charge and the electric field
q
E
s
ds
W  Fs
U  W  F  s
•If the path or the electric field are not straight lines,
we can get the change in energy by integration
•Divide it into little steps of size ds
•Add up all the little steps
F  qE
U   qE  s
dU  qE  ds
U  q  E  ds
The Electric Potential
U  q  E  ds
•Just like for electric forces, the electric potential
energy is always proportional to the charge
•Just like for electric field, it makes sense to divide
by the charge and get the electric potential V:
V U q
U  qV
V    E  ds
•Using the latter formula is a little tricky
•It looks like it depends on which path you take
•It doesn’t, because of conservation of energy
•Electric potential is a scalar; it doesn’t have a direction
•Electric potential is so important, it has its own unit, the volt (V)
•A volt is a moderate amount of electric potential
•Electric field is normally given as volts/meter
N V

Nm C  J C  V
V Es
C m
Calculating the Electric Potential
B
V    E  ds
VB  VA    E  ds
A
To find the potential at a general point B:
•Pick a point A which we will assign potential 0
•Pick a path from A to B
•It doesn’t matter which path, so pick the simplest possible one
•Perform the integration
Example: Potential from a uniform electric field E:
V high
•Choose r = 0 to have potential zero
r
r
V  r   0    E  d s   E   ds   E  r
0
V low
0
•Equipotential lines are perpendicular to E-field
•E-field lines point from high potential to low
•Positive charges have the most energy at high potential
•Negative charges have the most energy at low potential
+
E
-
Why Electric Potential is useful
1. It is a scalar quantity – that makes it easier to calculate and work with
2. It is useful for problems involving conservation of energy
A proton initially at rest moves from an initial point with V = 0 to a
point where V = - 1.5 V. How fast is the proton moving at the end?
V =0
•Find the change in potential energy
U  qV  e  1.5 V   1.602 1019 C   1.5 V 
19
 2.4 10 J
•Since energy is conserved, this must be counterbalanced by a corresponding increase in kinetic energy
V = -1.5 V
E
+
T  U  12 mv 2
19
2

T
2
2.4

10
J

v2 

m
1.67 1027 kg
1.5 V
 2.87 108 m 2 / s 2
v  17 km/s
The Zero of the Potential
V    E  ds
We can only calculate the difference between the electric potential in two places
•This is because the zero of potential energy is arbitrary
•Compare U = mgh from gravity
•There are two arbitrary conventions used to set the zero point:
•Physicists: Set V = 0 at 
•Electrical Engineers: Set V = 0 on the Earth
•In circuit diagrams, we have a specific symbol
V=0
to designate something has V = 0.
Anything attached
here has V = 0
Potential From a Point Charge
r
ke q
E  2 rˆ
r
q
•Integrate from infinity to an arbitrary distance
r
r
r
U  r    0
r
dr ke q
ke q
V  r     E  ds    Edr  ke q  2 

r
r
r 



ke q
V
r
•For a point charge, the equipotential surfaces are spheres
centered on the charge
•For multiple charges, or for continuous charges, add or integrate
ke qi
V 
ri
i
ke  dl
V 
r
ke dA
V 
r
ke  dV
V 
r
Sample Problems
What is the potential V a distance z above a disk of radius
R if the disk has surface charge density ? A charge q at
rest with mass m moves from the center of the disk to
infinity. What is the final speed of this charge?
z
r
s
R
•Divide the disk into little circles of radius s and thickness ds
•Find the distance r for each of these circles
r  s2  z 2
2 sds
ke dA

k

V 
e 
r
s2  z2
0
R
 2 ke s  z
2
2
R
0
V  2 ke

R2  z 2  z
U  qV  2 ke Rq
•The initial energy of the charge q is:
•At infinity there is no potential energy
•This energy must become kinetic energy:
4 ke Rq
2U
v
v
U  12 mv 2
m
m

Getting Electric Field from Electric Potential
V    E  ds
•To go from electric field to potential, we integrate
•Can we go from electric potential to electric field?
•Consider a small motion in one dimension, say the z-direction
ds  kˆz
V    Ez dz   Ez z
Ez  
V
z
•For sufficiently small distances, this becomes a derivative
V
V  x, y, z  z   V  x, y, z 
V

Ez   lim
Ex  
z 0
z
z
x
•This is a partial derivative – a derivative that treats
x and y as constants while treating z as a variable
•Generalize to three dimensions:
V
Ey  
y
V
Ez  
z
E  V
Gradients
•Fancy notation: Mathematically, it is
useful to define the operator
 ˆ  ˆ 
ˆ
  i  j k
x
y
z
E  V
•When this derivative operator is used this way (to make a
vector out of a scalar) it is called a gradient
•“The electric field is minus the gradient of the potential”
Yellow boxes mean a more mathematically
sophisticated way to write the same thing. You
don’t need to know or use it if you don’t want to.
V
Ex  
x
V
Ey  
y
V
Ez  
z
Equipotential Lines Are Topographical Maps
E  V
•Regions of high potential are like “mountains”
•For positive charges, they have a lot of
energy there
•Regions of low potential are like “valleys”
•For positive charges, they have minimum
energy there
•Electric fields point down the slope
•Closely spaced equipotential lines means
big electric field
Conductors and Batteries
V    E  ds  0
•A conductor has zero electric field inside it
•Therefore, conductors always have constant potential
•A wire is a thin, flexible conductor: circuit
diagram looks like this:
•A switch is a wire that can be connected or disconnected
open switch
closed switch
A battery or cell is a device that creates a fixed potential difference
•The circuit symbol for a battery looks like this:
1.5 V
•The long side is at higher potential
•It is labeled by the potential difference
The potential difference E
across the battery is called
electromotive force (emf)
Conducting Spheres
•Given the charge q on a conducting sphere of radius
R, what is the potential everywhere?
•Outside the sphere, the electric field is the same as
for a point charge
ke q
Vout 
ˆ
E

r
2
•Therefore, so is the potential
r
•Inside, the potential is constant
ke q r
•It must be continuous at the boundary V  

ke q R
VSphere 
ke q
R
ke q
r
for r  R
for r  R
q
R
Sample Problem
q1
q2
Two widely separated conducting spheres, of radii R1 =
1.00 cm and R2 = 2.00 cm, each have 6.00 nC of charge
put on them. What is their potential? They are then
joined by an electrical wire. How much charge do they
each end up with, and what is the final potential?
V1 
9
2
2
9
8.988

10
N

m
/C
6

10
C


10
2
m
 5390 V
Vi 
ke qi
Ri
V2  2700 V
•After connections, their potentials must be equal
2q1  q2
q1  4 nC
ke q1
k e q2

q1  q2  12 nC
1 cm 2 cm
q2  8 nC
V1  V2 
 3600 V
Electric Fields near conductors
•The potential for the two spheres ended up the same
•The electric fields at the surface are not the same
ke q1 ke q2

R1
R2
q1
ke q1
R12
kq
E2  R2   e 22
R2
E1  R1  
q2
E1  R1  ke q1 R22
R2
 2

E2  R2  R1 ke q2
R1
•The more curved the surface
Very strong electric field here
is, the higher the electric
field is there
•A sharp point can cause charged particles to spontaneously be shed into air,
even though we normally think of air as an insulator.
•Called “Corona discharge”
The Millikan Oil Drop experiment
•Atomizer produced tiny drops of oil;
gravity pulls them down
•Atomizer also induces small charges
•Electric field opposes gravity
•If electric field is right, drop stops falling
-
FE  Fg
•Millikan showed that you always got integer
multiples of a simple fundamental charge
e  1.60 1019 C
The Lightning Rod
•Rain drops “rubbing” against the air can cause a separation of charge
•This produces an enormous electric field
•If electric field gets strong enough, it can cause breakdown of atmosphere
•Put a pointy rod on top of the
building you want to protect
•Coronal discharge drains away
the charge near the protected
object
•Lightning hits somewhere else
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
The Van de Graff Generator
•Hollow conducting sphere, insulating belt, source of electric charge
•Source causes charge to move to the belt
•Belt rotates up inside sphere
•Charge jumps to conductor inside sphere
•Charge moves to outside of sphere
•Since all the charge is on the outside of the sphere,
process can be repeated indefinitely.
-
Electrostatic Precipitator
•Hollow conducting tube with a thin wire hanging
down inside it
•Dirty air enters at the bottom
•Coronal discharge from wire produces lots of O2ions
•O2- ions hit dust particles, giving them charge
•Charged dust now flows towards walls
•Clean gas flows out the top
•Gravity (shaking helps) causes dust to fall
to the bottom of the container
50 kV
Dirty
air
Clean
air