Transcript electric potential V

Ch. 25
Electric Potential Energy
Warmup 05
Electric Potential
Energy
•Electric fields produce forces; forces do work
W  Fs
•Since the electric fields are doing work,
U  W  F  s
they must have potential energy
•The amount of work done is the change in
q
E
the potential energy
•The force can be calculated from the charge
s
and the electric field F  qE
ds
U   qE  s
•If the path or the electric field are not straight lines,
we can get the change in energy by integration dU  qE  ds
•Divide it into little steps of size ds
•Add up all the little steps
U   q E  ds

Warmup 05
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
[V] = [E][s]=N●m/C=J/C=volt=V
N V

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 low
V
high
•Choose r = 0 to have potential zero
r
r
V  r   0    E  d s   E   ds   E  r
0
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
-
Example - Serway 25-3. (a) Calculate the speed of a proton that is
accelerated from rest through a potential difference of 120V. (b) Calculate
the speed of an electron that is accelerated through the same potential
difference.
Solve on Board
Ex-Serway 25-7. An electron moving parallel to the x-axis has an initial
speed of 3.7 x 106 m/s at the origin. Its speed is reduced to 1.4 x 105 m/s at
the point x = 2.0 cm. Calculate the potential difference between the origin
and this point. Which point is at a higher potential?
Solve on Board
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?
•Find the change in potential energy
U  qV e  1.5 V 
 1.602 1019 C   1.5 V 
V =0
V = -1.5 V
+
 2.4 1019 J
•Since energy is conserved, this must be counterbalanced by a corresponding increase in kinetic energy
K  U  12 mv 2
1.5 V
19
2 U 2  2.4 10 J 
8
2
2
2

2.87

10
m
/
s

v 
1.67 1027 kg
m
v  17 km/s
E
Warmup 05
Warmup 06
The Zero of the Potential
V    E  ds
We can only calculate the difference between the electric potential
between 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
V=0
•In circuit diagrams, we have a specific symbol
to designate something has V = 0.
Anything attached
here has V = 0
Potential rFrom a Point Charge
ke q
E  2 rˆ
r
q
•Integrate from infinity to an arbitrary distance
r
r
r
U  r    0
dr
ke q
V  r     E  ds   | E | dr  ke q  | 2 | 
r
r



ke q
V 
r
r

ke q

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
ke  dl
ke dA
V 
V 
V 
ri
r
r
i
ke  dV
V 
r
Calculating Potentials is Straight-Forward
q
q
q
q
Four charges q are each arranged symmetrically
around a central point, each a distance a from that
point. What is the potential at that point?
A) 0 B) 2keq/a C) 4keq/a D) None of the above
CT-1 - Two test charges are brought separately into the vicinity of a charge
+Q. First, test charge +q is brought to point A, a distance r from +Q. Next,
+q is removed and a test charge +2q is brought to point B a distance 2r from
+Q.
Compared with the electrostatic potential of the charge at A due to Q, that of
the charge at B is
A. greater.
B. smaller.
C. the same.
CT - 2 - Two test charges are brought separately into the vicinity of a charge
+Q. First, test charge +q is brought to a point a distance r from
+Q. Then this charge is removed and test charge –q is brought to the same
point. The electrostatic potential energy of which configuration is greater:
A. +q
B. –q
C. It is the same for both.
Getting Electric Field from Potential
V    E  ds
•To go from electric field to potential, we integrate
•Can we go from potential to electric field?
•Consider a small motion in one dimension, say the z-direction
ds  kˆ z
V
Ez  
z
V    Ez dz   Ez z
•For sufficiently small distances, this becomes a derivative
V  x, y, z  z   V  x, y, z 
V
Ez   lim

z 0
z
z
•This is a partial derivative – a derivative that treats x and y as constants
V
while treating z as a variable
E 
A(x2
y2
2z2),
If V =
+ what is Ez?
A) 2Ax + 2Ay – 4Az B) – 2Ax – 2Ay + 4Az
C) – 4Az
D) 4Az E) None of the above
z
z
  A 0  0  4z 
 4 Az
Getting Electric Field from Potential (2)
•The other components can be found the same way:
•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
Ex- (Serway 25-39). Over a certain region of space, the electric potential is
V = 5x-3x2y+2yz2 V. Find the expressions for the x, y, and z components of
the electric field over this region. What is the magnitude of the electric field
at point P, which has coordinates (1.00, 0, -2.00) m?
Solve on
Board
Equipotential Lines Are Like
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
Understanding Equipotential Lines
In the graph below, what type of charge is at X, and what at Y?
A) Positive, both places
B) Positive at X, negative at Y
C) Negative at X, positive at Y
D) Negative, both places
•Positive charges don’t want to climb the high
potentials in kV
-1
+1
0
mountain at Y
-2
+2
•Must be positive charge repelling
-3
them!
X-4
Y
+4 +3
•Positive charges want to flow into
low valley at X
•Must be negative charge attracting them!
•Electric fields are perpendicular to equipotential surfaces
Conductors and Batteries

•A conductor has zero electric field inside it V   E  ds  0
•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
closed switch
open switch
A battery or cell is a device that creates a fixed
potential difference
•The circuit symbol for a battery looks like this:
•The long side is at higher potential
•It is labeled by the potential difference
What is the potential at point X?
A) 11 V
B) -11 V
C) +10 V
D) – 10 V
E) +8 V
F) -8 V
1.5 V
0V
1V
–1V
+8V X
3V
9V
Conducting Spheres
•Given the charge q on a conducting sphere of radius
q
R, what is the potential everywhere?
•Outside the sphere, the electric field is the same as
ke q
for a point charge
E  2 rˆ Vout  ke q
•Therefore, so is the potential
r
r
R
•Inside, the potential is constant
•It must be continuous at the boundary V   ke q r for r  R
VSphere
ke q

R

ke q R
for r  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
ke qi
Vi 
Ri
V2  2700 V
 5390 V
•After connections, their potentials must be equal
ke q1
k e q2

1 cm 2 cm
2q1  q2
q1  q2  12 nC
q1  4 nC
q2  8 nC
V1  V2 
 3600 V
Warmup 07
25.19
Electric Fields near conductors
q1
q2
•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
ke q1
E1  R1   2
R1
E1  R1  ke q1 R22
R2
 2

E2  R2 
R1 ke q2
R1
ke q2
E2  R2   2
R2
•The more curved the surface is, the higher the electric field is there
Very strong electric field here
•A sharp point can cause charged particles to spontaneously be shed into
air, even though we normally think of air as an insulator [ionize air]
•Called “Corona discharge”
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
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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.
-