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Lectures 5,6 (Ch. 23)
Electric Potential

1. Work in E
2. Electric Potential Energy, U
3. Electric Potential, V
4.V of a point charge
5. A set of the point charges
6. Distributed charges
a) superposition principle
b) finding V via E
7. Inverse problem: E via V.
 
F E



Work in E
 
  F  dl 
b
dl
b
Wab
a
q
b
 F cos  dl
a


Gravity F  mg is a
conservative force
a
Conservative force:
Wab does not depend on pass

dl
Wab
 Wab

It is reversible Wab  Wba  F  dl  0


Its circulation is zero:

  F  dl  0
Electric force F  qE
is a conservative force
 
 q  E  dl
b
Wab
a
Potential energy
  b
Wab   F  dl   F cos   dl  U a  U b  U
b
a
a
   / 2  Wab  0  U 
   / 2  Wab  0  U 


 E
dl
F


F  qE
 
Wab  q  E  dl  U a  U b  U
b
a
a
Electric Potential (Voltage) A work per unit charge
b 
 U U
Wab
Vab 
  E  dl  a  b  Va  Vb  V
q
q
q
a
b
SI units of Voltage
[V]=[W]/[q]=1J/1C=1V (Volt)
[E]=V/m=J/Cm=N/C
1ev (electron-volt) is a unit of
energy (not a unit of V!)
It is defined as an energy gained by an
electron accelerated between two points→
with voltage =1v
ΔU=qΔV→1ev=1.6x10-19Cx1V=1.6x10-19J
Alessandro Volta
(1745-1827)
Inventor of the first
battery
General properties of electric potential
Vab
Wab


q


Ua
Ub
a E  dl  q  q  Va  Vb  V
b


Wab  q  E  dl  q (Va  Vb )  U a  U b   U


a
b
1. V always decreases along Elines: a E
Vb < Va

b
E
is always directed from higher to lower potential



( Indeed , if dl  E (or   )  V  0)
2
b
b
a
a
b
a

E

E
+++++
a
b

E
2.Equipotential surfaces
(the surfaces where V=const)
a) No work can be done moving along such surface.

┴ E

Indeed, for arbitrary a and b: Va = Vb→W ab=0
b) Equipotential surfaces are

Indeed , on such surface arbitrary dl  E sin ceWab  0
c) Surface
 of conductor is equipotential
(since
E
┴ to this surface)
E=0,
V=Vsurface
d) If E=0 in some region arbitrary two points
b 

in this region have the same potential. Indeed, Vab   E  dl  0
a
f) V=const inside the volume of the conductor =Vsurface
Electric potential and potential energy in
the uniform electric field
+ + + + + + + +
 y
E
V
tan ὰ=V/y=E
- - - - - - - 0y
 
V (0)  V ( y )   E  dy   Ey
ὰ
y
W0 y  q[V (0)  V ( y )]  qEy
0
Choose V (0)  0  V ( y )  Ey U ( y )  qEy
V=const→y=const (Equipotential surfaces are planes)
Analogy between potential energy in the uniform
electric and gravity fields
h
Vab  Eh
Wab   U  qVa b  qEh
Energy conservation law: K+U=const
Example. An electron was accelerated in a uniform
field
E  900V / m
from v0=0 to vf=106m/s.
I.
vf=106m/s
h
Find :a) change of potential energy;
b) work done by the field;
c) change of potential;
d) displacement of the electron, h.
2. How the answers would change if v0=106m/s in a
horizontal direction and vf=√2x106m/s?
a) Ka+Ua= Kb+Ub; ΔU=Ub-Ua
ΔU=Ka-Kb=-m vf2/2
ΔU=-9x10-31kgx1012m2/2sec2
ΔU =-4.5x10-19J
b) Wab=-ΔU= 4.5x10-19J
c)ΔV= ΔU/q=-4.5x10-19J/(-1.6x10-19C)=2.8V
d)ΔV=Eh→h=ΔV/E= 2.8Vm/900V≈3mm
II. The answers do not change since x component of the velocity
does not change and change of the y component is the same
as in part I. Indeed, v0y=0, vfy=√(vf2-v02)=106m/s
v0=0
Cathode ray tube
Cathode (C)
Deflecting plates (DP)
z
y
Anode (A)
Fluorescence
screen
ΔV
electrons emitted by cathode
are accelerated by voltage between C and A,
entering the region between DP with
v0 
2eV .Then electrons are deflected by E.
m
E
x
In DP region
along x: motion
with v=v0;
along y: motion
with acceleration
in uniform E
see the previous example and an example in Lect.2
V of a point charge
 kq 
E  2 r0
r b 
b
a
b

dr
1 1
V aVb   E  dr  kq 2  kq(  )
r
a b
a
a
Choose Vb  0 when b  
kq
denote a  r  V (r ) 
r
Equipotential surfaces:V(r)=const
→

r=const → Spheres ┴ E
Take ΔV=const between the neighboring
equipotenttial surfaces then density of
equipotential surfaces shows the steepness
of V(r).
V
70V
50V
30V
r
Topographic map analogous to equipotential surfaces
V=kq/r
h
Δh=const
: Larger density of the lines →steeper the hill
Positive q vs. negative q
V decreases
as you move
outward
V=-30V
V=-50V
V
V
0
0
r
V=-70V
r
r
ὰ
d
r>>d
 
1
1
kqd cos  kp  r0
V (r )  kq(

)
 2
2
r  (d / 2) cos  r  (d / 2) cos 
r
r
V of a dipole
Figure 23.24c
r
r>>d
V=2kq/r
V of a ring of charge
kdq
kdq
dV (r ) 

,
2
2
r
a x
k
kQ
V   dV 
dq 

2
2
2
2
a  x ring
a x
ring
kQ
V ( x  0) 
a
Example. A uniformly charged thin ring has a radius10 cm and a
total charge 12 nC. An electron is placed on the ring axis a distance 25 cm
from the center of the ring. The electron is then released from rest.
1. Find the speed of the electron when it reaches the center of the ring.
2. Describe the subsequent motion of the electron.
1.U ( x)  U (0)  K (0)
K (0)  U ( x)  U (0)  q[V ( x)  V (0)]
kQ
kQ
V ( x) 
; V ( x  0) 
a
a2  x2
2keQ 1
1
v
[ 
]
2
2
m a
a x
mv 2
1
 keQ[ 
2
a
1
a x
2
2
]
1
1

]
2
2
0.1m
(0.1m)  (0.25m)
7

1
.
54

10
m/s
31
2 2
9 10 kg C
2  9 109 Nm2 1.6 10 19 C 12 10 9 C[

2. Periodic motion back and forth through the center of the ring.
V of the conducting sphere
1. r>R (as for the point charge)
 kq 
E  2 r0
r b
b
 
dr
1 1
V aVb   E  dr  kq 2  kq(  )
r
a b
a
a
Choose Vb  0 when b  
kq
denote a  r  V (r ) 
r
2. r<R
R
V (r )  V ( R)   0  dr  0  V (r )  V ( R) 
r
V ( R)
V ( R)  R  E ( R)  E ( R) 
R
kQ
R
Lightening
electric discharge
in atmosphere
----V
+++
Benjamin Franklin
(1706-1790)
Lightening rod
controls
conducting pass:
E is stronger near
the surface with
small radius:
V ( R)
E ( R) 
R
+
+
E=Ecr dielectric strength (dielectric
material becomes
conducting due to ionization). In air
Ecr=3x106V/m.
When E>3MV/m accelerated
electrons collide with
molecules in air producing
avalanche ionization (conducting
E is also stronger closer to the cloud
pass).
Van der Graaf Generator
It produces high charge or
voltage, limited by Ecr=3x106V/m.
Vmax ( R)  R  Ecr ( R) 
V ( R  1cm)  102 m  3 106V / m  3 104V
V ( R  1m)  1m  3 106V / m  3 106V
Scanning tunneling microscope (STM),
Nobel Prize 1986
Very sharp tip
E ( R) 
V ( R)
R
-e
Dielectric breakdown of vacuum
In very strong fields e-p pairs are produced and vacuum becomes a conductor
Recombination of e and p (annihilation) results in emission of
electromagnetic radiation (0.5 MeV annihilation line from the center og our
Galaxy): e+p=2mc2 → E=mc2/e=2x9x10-31kgx9x1016m/s/1.6x10-19 C=0.5Me
V of the dielectric sphere
1. r<R
q
R
 r
Q
E
;
3 0
4R 3 / 3
rdr  ( R 2  r 2 )
V (r )  V ( R)  

3 0
2  3 0
r
R
kQ kQ( R 2  r 2 ) 3kQ kQr 2
V (r ) 




3
3
R
2R
2R
2R
kQ
r2
3kQ

(3  2 ); V (r  0) 
2R
R
2R
2. r>R , the same as for
conducting sphere:
kq
V (r ) 
r
Very long conducting cylinder, L>>R, r

L
1. r<R: E(r)=0→V(r)=V(R)
2. r>R:
E
2k
r
b
dr
b
Va  Vb  2k 
 2k ln ; If we takeVb  0 when b  
r
a lnx
a
x
we get Va  2k ln   !
1
For infinitely long charged objects never choose V(r→∞)=0 !!!
Choose V=0 at some finite distance, for example, V(R)=0.
V(r)
R
dr
R
r
V (r )  2k 
 2k ln
 2k ln
r
r
r
R
r
Long conducting cylindrical shell
a


Choose V(a)=0. Then 1. r<a: V(r)=0.
b 2. a<r<b:
r
V
a
b
V ( r )  2k ln
a
b
V (b)  2k ln
a
b
Vab  2k ln
a
Vab 1
2k
E (r ) 
, E (r ) 

b
r
r
ln
a
b
3. r>b: V ( r )  2k ln a
A Geiger counter counts the particles which
ionize the air . The emitted electrons
accelerated by high E lead to avalanche
ionization resulting in detected current.
r
b
V (r )  2k ln
a
Conducting slabs
E=const
Choose V(a)=0
1.x<a →V(x)=0
2. x>a →
V(x)=E(x-a)
v
a
x
-a
-
E=const
Choose V(a)=0
1.x<a →V(x)=0
2. x>a →
V(x)=E(a-x)
+
+
+
+
+
+
+
+
+
+
v
-a
V=Ed
V=Ex
x
-
d
+
+
+
+
+
x
-
a
Conclusions
1. For a finite size objects:
Choose V(∞)=0
Calculate the potential first outside the object
then inside the object
2. For infinite size objects:
Choose V on some surface=0
Calculate potential inside and outside
3. Potential varies continuously on the surface of
both conductor and insulator
4. Inside
conductor V=const=V on the surface
Insulator V ~ r2
5. Outside both conductor and insulator
sphere ~ 1/r
Cylinder ~ ln r
Plane ~ r
Electric potential energy of two charges
→
 U a  U b  q0 (Va  Vb )
Choose Vb=0 (and hence Ub=0) when rb→∞ then U (r ) 
kqq0
 Vq q0
r
Thus U is = the work done by electric force produced by charge q
on charge q0→to move q0→from r to infinity.
kqq0
U (r ) 
r
The same sign of q and q : U>0
0
The work by electric force to bring q0
to ∞ is positive: (force is in the
direction of motion)
Opposite sign of q and q0 : U<0
The work by electric force to bring q0 to
∞ is negative: (force opposes motion)
Ionization energy
(disassembling H atom)
F
r=0.5x10-10m
Ionization energy=E=13.6eV=
Wexternal force to move e from
r=0.5x10-10m to ∞
ke2
U 
 27.2eV
r
mv 2
K
; F  ma
2
2
2
ke
v
F  2 ,a
r
r
2
2
ke
v
m 
2
r
r
mv 2 ke2

 13.6eV
2
2r
E  K  U  13.6eV
Nuclear energy
He4
r=10-10m
2
ke
U
 1.4MeV
r
Nuclear radius
Au, Z=79
ὰ particle (4He2+), Z=2
K0=6MeV
r
Head on collision:
Ernest Rutherford
(1871– 1937), Nobel Prize 1908
K 0  U 0  K f  U f ;U 0  0 (initially r  )
kqq0
K f  0 ( particle stopped ); K 0  U f  K 0 
r
kqq0 9 109 Nm 2  2  79  (1.6 10 19 ) 2 C 2
14
r


4

10
m
2
6
19
K0
C  6 10 1.6 10 J
Some ὰ particles were
scattered back which
contradicted the “plum
pudding” model of
atom and led to the
planetary model with
tiny nucleus in the
center.
Several point charges
q0
r01
q0 qi
U q0  q0V  k 
i 1 r0 i
= work which was done by external force
to bring an electric charge q0 from infinity to a
given position in the presence of all other charges
q1
q2
N
qN
U tot
q0 qi
 k
i  j rij
= work which was done by external force
to bring all electric charges from infinityto given
positions
Example. Three point charges
r01
q0
r02
q1
r12
q2
kq0 q1 kq0 q2
U q0  q0V 

r01
r02
U tot
kq0 q1 kq0 q2 kq1q2



r01
r02
r12
Inverse problem:
Finding E via V
b


 E  dl  V    dV
b
a


dV
if dl  E then E  
dl
Example 1.
q
kq
V (r ) 
r
dV
kq
E
 2
dr
r
a
Example 2.
Example 3.


r
V  2k ln
R
dV 2k
E

dr
r

V 
x
2 0
dV

E
E
dx
2 0
General case:
b
 
 E  dl  V   dV
b
a
a

 

E  i Ex  j E y  kEz
 


dl  i dx  j dy  k dz
 
E  dl  E x dx  E y dy  E z dz
dV
V
Ex  
| y const, z const  
dx
x
V
V
Ey  
; Ez  
y
z

 V  V  V
E  (i
j
k
)
x
y
z
     
V  (i
j
 k )V
x
y
z


E   V
Finding E via V
Example. Knowing potential of the ring on x axis
find E at P.
V
kQ
a2  x2
V
kQx
Ex  
 2
x (a  x 2 )3 / 2
NB: the above expression for V is valid only on the x-axis.
It does not allow to derive Ey and Ez.
The obtained expression for Ex is valid also only on the x-axis.

Example. V  y  2 xy  4 xyz, E  ?

 

E  i Ex  j E y  kEz
 


dl  i dx  j dy  k dz
 
E  dl  E x dx  E y dy  E z dz
2
V
Ex  
 (2 y  4 yz )
x
V
Ey  
 (2 y  2 x  4 xz)
y
V
Ez  
 4 xy
z




E  2 y (2 z  1)i  2( y  x  2 xz) j  4 xyk