56
Work, Potential Energy, Conservation of Energy
the electric forces are conservative:
0Fd r =ur r∫
Work, Potential Energy, Conservation of Energy
the electric forces are conservative
0Fd l =ur r∫( ) [ ] ( )
b
a
r
a br
W Fdl F r r drr d l F r dr⊥→ = = + =∫ ∫ ∫b b
a a
ur r r$ $
Work, Potential Energy, Conservation of Energy
0q qa b b a
a b a b
a b b a
a a b b
W K K F dl
W U UU U K KU K U K
→→
→
= − =
= −− = −+ = +
∫b
a
ur r
0 0
0 04 4a ba b
qq qqU Ur rπε πε
= =0
0
a force with which charge q acts on q ;
a work performed when a charge is moved from point to point
q q
a b
F
W
→
→ −
notations :
a b
ur
Work, Potential Energy, infinity as a reference point
potential energy has to be measured with respect to …
a b a bW U U→ = −
0 0
0 04 4a ba b
qq qqU Ur rπε πε
= =
0
0
0
0
0
0
4
4
q qa aa
q qb bb
qqU U F dlr
qqU U F dlr
πε
πε
∞
→∞
∞
→∞
= = =
= = =
∫
∫
a
b
ur r
ur r
Here, potential energy is measured with respect to an infinitely remote point
0 0
0 04 4a ba b
qq qqU Ur rπε πε
= =
U = work which was done by a charge q when an electric charge qo is delivered to infinity
= work which was done (by …) to bring an electric charge q0 from infinity
0q qa bU U F dl→− = ∫b
a
ur r
Work, Potential Energy, Conservation of Energy
0. .
04
externala a
el field el field
a
U U F dl
qqF dl F dlrπε
∞∞
∞
∞
= = =
− = =
∫
∫ ∫
a
a
a
ur r
ur r ur r
Here, potential energy is measured with respect to an infinitely remote point
0
04aa
qqUrπε
=
U = work which was done to bring an electric charge qo from infinity= the kinetic energy which this charge will acquire at infinity
if it will be released
0.el field q qF F →=ur ur
Here (!!!), potential energy is measured with respect to an infinitely remote point
0 0
0 04 4a ba b
qq qqU Ur rπε πε
= =
0 0
0 04 4a ba b
qq qqU Ur rπε πε
= =
Positron q=e, alpha particle Q=2e
Potential Energy
Electric Potential Energy with Several Charges
U= work which was done to bring an Electric Charge qo from infinity
along along 1 1 along 1contour contour contour
N N N
i ij j j ii i i
U Edl d l E E dl U= = =
= = = =∑ ∑ ∑ ∑ ∑ ∑ur r r ur ur r
Electric Potential Energy with Several Charges
U= work which was done to bring an electric charge qo from infinity= the kinetic energy which this charge will acquire at infinity if it will be released
potential energy U(x,y,z) and force F:
gradient of the potential energy is equal to minus force
( ) ( ) ( )x y z
U U UF F Fx y z
∂ ∂ ∂= − = − = −
∂ ∂ ∂r r r
( )F U∇= − rur ur
/ / /x y z∇ = ∂ ∂ + ∂ ∂ + ∂ ∂i j kur
2
1
r
r
U U Fdl− = −∫2 1r r
r
r
ur r
differentiation of this integral with works like differentiation of an ordinary integral:
∇ur
( )
( )
r
r
Fd l F r
Fdl F r
∇ =
∇ = −
∫
∫
r
r
ur ur r ur r
ur ur r ur r
The operation is called “taking gradient”; result – gradient.
( ) ( ) ( )U U UFx y z
⎛ ⎞∂ ∂ ∂= − + +⎜ ⎟∂ ∂ ∂⎝ ⎠
r r ri j kur
Electric Potential
( )
( )
F U
E V
= −
=
∇
∇−
r
r
ur ur
ur ur
( )
( )
F U
E V
⇔
⇔
r
r
ur
ur
Electric field is equal to electric force per unit charge
Electric potential is equal
to potential energy per unit charge
0
( )( ) UVq
=rr
CAUTION:potential energy has to be measured with respect to …
here, electric potential is measured with respect to an infinitely remote point, while r is the distance from the location of charge q
0
( )4q
qV rrπε
=
Electric Potential
Electric potential is equal to potential energy
per unit charge of a probe charge
( )
( )
F U
E V
= −
=
∇
∇−
r
r
ur ur
ur ur
( )
( )
F U
E V
⇔
⇔
r
r
ur
ur0
( )( ) UVq
=rr
0
( )4
a
a a
qVrπε
= ∑r
0 00
( ) ( )4
a
a a
qU q V qrπε
= = ∑r r
Units:
1V=1 volt= 1 J/C= 1 joule/coulomb
1V=1 (N/C)m
electric field is, thus, measured in volts per meter: V/m
xExVConstE ||)( ==r
rqrV
041)(πε
=
ε0 = 8.85×10-12 C/(Vm) CVm109
4π1 9
0
×=ε
This implies that the difference between the electric potentials of two points does not depend on the trajectory connecting these points:
if V(x,y,z) is known one can find the components of the electric field and vice versa
In electrostatics, the electric fields are conservative:
∫ = 0rdE rr
2
1
2 1[ ( ) ( )]r
r
Edr V r V r= − −∫r
r
r r r r
( )E V∇= − rur ur
Similarly to U(x,y,z), one can introduce the electric potentialV(x,y,z) such that:
( ) ( ) ( )x y z
V V VE E Ex y z
∂ ∂ ∂= − = − = −
∂ ∂ ∂r r r
( )E V∇−⇔ = rur ur
a aV V Edl∞
∞= = ∫a
ur r
( )r
r
Ed l Ed l E r∇ = −∇ = −∫ ∫r
r
ur ur r ur ur r ur r
( ) ( ) ( )V V VEx y z
⎛ ⎞∂ ∂ ∂= − + +⎜ ⎟∂ ∂ ∂⎝ ⎠
r r ri j kur
( ) ( ) ( )x y z
V V VE E Ex y z
∂ ∂ ∂= − = − = −
∂ ∂ ∂r r r
Elements of math
( )E V∇= − rur ur
partial derivative (consider and as constants)y zx∂∂
( ): (| |)
( ) ( )(| |)| |
is a unit vector a
a useful formula
long ( )| |
aa
a
aa
a
dV rVdr
dV lV r rdl
∇ =
−∇ − =
−−
−−
r r
r rr r
r r r rr r
ur$
ur r r
100eV =1.60217653(14)×10-17 J
Moving through a potential difference:change of the potential energy of a charge q is equal to the product of the electric potential change and the charge
Moving through a potential difference
1V =1J/C
• There is no electric field inside a conductor• Net charge can only reside on the surface of a conductor • Any external electric field lines are perpendicular to the
surface (there is no component of electric field that is tangent to the surface).
• The electric potential within a conductor is constant (valid only in the absence of currents)
Owing to the fact that inside the conductor, the difference
0=Er
For any two points and inside the conductor 1rr
2rr
21 rr VV =
2
1
r
r
V V Edl− = −∫2 1r r
r
r
rr
0V V− =2 1r r
Definition:
the electric potential difference is equal to an integral along line connecting the two points (any line!)
In equilibrium (i.e., without currents): 1) the conductor’s surface is equipotential.2) the whole body of the conductor is equipotential.
Equipotential Surfaces
An equipotential surface is a surface on whichthe electric potential V is the same at everypoint.Conversely, the electric field can do no work on a charge moving alongan equipotential surface. Electric field must be perpendicular to thesurface at every point so that the electric force is always perpendicular tothe displacement of a charge moving on the surface.
Field lines and equipotential surfaces are alwaysmutually perpendicular.
2
01
2 1 2 100
1[ ( ) ( )] lim [ ( ) ( )]r
qr
Ed l V r V r U r U rq→
= − − = − −∫r
r
rr r r r r
Gradient of the electric potential is equal to minus electric field.
It means that along the electric field line the electric potential goes DOWN, down, down!
( ) ( ) ( )r
E V Edl E r= − ∇ =∇ ∫r rr
ur ur ur rr r r
So, what is the use in one more quantity?
Because to draw a map of a scalar quantity representing the shape (relief) of the potential energy is much easier than to draw a map of the vector field.
Look on a topographic map of mountains ! It is a map of the gravitational potential:
Question: what is an analogue of a lake?
Examples
rqrV
041)(πε
=
constxExVConstE
+==
||)(
r
)]()([)]()([1lim 12120
00
2
1
rVrVrUrUq
rdEq
r
r
rrrrrrr
r−−=−−=
→∫
For a point charge at the origin (we already know the answer):
1 1
1 20 0 1
[ ( ) ( )]4 4r r
q dr qV V r Edrr rπε πε
∞ ∞
− ∞ − = = =∫ ∫r
rr r
( ) 0V ∞ =
( )( ) dV rV rdr
∇ = r$
( )a useful formula: (| |)
( ) ( )(| |)| |
( ) is a unit vector pointing from toward | |
a
dV rVdr
dV lV r adl
∇ =
−∇ − =
−−−
r r
r ar r
r a a rr a
ur$
ur r r
Example
r
RQ
041πε
?
Non-conducting with a homogenously distributed charge Q
V-V(∞)
R
r
RQ
041πε
?
Non-conducting sphere of radius A with a homogenously distributed charge Q
204
1rQERr
πε=>
304
QrR
r ERπε
=<
! with respect to what poin
( )t?
? becarefulV r =V-V(∞)
R
E
r
( ) ( ) ( )r
V r V E y dy∞
− ∞ = ∫
R
r
RQ
041πε
?
Non-conducting sphere of radius A with a homogenously distributed charge Q
204
1)(rQrERr
πε=>
30
( )4
QRR
r rE rπε
=<
R
E
r2
20
0
( ) ( ) ( )
1 3( ) ( )4 2 2
1( ) ( )4
r
V r V E y dy
Q rr R V r VR RQr R V r Vr
πε
πε
∞
− ∞ =
⎛ ⎞< − ∞ = − +⎜ ⎟
⎝ ⎠
> − ∞ =
∫
V-V(∞)
R
Non-conducting sphere of radius A with a homogenously distributed charge Q
204
1rQERr
πε=>
304
QrR
r ERπε
=<
R
E
r 0
02
30
0
( ) (0) ( ) ( )
( ) (0)8
1 3( ) (0)4 2
r
r
V r V E y dy E y dy
r Qr R V r VRQ Rr R V r VR r
πε
πε
− = = −
−< − =
⎛ ⎞> − = −⎜ ⎟⎝ ⎠
∫ ∫V(r)-V(0) is like V(r)-V(∞), but is shifted down on a constant
equal to
V(r) is parabolic at small r0
38
QRπε
constxExVConstE
+−==
||)(
r
For a point charge at the origin:
1
1
1
20 0 1
[ ( ) ( )]
4 4
r
r
V V r Edr
q dr qr rπε πε
∞
∞
− ∞ − = =
= =
∫
∫
r
rr r
( ) 0 ???( ) ???
VV
∞ =∞ = −∞
Potential of a charged plate
σ+
x
Potential of a charged plate
+
+
+
+
+
+σ+
SS
constxExVConstE +−== ||)(||r
σ+
x
Caution: “it is convenient to place the initial point, i.e., “b”, to infinity and set V(b)=0. “Never do it when you deal with “infinitely large/extended” objects!
+++
+
+
+
-
-
-
σ+
σ−-
-
-
-
-
-
S σ+σ−
x
Two parallel conducting plates
||)( xEblueV −=
||)( axEredV +=
Field between the plates is constant,potential is linear
0/εσ=E
+++
+
+
+
-
-
-
σ+
σ−-
-
-
-
-
-
S
Two parallel conducting plates
Field between the plates is constant,potential is linear
0/εσ=E
σ+σ−
x
σ+σ−
Potential at infinity is not zero!
( ) 0 ???V ∞ =
0 0
[ ( ) ( )]
ln | ???2 2
r
rr
V V r Edr
dy yy
λ λπε πε
∞
∞∞
− ∞ − = =
= =
∫
∫
r
rr r
0( ) 0V r = 0 0
0
0 0
0
0
( )] ln |2 2
ln2
r rrr
r r
dyV r Edr yy
rr
λ λπε πε
λπε
= = =
=
∫ ∫r
rr r
Again the problem with infinityResolution: only potential difference matters, measure potential difference with respect to
an arbitrary point r0
Method of images: What is a force on the point charge near a conducting plate?
-
-
-
-
Equipotential surface
The trick with imaging can be done not only with a flat surface
Additional material 1
Physics of Lightings
Benjamin Franklin
mm E
RV
=mVEm
6103 ⋅=
There is a “maximum potential to which a conductor in air can be raised because of ionization.”
ionization and corona discharge
The metal mast at the top of the Empire State Building acts as a lightning rod. It is struck by lightning as many as 500 times each year.
A lightning rod has a sharp end so that lightning bolts will pass through aconducting path in the air that leads to the rod; a conducting wire leads from thelightning rod to the ground.
Caution: what can be misleading in the above quotation?
“Even relatively small potentials applied to sharp points in air produce sufficiently high fields just outside the point to ionize the surrounding air”.
mVEm
6103 ⋅=
“Even relatively small potentials applied to sharp points in air produce sufficiently high fields just outside the point to ionize the surrounding air. “
ionization and corona discharge
There is a “maximum potential to which a conductor in air can be raised because of ionization.”
A sharp edge leads to the deformation of equipotential lines and, hence, to a high electric field near the edge
The taller the edge the stronger deformation of the lines and, therefore, the more probable discharge by a lighting
Method of images: What is a force on the point charge near a conducting plate?
-
-
-
-
Equipotential surface
The trick with imaging can be done not only with a flat surface
The force acting on the positive charge is exactly the same as it would be with the negative image charge instead of the plate.
2
2
0 )2(41||
aqF
πε=
The point charge feels a force towards the plate with a magnitude:
a-
-
-
-
Caution:2a rather than a !
Additional material 2
2
01
2 1 2 100
1[ ( ) ( )] lim [ ( ) ( )]r
qr
Ed l V r V r U r U rq→
= − − = − −∫r
r
rr r r r r
2
1
2 1[ ( ) ( )]r
r
Fd l U r U r= − −∫r
r
ur r r r 1 ( )g Um∇= − r
ur ur
electric potential V=U/q is similar to U/m in the case of the Earth gravitation
[ ( ) ( )] ( ) ( )a
b
r
a br
Ed l V r V r V a V b= − − = − +∫r
r
rr r r
arur
brur it is convenient to place the initial point,
i.e., “b”, to infinity and set V(b)=0.
dlr
What potential V(a) creates a positive charge Q? It is positive!
Note that when a positive charge is approached, is directed against , and therefore is generally negative. Hence V(a) is positive.
dlr
Er
Edlrr
( ) ( ) [ ( ) ( )] 0ar
aV a E dl V r V∞
= − = − ∞ >∫r
rr r
qV(a) is equal to a work W needed to perform in order to deliver the charge q from infinity to the point a. The work W is done against the electric field.
arur
∞dlr
( ) ( ) [ ( ) ( )]ar
aqV a qE dl q V r V∞
= − = − ∞∫r
rr r
For a point-like charge: If qQ>0 this work is positive; W>0. Correspondingly V(a)=W/q is positive if Q>0.
Now qV(a) can be interpreted as a work W which performs the electric field of the charge Q when the charge q is moving away.In this formulation, the work W is done by the electric field.
arur
∞dlr
( ) ( ) [ ( ) ( )]a
a
r
a
r
qV a qE dl q V r V
qEdl
∞
∞
= − = − ∞
=
∫
∫
r
r
rr r
rr
For a point-like charge. We are moving away from a positive charge: now and have common direction toward infinity.Hence , If qQ>0, this work is positive; W>0. Correspondingly V(a)=W/q is positive when Q>0.
dlr
Er
0Edl >rr
arur
∞dlr
Additional material 3
ay
23220 )(4
1ya
aqE+
−=+ πε
23220 )(4
1ya
aqE+
−=− πε
2 2 3 20
12 ( )
aqEa yπε
= −+
Calculation of the electric field near the conducting surface is possible, but not the shortest way, to find a force acting on the metallic plane
-
-
-
-
?F =ur
ay
2 2 3 20
2
2 2 3 20
2 2 2 2 2
2 2 3 4 20 0 00
12 ( )
1( ) 22 2 ( )
1 1 14 ( ) 4 4 4 (2 )
aqEa y
E aqdF ydy ydya y
a q a q qF ydya y a a
πε
πσ πε π
πε πε πε
∞
= −+
⎡ ⎤= = ⎢ ⎥+⎣ ⎦
= = =+∫
!
Calculation of the electric field near the conducting surface is possible, but not the shortest way, to find a force acting on the metallic plane
-
-
-
-
?F =ur
E0εσ =
Additional material 4
•In a recorded lecture from 1961, Richard Feynman explained to his students why physicists use electron volts to measure energy instead of some multiple of the joule:•A single atom is such a small thing that to talk about its energy in joules would be inconvenient. But instead of taking a definite unit in the same system, like 10−20 J, [physicists] have unfortunately chosen, arbitrarily, a funny unit called an electronvolt (eV) ... I am sorry that we do that, but that's the way it is for the physicists.
• Electron-volt• The electron-volt (eV) is a unit of energy.
By definition, it is equal to the amount of energy gained by a single unbound electron when it accelerates through an electrostatic potential difference of one volt.
• 1 eV = 1.60217653(14)×10−19 J. • So an electron volt is 1 volt (1 joule divided
by 1 coulomb) multiplied by the electron charge (1.60217653(14)×10−19 coulomb).
• The electron-volt is now accepted within SI.
1eV =1.60217653(14)×10-19 J
100eV =1.60217653(14)×10-17 J
