Physics 121 - Electricity and Magnetism
Lecture 05 -Electric Potential
Y&F Chapter 23 Sect. 1-5
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Electric Potential Energy versus Electric Potential
Calculating the Potential from the Field
Potential due to a Point Charge
Equipotential Surfaces
Calculating the Field from the Potential
Potentials on, within, and near Conductors
Potential due to a Group of Point Charges
Potential due to a Continuous Charge Distribution
Summary
1
Copyright R. Janow Fall 2015
Electrostatics: Two spheres, different radii, one with charge
Initially
Connect wire between spheres,
then disconnect it
Q10= 10 C
r1= 10 cm
wire
Q1f= ?? Q2f= ??
•Are final charges equal?
•What determines how charge
redistributes itself?
Q20= 0 C
r2= 20 cm
Mechanical analogy: Water pressure
Open valve, water flows
What determines final
water levels?
gy = PE/unit mass
P2 = rgy2
P1 = rgy1
X
Copyright R. Janow Fall 2015
ELECTRIC POTENTIAL

V(r )
Define: Electrostatic Potential … equals …
Potential Energy per unit (test) charge
Related to Electrostatic Potential Energy……but……
• DPE: ~ work done ( = force x displacement = charge x field x displacement )
• DV: ~ work done/unit charge ( = field x displacement )
Potential:
• summarizes effect of charge at some point in space without specifying
a test charge there (like E field, unlike PE)
• is a scalar field  easier to use than E (vector). E = Gradient (Potential)
Both DPE and DV:
• imply a reference level
• are conservative forces/fields, like gravity
• can determine motion of charged particles using: Second Law, F = qE
or PE, Work-KE theorem &/or mechanical energy conservation
Units, Dimensions:
• Potential Energy U: Joules
• Potential V: [U]/[q] Joules/C. = VOLTS
• Units of field are [V]/[d] = Volts / meter – same as N/C.
Copyright R. Janow Fall 2015
Reminder: Work Done by a Constant Force
5-1: In the four examples shown in the sketch, a force F
an object and does work. In all four cases, the force has
same magnitude and the displacement of the object is to
right and has the same magnitude.
Rank the cases in order of the work done by the force on
object, from most positive to the most negative.
acts on
the
the
the
Ds
A.
B.
C.
D.
E.
I, IV, III, II
II, I, IV, III
III, II, IV, I
I, IV, II, III
III, IV, I, II

F
I
III

F

F
II

F
IV
Copyright R. Janow Fall 2015
Work Done by a Constant Force (a reminder)
 
DW  F  Ds = FDs cos 
The work DW done by a constant external
force on it is the product of:
• the magnitude F of the force
• the magnitude Δs of the displacement
of the point of application of the force
• and cos(θ), where θ is the angle
between force and displacement vectors:
Dr

F

F

Dr

Dr
II
I
WII =  FDr
WI = 0

F

F

Dr

Dr
III
In general, force may vary in direction
and magnitude along the path:


DW   F  ds
WIII = FDr
IV
WIV = FDr cos 
f
i
Result of “Path Integration”
may depend on path taken
Copyright R. Janow
Fall 2015
Definitions: Electrostatic Potential Energy versus Potential
E field is “conservative”, like gravity
• Work done BY THE FIELD on a test charge moving
from i to f does not depend on the path taken.
• Work done around any closed path equals zero.


dU = dW = Fe  ds


Fe = q0E
(work done by field
dV  dU / q0
POTENTIAL ENERGY DIFFERENCE:
Charge q0 moves
from i to f
along ANY path
POTENTIAL
DIFFERENCE:
Potential is
potential energy
per unit charge
f

 

Uf  Ui  DU   DW    Fe  ds = q0  E  ds
f
i
Path Integral
i


DU
DW
DV 
=
= E  Ds
q0
q0
(from basic definition)
 
Vf  Vi  DV   DW / q0 =   E  ds
f
i
( Evaluate integrals
on ANY path from
i to f )
Copyright R. Janow Fall 2015
Some distinctions and details
D U = q0 D V
•
•
•
•
•
The field is created by a charge
distribution elsewhere).
A test charge q0 moved
between i and f gains or loses
potential energy DU.
DU and DV do not depend on
path. DU also does not
depend on |q0| (test charge).
Use Work-KE theorem to link
potential differences to motion
Only differences in electric potential and
PE are meaningful:
– Relative reference: Choose arbitrary zero
reference level for ΔU or ΔV.
– Absolute reference: Set Ui = 0 with all
charges infinitely far apart
– Volt (V) = SI Unit of electric potential
• 1 volt = 1 joule per coulomb = 1 J/C
• 1 J = 1 VC and 1 J = 1 N m
•
Electric field units – new name:
– 1 N/C = (1 N/C)(1 VC/1 Nm) = 1 V/m
•
“Electron volt” is an energy unit:
–
1 eV = work done moving charge e
through a 1 volt potential difference
= (1.60×10-19 C)(1 J/C) = 1.60×10-19 J
Copyright R. Janow Fall 2015
Work and PE : Who/what does positive or negative work?
5-2: In the figure, suppose we exert a
force and move the proton from point i
to point f in a uniform electric field
directed as shown. Which statement of
the following is true?
A.
B.
C.
D.
E.
f
i
E
Electric field does positive work on the proton.
Electric potential energy of the proton increases.
Electric field does negative work on the proton.
Electric potential energy of the proton decreases.
Our force does positive work on the proton.
Electric potential energy of the proton increases.
Our force does positive work on the proton.
Electric potential energy of the proton decreases.
The changes cannot be determined.
•Hint: which directions pertain to displacement and force?
Copyright R. Janow Fall 2015
EXAMPLE: Find change in potential as test charge +q0
moves from point i to f in a uniform field
i
E
f
o
Dx
DV or DU depend only on the endpoints
ANY PATH from i to f gives same results
uniform field
To convert potential to/from PE just multiply/divide by q0
 
DVfi =   E  ds
path


Fe = q0E
 
DUfi   DWfi =   F  ds
path
DVfi  DUfi / q0
DUfi = q0DVfi
EXAMPLE: CHOOSE A SIMPLE PATH THROUGH POINT “O”
DVf ,i = DVo,i  DVf ,o
DVo,i = 0
Displacement i  o is normal to field (path along equipotential)
 DVf ,i = DVf ,o


= - E  Dx = E | Dx |
• External agent must do positive work on
positive test charge to move it from o  f
• E field does negative work
• units of E can be volts/meter
What are signs of DU and DV if test charge isCopyright
negative?
R. Janow
Fall 2015
Potential Function for a Point Charge
•
•
•
•
Infinitely far away choose V(R = infinity) = 0 (reference level)
DU = work done on a test charge as it moves to final location
DU = q0DV
Field is conservative  choose most convenient path = radial
Find potential V(R) a distance R from a point charge q :
 

V(R)  V  VR =   E  ds along radial path from r = R to  ds = r̂dr
R



q

dr
1
q
E(r ) = k 2 r̂
  E  ds =  V(R) = kq  2 = kq
= k
r
rR
R
r

q
 V(R ) =  k
R


R
R
(23.14)
• Positive for q > 0, Negative for q<0
• Inversely proportional to r1 NOT r2
Similarly, for potential ENERGY: (use same method but integrate force)
U(r )  QV (R ) = k
q.Q
R
(23.9)
• Shared PE between q and Q
• Overall sign depends on both signs
Copyright R. Janow Fall 2015
Visualizing the potential function V(r)
for a positive point charge (2 D)
For q negative
V is negative
(funnel)
V(r)
1/r
r
r
Copyright R. Janow Fall 2015
On Equi-potential Surfaces:
Voltage and potential energy are constant; i.e. DV=0, DU=0
Equipotentials are perpendicular to the electric field lines
DV = 0
Vi > Vf
DV = 0
Vfi

E is the gradient
of V


DV  E  Ds
DV = 0
• Zero work is done moving charges along an equi-potential
DV = - E Ds cos() = 0 for Ds along surface equi - potential


and
DU = DW = Fe  Ds = 0
CONDUCTORS ARE ALWAYS EQUIPOTENTIALS
- Charge on conductors moves to make Einside = 0
- Esurf is perpendicular to surface
so DV = 0 along
any path on or
in a conductor
Copyright R. Janow Fall 2015
Examples of equipotential surfaces
Uniform Field
Point charge or
outside sphere of
charge
Dipole Field
Equipotentials
are planes
(evenly spaced)
Equipotentials
are spheres
(not evenly spaced)
Equipotentials
are not simple
shapes
Copyright R. Janow Fall 2015
The field E(r) is the gradient of the potential


dV  E  ds = - E ds cos()
•Component of ds on E produces potential change
•Component of ds normal to E produces no change
•Field is normal to equipotential surfaces
•For path along equipotential, DV = 0

ds

• Gradient = spatial rate of change



dV
V  V
V
 E=-  =
i
ĵ 
k̂  - V ds is  to equipotential
ds
x
y
z
f (x, y, z )
Math note :
is a " partial" derivative
x
EXAMPLE: UNIFORM FIELD E – 1 dimension


DV  E  Ds =  E Ds
Ds
E
DU = q0 DV = q0EDs = FDs
Copyright R. Janow Fall 2015
Potential difference between oppositely charged
conductors (parallel plate capacitor)
+
-
• Assume Dx << L (infinite sheets)
• Equal and opposite surface charges
• All charge resides on inner surfaces
(opposite charges attract)

   = 


E=
0
constant
L
Dx


DV  Vf  Vi = E  Dx
Vf
E=0
Example:
Find the potential difference DV across the capacitor, assuming:
•  = 1 nanoCoulomb/m2
• Dx = 1 cm & points from negative to positive plate
• Uniform field E


1 10-9
DV = E  Dx =  E Dx =
 10-2
0
Vi
DV =  1.13 volts
A positive test charge +q gains potential energy DU = qDV as it moves
from - plate to + plate along any path (including external circuit)
Copyright R. Janow Fall 2015
Conductors are always equipotentials
Example: Two spheres, different radii, one charged to 90,000 V.
Connect wire between spheres – charge moves
Conductors come to same potential
Charge redistributes to make it so
V1f = V2 f
r1= 10 cm
V10= 90,000 V.
wire
Q1f  Q2f = Q10
r2= 20 cm
Initially:
V10
kQ10
= 9  104 Volts =
 Q10 = 1.0 C.
r1
V20= 0 V.
Q20= 0 V.
Find the final charges:
V1f =
k[Q 10  Q1f ]
kQ1f
= V2 f =
r1
r2
Find the final potential(s):
kQ1f
V1f =
=
r1
r2 -1
Q1f = Q10 (1  ) = 0.33 C.
r1
Q2f = Q10  Q1f = 0.67 C.
9  109  0.33  x10 6
= 30,000 Volts = V2f
Copyright R. Janow
0.1
Fall 2015
Potential inside a hollow conducting shell
Vc = Vb (shell is an equipotential)
= 18,000 Volts on surface
R = 10 cm
Shell can be any closed surface (sphere or not)
d
c
a R
Find potential Va at point “a” insidea shell
b
Definition:
DVab
 
 Va  Vb =   E  ds
b
Apply Gauss’ Law: choose GS just inside shell:
qenc = 0  E = 0 everywhere inside  DVab = 0
 Va = Vsurface = Vb = Vc = Vd = 18,000 Volts
Potential is continuous across surface – field is not
E=
E(r)
dV
dr
Eoutside= kq / r2
Einside=0
R
r
V(r)
Vinside=Vsurf
Voutside= kq / r
r
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Copyright
Potential due to a group of point charges
• Use superposition for n point charges
• Potential is a scaler… so …
• The sum is an algebraic sum, not a vector sum.

V(r ) =
n
1
V
=
 i 4 
0
i =1
n
 r
i =1
 
r  r2
 
r  r1

r
qi

 ri

r1

r2
• Comparison: For the electric field, by superposition, for n point charges
 
E(r ) =

 Ei 
n
i =1
1
4 0
n

i=1
qi
  2 r̂i
r  ri
• E may be zero where V does not equal to zero.
• V may be zero where E does not equal to zero.
Copyright R. Janow Fall 2015
Use Superposition
Examples: potential due to point charges
Note: E may be zero where V does not = 0
V may be zero where E does not = 0
TWO EQUAL CHARGES – Point P at the midpoint between them
EP = 0
d
+q
+q
P
VP =
by symmetry
kq kq
kq

=4
d/2 d/2
d
obviously not zero
F and E are zero at P but work would have
to be done to move a test charge to P from infinity.
DIPOLE – Otherwise positioned as above
EP  0
d
+q
-q
P
but
VP =
obviously
EP = 2
kq
kq

=0
d/2
d/2
kq
d2 /4
= 8
kq
d2
Copyright R. Janow Fall 2015
Examples: potential due to point charges - continued
Another example: square with charges on corners
a
q
-q
d
a
P
a
d
-q
a
q
Find E & V at center point P
d= a 2/2
EP = 0 by symmetry
kq k
k
VP =  i =  qi = [q  q  q  q]
d i
d
i ri
VP = 0
Another example: same as above with all charges positive
EP = 0
VP =
by symmetry, again
kqi k
4kq
8kq
=
q
=
=
i r d i i a 2 / 2 a 2
i
Copyright R. Janow Fall 2015
Electric Field and Electric Potential
5-3: Which of the following figures have V=0 and E=0 at
the red point?
q
-q
q
q
A
B
q
q
q
q
q
-q
C
-q
q
D
q
-q
E
Copyright R. Janow Fall 2015
Method for finding potential function V at a point P due to a
continuous charge distribution
1. Assume V = 0 infinitely far away from charge distribution (finite size)
2. Find an expression for dq, the charge in a “small” chunk of the distribution, in
terms of l, , or r
 ldl for a linear distributi on



dq =  d2 A for a surface distributi on

 rd3 V for a volume distributi on



Typical challenge: express above in terms of chosen coordinates
3. At point P, dV is the differential contribution to the potential due to a pointlike charge dq located in the distribution. Use symmetry.
dq
dV =
scalar, r = distance from dq to P
4  0r
4. Use “superposition”. Add up (integrate) the contributions over the whole
charge distribution, varying the displacement r as needed. Scalar VP.
1
dq
VP =  dVP =
(line, surface, or volume integral)
4 0  r
dist
dist
5. Field E can be gotten from potential by taking the “gradient”:


 
V
Rate of potential change
E =     V
dV  E  ds
perpendicular to equipotential
s
Copyright R. Janow Fall 2015
Example 23.11: Potential along Z-axis of a ring of charge
z
Q = charge on the ring
l = uniform linear charge density = Q/2a
r = [a2 + z2]1/2 = distance from dq to “P”
ds = arc length = adf
P
dq = lds = ladf
dq
dV = k
r
2
kal
kQ
V =  dV =
d
f
=
r 0
r
ring
r
z
f
y
a
dq
x
FIND ELECTRIC FIELD
USING GRADIENT
(along z by symmetry)
As
Before
 V=
All scalars - no need to
worry about direction
Looks like a point
charge formula,
but r is on the ring
kQ
[ z 2  a 2 ]1 / 2
• As z  0, V  kQ/a
• As a  0 or z  inf, V  point charge
V
kQ  (z 2 )
kQz
Ez = 
k̂ = 3
k̂ = 2
k̂
2 3/2
z
2r
z
[z  a ]
• E  0 as z  0 (for “a” finite)
• E  point charge formula
forR. Janow
z >> Fall
a 2015
Copyright
Example: Potential Due to a Charged Rod
•
A rod of length L located parallel to the x axis has a uniform linear charge density  . Find
the electric potential at a point P located on the y axis a distance d from the origin.
•
Start with
r  [x 2  d2 ]1 / 2
dq = ldx
1 dq
1
ldx
dV =
=
4 0 r
4 0 (x 2  d2 )1 / 2
•
Integrate over the charge distribution on interval 0 < x < L
 
L
l
dx
l
=
ln x  ( x 2  d2 )1 / 2
2
2
1
/
2
4 0 (x  d )
4 0
0
V =  dV = 
=
•
 

l
ln L  (L2  d2 )1 / 2  ln(d)
4 0
 L0

Check by differentiating
d
log(x  r )
dx
for
r = [x 2  d2 ]1 / 2
d
1 d(x  r )
1
dr
1
x
1 rx
1
log(x  r ) =
=
(1 
)=
(1  ) =
(
)=
dx
x  r dx
xr
dx
xr
r
xr r
r
•
Result
 L  (L2  d2 )1 / 2 
l
V=
ln 

4 0 
d

Copyright R. Janow Fall 2015
Example 23.10: Potential near an infinitely long charged
line or a charged conducting cylinder
E=
2kl
r
Near line or outside cylinder r > R
E is radial. Choose radial integration path if
DVfi = Vf - Vi =  
i
f

f dr

E  d r =  2kl 
i r
DVfi =  2kl ln[ri / rf ]
Above is negative for rf > ri with l positive
E=0
Inside conducting cylinder r < R
E is radial. Choose radial integration path if
DVinside =  
i
f


E  dr = 0
Potential inside is constant and equals surface value
Copyright R. Janow Fall 2015
Example 23.12: Potential at a symmetry point near a
finite line of charge
l = Q/2a Uniform linear charge density
dq = ldy Charge in length dy
dq Potential of point charge
dV = k
r
a
a
dy
VP =  dV = kl 
-a
- a (x 2  y2 )1/ 2
Standard integral from tables:

(x
2
dy
=
 y2 )1/ 2
ln [ y  r ] = ln [ y  (x 2  y2 )1/ 2 ]
VP =
VP
 a  (x 2  a2 )1/ 2 
kl ln
2
2 1/ 2 
  a  (x  a )

 (x 2  a2 )1/ 2  a 
kQ
=
ln

2
2 1/ 2
2a
(x

a
)

a


Limiting cases:
• Point charge formula for x >> 2a
• Example 23.10 formula for near
field limit x << 2a
Copyright R. Janow Fall 2015
Example: Potential on the symmetry axis of a charged disk
P
• Q = charge on disk whose radius = R.
• Uniform surface charge density  = Q/4R2
• Disc is a set of rings, each of them da wide in radius
• For one of the rings:
dq   dA =  a da df
VP,z
1
=
4 0
R
2
 
0
0
dVP ,z
 a da d
[a2  z2 ]1/ 2

r
z
R
dA = adfda
cos() = z/r
r 2 = a2  z2
dA=adfda
k dq
=
r
f
a
Double integral
x
• Integrate twice: first on azimuthal angle f from 0 to 2 which yields a factor of 2
then on ring radius a from 0 to R
(note: (1  x)1/ 2  1  21 x  21 41 x2..... for x2  1 )
2 
a da
4  0 0 [ a 2  z 2 ]1/ 2
R
VP,z =
“Far field” (z>>R): disc looks like point charge
a
d 2
Use Anti=
[ a  z 2 ]1/ 2
2
2 1/ 2
derivative: [ a  z ]
da
Vdisk =

2 0
 z
2
 R2

1/2
 z

Vdisk


2 0


1 R2
1 Q
 z =
z 
2 z
4 0 z


“Near field” (z<<R): disc looks like infinite
non-conducting sheet of charge
Vdisk 
R
20
z 

1  R 
E
dV

=
dz
2 0
Copyright R. Janow Fall 2015