# Electric Potential

```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
Electrostatics: Two spheres, different radii, one with charge
Connect wire between spheres,
then disconnect it
Initially
Q10= 10
C cm
r = 10
1
Q1f= ??
wire
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
ELECTRIC POTENTIAL

V(r )
Potential Energy due to an electric field
per unit (test) charge
• Closely related to Electrostatic Potential Energy……but……
• DPE: ~ work done ( = force x displacement)
• DV: ~ work done/unit charge ( = field x displacement)
• Potential summarizes effect of charge on a distant point without specifying
a test charge there (Like field, unlike PE)
• Scalar field  Easier to use than E (vector)
• Both DPE and DV imply a reference level
• Both PE and V 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
• Synonyms for V: both [F][d]/[q], and [q][E][d]/[q] = N.m / C.
• Units of field are [V]/[d] = Volts / meter – same as N/C.
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
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
If the force varies in direction and/or
magnitude along the path:


DW   F  ds
WIII = FDr
IV
WIV = FDr cos 
f
i
Example of a “Path Integral”
Result may depend on path
Spring 2014
Definitions: Electrostatic Potential Energy versus Potential
Recall: Conservative Fields definition
• 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
dV  dU / q0
(basic definition)


Fe = q0E
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
 
Vf  Vi  DV   DW / q0 =   E  ds
f
i
(from basic definition)
( Evaluate integrals
on ANY path from
i to f )
Some distinctions and details
D U = q0 D V
•
•
•
•
•
•
The field depends on a charge
distribution elsewhere).
A test charge q0 moved
between i and f gains or loses
potential energy DU.
DU does not depend on path
DV also does not depend on
path and also does not depend
on |q0| (test charge).
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
•
A convenient energy unit: electron volt
–
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
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?
EXAMPLE: Find change in potential as test charge +q0
moves from point i to f in a uniform field
i
E
f
Dx
o
DU and DV 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
- units of E can be volts/meter
• E field does negative work
What are signs of DU and DV if test charge isCopyright
negative?
R. Janow
Spring 2014
Potential Function for a Point Charge
•
•
•
•
Charges are infinitely far apart  choose Vinfinity = 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
Equi-potential surfaces:
Voltage and potential energy are constant; i.e.
DV=0, DU=0
• No change in potential energy along an equi-potential
• Zero work is done moving charges along an equi-potential
• Electric field must be perpendicular to tangent of equipotential


DV  E  Ds = -E Ds cos() = 0

and DU = DW = Fe  Ds = 0

for Ds along surface
• Equipotentials are perpendicular to the electric field lines
DV = 0
Vi > Vf
DV = 0
Vfi

DV = 0
CONDUCTORS ARE ALWAYS EQUIPOTENTIALS
- Charge on conductors moves to make Einside = 0
- Esurf is perpendicular to surface
of V
so DV = 0 along
any path on or
Spring 2014
inR.aJanow
conductor
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
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
Potential difference between oppositely charged
conductors (parallel plate capacitor)
+
-
• Equal and opposite surface charges
• All charge resides on inner surfaces
(opposite charges attract)
Dx  L
   = 

E=

 0
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)
Comparison of point charge and mass formulas
for vector and scalar fields
FIELD
FORCE
VECTORS
 
 
M
mM
force/unit mass
g
(
r
)
=
G
r̂
F(r ) = G 2 r̂
Gravitation
(acceleration)
r
r2
 
 
1 Q
force/unit charge
1 qQ
Electrostatics F(r ) =
E(r ) =
r̂
r̂
(n/C)
4 0 r 2
4 0 r 2
SCALARS
Gravitation
POTENTIAL ENERGY
Ug (r ) = G
mM
r
1 qQ
Electrostatics Ue (r ) =
4 0 r
POTENTIAL
Vg (r ) = G
M
r
1 Q
Ve (r ) =
4 0 r
PE/unit mass
(not used often)
PE/unit charge
Fields and forces ~ 1/R2 but Potentials and PEs ~ 1/R1

U
F= 
s

V
E= 
s
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
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
0.1
Spring 2014
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
V(r)
E(r)
Vinside=Vsurf
Voutside= kq / r
Eoutside= kq / r2
Einside=0
R
r
R
r
Potential due to a group of point charges
• Use superposition for n point charges

V(r ) =
n
1
V
=
 i 4 
0
i =1
n
 r
i =1
qi

 ri
• The sum is an algebraic sum, not a vector sum.
 
r  r2
 
r  r1

r

r1

r2
• Reminder: 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.
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
VP =
+q
P
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.
Let q = 1 nC, d = 2 m:
9  10 9  10 9
VP = 4
= 18 Volts
2
DIPOLE – Otherwise positioned as above
EP  0
d
+q
-q
P
but
Let q = 1 nC, d = 2 m:
VP =
obviously
EP = 2
kq
kq

=0
d/2
d/2
kq
d2 /4
= 8
kq
d2
9  109  10 9
EP = = 8
= 18 V/m (or N/C)
4
Another example: square with charges on corners
a
q
-q
d
a
P
a
d
-q
q
a
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
by symmetry, again
kqi k
4kq
8kq
VP = 
=  qi =
=
= 510 Volts
r
d
a 2 /2 a 2
i
i
i
Another example: find work done by 12 volt battery in 1 minute
as 1 ampere current flows to light lamp
i
E
+
-
DW  work done = - DU = - QDV
Q = charge moved from  to - by current
= i Dt = 1 amp x 60 sec = 60 C.
DU = QDV = 60 x DV
DV = 12 Volts
 DU =  720 Joules
R. Janow Spring 2014
DW =  DU =  720 Joules (from
battery)
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
q
q
q
q
q
-q
C
B
-q
q
D
q
-q
E
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
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 = distance from dq to “P” = [a2 + z2]1/2
ds = arc length = adf
P
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
(along z by symmetry)
As
Before
 V=
All scalars - no need to
Almost point charge
formula
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 >> Spring
a
2014
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.
•
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
 
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

Example 23.10: Potential near an infinitely long charged
line or charged conducting cylinder
E=
2kl
r
Near line or outside cylinder r > R
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
DVinside =  
i
f


E  dr = 0
Potential inside is constant and equals surface value
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
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
2
R
 
0
0
dVP ,z
 a da d
[ a 2  z 2 ]1/ 2

r
z
R
cos() = z/r
r 2 = a2  z2
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
sheet of charge
Vdisk 
R
20
z 

1  R 
E
dV

=
dz
2 0
```