1. Physics 121 - Electricity and Magnetism Lecture 05 -Electric Potential Y&F Chapter 23 Sect. 1-5
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

2. Electrostatics: Two metal spheres, different radii, one charged
Initially
Connect wire between spheres,
then disconnect it
wire
Q10= 10 C
r1= 10 cm
Q1f = ??
Q2f = ??
Are final charges equal?
What determines how the
charge redistributes itself?
What if spheres are identical?
Q20= 0 C
r2= 20 cm X
P1 = rgy1
P2 = rgy2
Open valve, water flows. What determines final water levels?
Mechanical analogy:
Water pressure
PE/unit mass = gy
Pressure = rgy
Potential energy of water molecule at the top surface = mgy

3. ELECTROSTATIC POTENTIAL
Definition: Electrostatic Potential equals
Electrostatic Potential Energy per unit (test) charge
Potential Energy DU:
DU = Work done ~ force x displacement or charge x field x displacement.
Potential DV:
DV = - Work done on a test charge / test charge ~ - field x displacement.
= Change in PE of a test charge as it is displaced, per unit test charge.
Units, Dimensions:
Potential Energy DU: Joules
Potential DV: [DU] / [q] are Joules/C. = VOLTS
E field units: [V]/[d] are Volts / meter – same as N/C.
Both DU and DV:
use a reference level (possibly at infinite distance) & are functions of position.
represent conservative forces via scalar fields (does gravity).
can describe the motion of objects in conjunction the Work-KE theorem &/or mechanical
energy conservation, as an alternative to using the Second Law (F = qE = ma).

4. Reminder: Work Done by a Constant Force
5-1: In each example shown in the sketch, a force F has the same magnitude but the direction varies. The displacements of the object are all to the right with the same magnitude.
Rank the cases in order of the work done by the force on the object, from most positive to the most negative. I II
III IV Ds
I, IV, III, II
II, I, IV, III
III, II, IV, I
I, IV, II, III
III, IV, I, II
Force may vary in direction and magnitude along the path:
Result of a “Path Integration” in general
can depend on integration path taken

5. (from basic definition)
Definitions: Electrostatic Potential versus Potential Energy
E field is “conservative”, like gravity
Work dW done BY THE FIELD on a test charge
moving from i to f does not depend on path taken.
Work done changes potential energy and potential.
Work done around any closed path equals zero.
POTENTIAL ENERGY DIFFERENCE:
Charge q0 moves from i to f along ANY path
path integral
POTENTIAL DIFFERENCE:
Potential is work done by field per unit charge
( Evaluate integral on ANY path from i to f )
(from basic definition)

6. CHOOSE THE SIMPLE PATH SHOWN THROUGH POINT “O”
EXAMPLE: Find change in potential DV as test charge +q0 moves from point i to f in a uniform field E i f o Dx
DV or DU depend only on the endpoints
ANY PATH from i to f gives same results
CHOOSE THE SIMPLE PATH SHOWN THROUGH POINT “O”
Displacement i  o is normal to field (path along equipotential)
External agent must do positive work on
positive test charge to move it from o  f
E field does negative work
Units of E are also volts/meter
What are signs of DU and DV if test charge is negative?
To convert potential to PE just multiply by q0

7. Work and PE : Who/what does positive or negative work?
5-2: In the figure, we exert an external force that moves a proton from point i to point f in a uniform electric field as shown. Which of the following statements is true? E i f
A. Electric field does positive work on the proton.
Electric potential energy of the proton increases.
B. Electric field does negative work on the proton.
Electric potential energy of the proton decreases.
C. Our external force does positive work on the proton.
D. Our external force does positive work on the proton.
E. The changes cannot be determined.
Hint: which directions pertain to displacement and force?

8. Summary and details Reference levels can be:
Relative: Choose arbitrary zero reference level for ΔU or ΔV. …or…
Absolute: Set Ui = 0 for all charges that started 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
New name for electric field units:
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
The field is created by a charge distribution somewhere else.
A test charge q0 moved between i and f gains or loses potential energy DU.
DU and DV do not depend on path. DV also does not depend on q0 (test charge).
Use Work-KE theorem to link potential differences to motion.

9. Find potential V(R) a distance R from a point charge q :
Example: Potential Function for a Point Charge
Reference level: V(r) = 0 for r = infinity (makes sense)
Field is conservative  choose radial integration path for E(r)
DU = q0DV = work done on a test charge as it moves to final location
without change in kinetic energy
Find potential V(R) a distance R from a point charge q :
Positive for q > 0, Negative for q < 0
Inversely proportional to r1 NOT r2
For potential ENERGY: same method but integrate force
Shared PE between q and q0
Overall sign depends on both signs

10. Electric field lines are perpendicular to the equipotentials
On an “Equipotential Surface” (Contour) voltage and potential energy are constant; i.e. DV=0, DU=0 q
Vfi
Vi > Vf
DV = 0
for Ds normal to E
Steepest Change
E = -Gradient of V
= -spatial rate of change of V
- zero work is done moving charge along an equipotential
and
Electric field lines are perpendicular to the equipotentials

11. so DV = 0 along any path in or on a conductor
CONDUCTORS ARE ALWAYS EQUIPOTENTIALS - Charge on conductors moves to make Einside = Charge is on the outer surface - Esurf is perpendicular to surface = s/e0
so DV = 0 along any path in or on a conductor
Demonstration Source: Pearson Study Area - VTD
Chapter 23
Charged Conductor with Teardrop Shape
Where on the surface is the field strongest?
Discussion:
Blunt and sharp ends are at the same potential
How do the potential, field, and surface charge densities differ for the blunt and sharp ends?
E = s / e0 just outside the surface and is outward (Gauss’ Law)
At the sharp end q = sA is larger, so s is larger and so is E

12. Examples of Equipotential Surfaces
Point charge or
outside a sphere of charge
Uniform Field
Dipole Field
Equipotentials
are planes
(evenly spaced)
Equipotentials
are spheres
(spaced for 1/r)
Equipotentials
are not simple shapes

13. E is a known, constant function, so
Example: Find potential difference between oppositely charged plates of a parallel plate capacitor s+ s- L Dx
E = 0 Vf Vi
Infinite sheets of uniform charge (Dx << L)
Equal and opposite surface charges
All charge moves to inner surfaces
(opposite charges attract)
E is a known, constant function, so
A positive test charge +q gains potential energy DU = qDV as it moves from - plate to + plate along any path (including an external circuit)
Assume:
s = 1 nanoCoulomb/m2
Dx = 1 cm & points from negative to positive plate
Uniform field E (from + to – plate)

14. Conductors are always equipotentials
Example: Two spheres, different radii, one charged to 90,000 V.
Connect wire between spheres – charge moves Conducting spheres come to same potential
Net charge on system stays constant, but redistributes to make the system an equipotential. Spherical symmetry.
r1= 10 cm
r2= 20 cm
V10= 90,000 V.
V20= 0 V.
Q20= 0 V.
wire
Initially (use shell theorem):
Find the final charges:
Find the final potential(s):

15. Potential is continuous across conductor surface – field is not
Example: Find potential Va at point “a” inside a hollow conducting shell R a c b d
Assume Vb = 18,000 Volts on surface.
Vc = Vb (conducting shell is an equipotential).
Shell can be any closed surface (sphere or not)
Definition:
Apply Gauss’ Law with GS just inside shell:
E(r) r
Einside=0 R
Eoutside= kq / r2
V(r)
Vinside=Vsurf
Voutside= kq / r
Potential is continuous across conductor surface – field is not

16. Potential due to a group of point charges
Use superposition for n point charges
Potential is a scalar… so … the sum is an
algebraic sum, not a vector sum.
Comparison: For the electric field…
Unit vector from
ri to r
E may be zero where V does not equal to zero.
V may be zero where E does not equal to zero.

17. Examples: potential due to point charges Use Superposition
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 +q P d
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 +q -q P d

18. Find E & V at center point P
Examples: potential due to point charges - continued
Another example: square with charges on corners q -q a d
Find E & V at center point P P
Another example: same as above with all charges positive

19. Electric Field and Electric Potential
5-3: Which of the following figures have both V=0 and E=0 at the red point? A q q -q B q C D q -q E -q q

20. Finding potential if E is known: use the definition of V (Example 23
Finding potential if E is known: use the definition of V (Example 23.10: Potential near an infinitely long, charged, conducting cylinder mimics a line of charge for points outside the cylinder)
Gauss Law result:
Field outside cylinder at r > R
Field inside = 0 for r < R
E is radial. Choose radial integration path if
Above is negative for rf > ri with l positive
inside conducting cylinder for r < R
Take E to be radial, integrate over radial path if
Potential inside is constant and equals surface value

21. Rate of potential change perpendicular to equipotential
Method for finding potential function V due to a continuous charge distribution: E not known
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, s, or r
Typical challenge: express above in terms of chosen coordinates
3. At point P, dV is the differential contribution to the potential due to a point-like charge dq located in the distribution. Use symmetry to simplify.
4. Use “superposition”. Add up (integrate) the contributions over the whole charge distribution, varying the displacement r as needed. Scalar VP.
5. Field E can be gotten from potential by taking the “gradient”:
Rate of potential change perpendicular to equipotential

22. Example 23.11: Potential along Z-axis of a ring of chargey z r P a f dq
Q = charge on the ring
l = uniform linear charge density = Q/2pa
r = [a2 + z2]1/2 = distance from dq to “P”
ds = arc length = adf
All scalars - no need to worry about direction
As a  0 or z  inf, V  point charge
As z  0, V  kQ/a
Looks like a point charge formula,
but r is on the ring
FIND ELECTRIC FIELD USING GRADIENT
(along z by symmetry)
E  0 as z  0 (for “a” finite)
E  point charge formula for z >> a
As Before

23. Supplementary Material

24. Visualizing the potential function V(r) for a positive point charge (2 Dimensions)
For q negative
V is negative
(funnel) r
V(r)
1/r

25. Distinction: Slope, Grade, Gradients in Gravitational Field
EXTRA TOPIC
Height contours portray constant gravitational potential energy DU = mgDh. Force is along the gradient, perpendicular to a potential energy contour.
Grade means the same thing as slope. A 15% grade is a slope of
100 15
15% q Dh Dl Dx
Gradient is measured along the path. For the case at left it would be:
The gradient of the potential energy
is the gravitational force component along path Dl:
The GRADIENT of height (or gravitational potential energy) is a vector field representing steepness (or gravitational force)
In General: The GRADIENTS of scalar fields are vector fields.

26. The field E(r) is the gradient of the potential
EXTRA TOPIC
Uniform field, normal to equipotential surfaces
Component of ds on E changes potential
Component of ds normal to E produces no change
For a path along equipotential, DV = 0 q
equipotentials E
E = -spatial rate of change of V = -Gradient of V

27. Example 23.12: Potential at symmetry point P near a finite line of charge
Uniform linear charge density
Charge in length dy
Potential of point charge
From standard integral tables:
Limiting cases:
Point charge formula for x >> 2a
Example formula for near
field limit x << 2a
Use infinite series approximation

28. Optional Example: Potential Due to a Charged Rod
A rod of length L located parallel to the x axis has a uniform linear charge density λ. Find the electric potential at a point P located on the y axis a distance d from the origin.
Start with
The following will be used below to integrate:
Now, let me give you an example to show how to do that.
Let’s assume the rod is lying along the x axis, dx is the length of one small segment, and dq is the charge on that segment. So we have dq=lamda times dx. The magnitude of the electric field at P due to this small segment is
The total field at P is the integration from a to l+a. Since lamda equals to Q/l, the electric field can be expressed like that.
Now, let’s see two extreme cases: (1) l=>0, means the rod has shrunk to zero size,, equation reduces to the electric field due to a point charge.
(2) a>>l, means P is far from the rod, then l in the denominator can be neglected. This is exactly the form we would expect for a point charge.
Integrate over the charge distribution using interval 0 < x < L
Result

29. Homework Example: Potential on the symmetry axis of a charged diskz q P a
dA=adfda r
Q = charge on disk whose radius = R.
Uniform surface charge density s = Q/pR2
Disc is a set of rings, radius a, each of them da wide in radius
For one of the rings:
Double integral
Integrate twice: first on azimuthal angle f from 0 to 2p which yields a factor of 2p then on ring radius a from 0 to R
Use Anti-derivative:

30. Limiting cases: Potential on the symmetry axis of a charged diskz q P a
dA=adfda r
An approximation using infinite series
“Far field”: assume z>>R Disc mimics a point charge far away
“Near field”: assume z<<R Disk mimics an infinite non-conducting sheet of charge