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 spheres, different radii, one with charge
r1= 10 cm
r2= 20 cm
Q10= 10 C
Q20= 0 C
Initially
Connect wire between spheres,
then disconnect it
Q1f= ??
Q2f= ??
wire
Are final charges equal?
What determines how charge
redistributes itself?
Mechanical analogy: Water pressure
Open valve, water flows
What determines final water levels? X
P1 = rgy1
P2 = rgy2
gy = PE/unit mass

3. ELECTRIC POTENTIAL 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 at a distant point without specifying
a test charge there (Like field, unlike PE)
Scalar field  Easier to use than E (vector). E = Gradient(Potential)
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.

4. Reminder: Work Done by a Constant Force
5-1: In the four examples shown in the sketch, a force F acts on an object and does work. In all four cases, the force has the same magnitude and the displacement of the object is to the right and has 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. Ds
I, IV, III, II
II, I, IV, III
III, II, IV, I
I, IV, II, III
III, IV, I, II I II
III IV

5. Work Done by a Constant Force (a reminder)II I
III IV Dr
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:
If the force varies in direction and/or magnitude along the path:
Example of a “Path Integral”
Result may depend on path

6. (from basic definition)
Definitions: Electrostatic Potential Energy versus Potential
Recall: Conservative Fields
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.
(basic definition)
POTENTIAL ENERGY DIFFERENCE:
Charge q0 moves from i to f along ANY path
Path Integral
POTENTIAL DIFFERENCE:
Potential is potential energy per unit charge
( Evaluate integrals
on ANY path from
i to f )
(from basic definition)

7. Some distinctions and details
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
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).
Use Work-KE theorem to link potential differences to motion

8. 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? 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 force does positive work on the proton.
D. Our force does positive work on the proton.
E. The changes cannot be determined.
Hint: which directions pertain to displacement and force?

9. DU and DV depend only on the endpoints
EXAMPLE: Find change in potential as test charge +q0 moves from point i to f in a uniform field E i f o Dx
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
EXAMPLE: CHOOSE A SIMPLE PATH 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 can be volts/meter
What are signs of DU and DV if test charge is negative?

10. Find potential V(R) a distance R from a point charge q :
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 :
Positive for q > 0, Negative for q<0
Inversely proportional to r1 NOT r2
Similarly, for potential ENERGY: (use same method but integrate force)
Shared PE between q and Q
Overall sign depends on both signs

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

12. so DV = 0 along any path on or in a conductor
Equi-potential surfaces: Voltage and potential energy are constant; i.e. DV=0, DU=0
Equipotentials are perpendicular to the electric field lines q
Vfi
Vi > Vf
DV = 0
E is the gradient
of V
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
and
CONDUCTORS ARE ALWAYS EQUIPOTENTIALS - Charge on conductors moves to make Einside = Esurf is perpendicular to surface
so DV = 0 along any path on or in a conductor

13. Examples of equipotential surfaces
Point charge or
outside sphere of charge
Uniform Field
Dipole Field
Equipotentials
are planes
(evenly spaced)
Equipotentials
are spheres
(not evenly spaced)
Equipotentials
are not simple shapes

14. EXAMPLE: UNIFORM FIELD E – 1 dimension
The field E(r) is the gradient of the potential q
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
Gradient = spatial rate of change
EXAMPLE: UNIFORM FIELD E – 1 dimension E Ds

15. Potential difference between oppositely charged conductors (parallel plate capacitor)Dx
E = 0 Vf Vi
Equal and opposite surface charges
All charge resides on inner surfaces
(opposite charges attract)
A positive test charge +q gains potential energy DU = qDV as it moves from - plate to + plate along any path (including external circuit)
Example:
Find the potential difference DV across the capacitor, assuming:
s = 1 nanoCoulomb/m2
Dx = 1 cm & points from negative to positive plate
Uniform field E

16. 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
r1= 10 cm
r2= 20 cm
V10= 90,000 V.
V20= 0 V.
Q20= 0 V.
wire
Initially:
Find the final charges:
Find the final potential(s):

17. 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) R a c b d
Find potential Va at point “a” inside shell
Definition:
Apply Gauss’ Law: choose GS just inside shell:
E(r) r
Einside=0 R
Eoutside= kq / r2
V(r)
Vinside=Vsurf
Voutside= kq / r
Potential is continuous across surface – field is not

18. Potential due to a group of point charges
Use superposition for n point charges
The sum is an algebraic sum, not a vector sum.
Reminder: For the electric field, by superposition, for n point charges
E may be zero where V does not equal to zero.
V may be zero where E does not equal to zero.

19. 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

20. 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

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

22. Rate of potential change perpendicular to equipotential
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, 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.
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

23. 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 = distance from dq to “P” = [a2 + z2]1/2
ds = arc length = adf
All scalars - no need to worry about direction
As z  0, V  kQ/a
As a  0 or z  inf, V  point charge
Almost point charge formula
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

24. 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
Integrate over the charge distribution
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.
Check by differentiating
Result

25. Example 23.10: Potential near an infinitely long charged line or a charged conducting cylinder
Near line or outside cylinder r > R
E is radial. Choose radial integration path if
Above is negative for rf > ri with l positive
Inside conducting cylinder r < R
E is radial. Choose radial integration path if
Potential inside is constant and equals surface value

26. Example 23.12: Potential at a symmetry point near a finite line of charge
Uniform linear charge density
Charge in length dy
Potential of point charge
Standard integral from tables:
Limiting cases:
Point charge formula for x >> 2a
Example formula for near
field limit x << 2a

27. 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/4pR2
Disc is a set of rings, 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
(note: )
“Near field” (z<<R): disc looks like infinite non-conducting sheet of charge
“Far field” (z>>R): disc looks like point charge
Use Anti-derivative: