Potential due to a constant E field
A
B
E
C
distance d
d is the distance along the direction of the E field. It is
not the distance from A to C. Only the distance along
the E field direction matters.
E Field and Potential:
Creating
A point charge q creates a field and potential around it:
Ek
q
e
r
2
r V k
ˆ;
e
q
r
Use superposition for
systems of charges
They are related:
B
E  V ; V  VB VA  
A
Ed
P05 - 4
E Field and Potential: Effects
If you put a cha
arged particle, q, in a field:
Work done by field: Wfield = -q(Vf-Vi) = -(Uf-Ui) = - U
F  qE
To move
a charged particle, q, in a field:
If KE doesn’t
change,
External work Wext=Uf-Ui = q(Vf-Vi) = - U
W

U

qV
SI Unit for W or U: J, Joule.
1J=1C x 1Volt
Another unit for W or U: eV (electron Volt)
1eV=1.6x10-19 C x 1Volte;
1eV=1.6x10-19 Joule
Superposition principle
Potential of ONE point charge.
V=V1+V2+V3+V4+…VN
Potential of MANY point charge.
Since V is a scalar, when we add V contribution from various
changes, we add them directly, regardless “directions”.
V itself has no direction.
It is much simpler than to add the E field contributions from
various change, since E field has directions and the sum of E
must be added as vectors.
Summary of Equations
1 point charge
creates
Continuous charges
create
Many point charge s
create
Scalar
integral
Vector
integral
Electrical Potential energy
One q at potential V
A pair of point charges
Many point charges
Work done by field: Wfield = -q(Vf-Vi) = -(Uf-Ui)
If you use all combinations of
i and j, you need to multiply ½
If KE doesn’t change External work Wext=Uf-Ui = q(Vf-Vi)
Potential difference
between two points A and B
External Work done to move charge q from A to B
This will be equal to the change of Potential Energy.
When a positive charge moves from A to B (if it is
somewhat opposite to the direction of E field, notice
that the Potential is higher at B point than A point), the
E field force on a positive charge is against the motion
from A to B. So the external work needed to move is
positive and U change is Positive.
From A to B Work done by E field is negative.
Work done by E field
A
B
C
How about
from A to C?
Or from B to C?
KE is like your checking account. PE is like your saving account.
When you consider the total amount of K+U,
no need to worry about transitions between K and U.
NO NEED TO calculate W done by and spring anymore,
When you consider K+U together.
Wexternal,
Work done by push, pull, wind, man, wave, elevator,
WEfeild
Kinetic Energy
K = ½mv2
…
U
Potential Energy
When only E field force (conservative forces) does work,
(No friction, no resistance (in air, on ice, not water…),
no external forces or energy source does work, (no human, animal,
motor, lift, push…)
K+U will stay unchanged.
Kinetic Energy
U
Potential Energy
K = ½mv2
The Change of PE is equal to the opposite of the KE change
When no external force does work
E is always in the direction of reducing V
Ex is V’s partial derivative along the x
direction, which describes how fast
V reduces along x direction with y
and z held constant.
E=dV/dx
Equipotential surfaces
E field always points in the direction
which has the most rapid potential REDUCTION
for the same amount of displacement.
The properties of equipotential follows:
(i) The electric field lines are perpendicular to the equipotentials and
point from higher to lower potentials.
(ii) By symmetry, the equipotential surfaces produced by a point charge
form a family of concentric spheres, and for constant electric field, a
family of planes perpendicular to the field lines.
(iii) The tangential component of the electric field along the
equipotential surface is zero, otherwise non-vanishing work would be
done to move a charge from one point on the surface to the other.
(iv) No work is needed to move a particle along an equipotential surface.
Continuous charge distribution
Taking infinity as our reference point with zero potential,
the electric potential at P due to dq is
Summing over contributions from all differential
elements, we have:
It is a scalar integral!
A Uniformly charged ring
Total charge Q
Radius a.
Find the potential
along the axis
at a distance of x
A uniformly charged rod
Total charge Q, Length L.
Find the potential along
the axis at a distance of
d from one end.
(2) Express dq in terms of charge density : dq=l dx’
(3) Substitute dq into expression for dV
(4) Rewrite r and the differential element in terms of the
appropriate coordinates
r=d+x’
(5) Rewrite dV
dV= ke l dx’/(d+x’)
(6) Integrate to get V
Six PRS Questions On
Pace and Preparation
P05 - 2
Potential for Uniformly Charged
Non-Conducting Solid Sphere
From Gauss’s Law
 Q r̂,
rR

4 r 2

E
0

 Qr r̂, r  R
3

4

R
0

B
Use
VB VA    E  d s
A
Region 1: r > a
r
VB V      
Point Charge!
Q
1
Q
2 dr 
4 0 r
4 0 r
P05 -43
Potential for Uniformly
Charged Non-Conducting
Solid
Region
2: Sphere
r<a
R
VD V      drE  r  R  
0
R
 
dr
Q
40 r

2
r

R
dr

r
R
drE  r  R
Qr
4 0 R

3
1 Q
1 Q 1 2
 4 R  4 3 2 r  R 2
0
0 R

2
1 Q
r


3 
8 0 R  R 2 
P05 -24
Potential for Uniformly Charged
Non-Conducting Solid Sphere
P05 -25
Group Problem: Charge Slab
Infinite slab with uniform charge density 
Thickness is 2d (from x=-d to x=d).
If V=0 at x=0 (definition) then what is V(x) for x>0?
x̂
P05 -26
Group Problem: Spherical Shells
These two spherical
shells have equal
but opposite charge.
Find E everywhere
Find V everywhere
(assume V() = 0)
P05 -27
Conductors and Insulators
A conductor contains charges that are free to move
(electrons are weakly bound to atoms)
Example: metals. Regular water (with impurity)
An insulator contains charges that are NOT free to
move (electrons are strongly bound to atoms)
Examples: plastic, paper, wood, DI water.
Extremely high voltage difference may cause a
regular insulator (such as air) to conduct. (lightening)
under equilibrium condition.
Use Gauss’ Law to calculate E field
near a Conductor’s surface
Charged conductor sphere
E field inside is zero everywhere!
V of the conductor is a constant everywhere.
E field on surface is
proportional to Q/(4R2)
E outside is
proportional to Q/(4r2)
V at infinity = 0.
Integrate and get
V outside is proportional to Q/(4r)
http://ocw.mit.edu/ans7870/8/8.02T/f04/visualizations/electrostatics/34pentagon/34-pentagon320.html
Conductors as Shields