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Electric Potential
Equipotentials and Energy
Today: Mini-quiz + hints for HWK
U = qV
Should lightening rods have a small
or large radius of curvature ?
V is proportional to
1/R. If you want a
high voltage to pass
through the rod
then use a small
radius of curvature.
Air is normally an insulator,
however for large E fields
(E>3 x 106 V/m) it starts
conducting.
Empire State Building, NYC
Electrical Potential
Review:
Wa → b = work done by force in going from a to b along path.
Wa → b = ∫
b
a
r r b r r
F • dl = ∫ qE • dl
a
∆U = U b − U a = − Wa → b
b
F
r r
= − ∫ qE • dl
b
a
θ
a
U = potential energy
dl
r
Wa → b
b r
∆U U b − U a
∆V = Vb − Va =
=
=−
= − ∫ E • dl
a
q
q
q
• Potential difference is minus the work done per unit
charge by the electric field as the charge moves from a to b.
• Only changes in V are important; can choose the zero at any point.
Let Va = 0 at a = infinity and Vb → V, then:
r r
V = − ∫ E • dl
r
∞
V = electric potential
allows us to calculate V
everywhere if we know E
Potential from charged spherical shell
• E-field (from Gauss' Law)
•
r < R: Er = 0
•
r >R:
V
q
4πε
πε0 R
1
q
Er =
4 πε 0 r 2
q
4πε
πε0 r
R
r
R
• Potential
R
• r > R:
r=r
V(r )= −
∫
r
r
r
E • d l = − ∫ E r ( dr ) =
r =∞
∞
1
4πε
0
q
Q
r
• r < R:
r
aR
r
r r
1
V ( r ) = − ∫ E • dl = −∫ Er ( dr ) = −∫ Er ( dr ) − ∫ Er ( dr ) =
4πε0
r =∞
∞
∞
a
r =r
R
q
Q
+0
aR
ELECTRIC POTENTIAL for Charged Sphere (Y&F, ex.23.8)
Suppose we have a charged
metal sphere with charge q.
What is the electric potential
as a function of radius r?
N.B. V is continuous but
E is not.
q
Clicker exercise
1
A point charge Q is fixed at the
center of an uncharged conducting
spherical shell of inner radius a
and outer radius b.
– What is the value of the potential Va at
a
Q
b
the inner surface of the spherical shell?
(a) Va = 0
(b)
Va =
1
4 πε
0
Q
a
(c)
Va =
1
4 πε
0
Q
b
Eout
Clicker exercise
1
A point charge Q is fixed at the
center of an uncharged conducting
spherical shell of inner radius a
and outer radius b.
– What is the value of the potential Va at
a
Q
b
the inner surface of the spherical shell?
(a) V a = 0 (b)
Va =
1
4 πε
0
Q
a
(c)
Va =
1
4 πε
0
Q
b
• How to start?? The only thing we know about the potential is
a r
r
its definition:
Va ≡ Va − V∞ = −∫ E • dl
∞
• To calculate
Va, we need to know the electric field E
• Outside the spherical shell:
• Apply Gauss’ Law to sphere:
• Inside the spherical shell:
r a r
r
b r
Va = − ∫ E • dl − ∫ E • dl
∞
b
r
Q rˆ
E=
4πε 0 r 2
E=0
1 Q
Va =
+0
4πε 0 b
Va =
1
4 πε
0
Q
b
Preflight 6:
Two spherical conductors are separated by a large distance.
They each carry the same positive charge Q. Conductor A has a
larger radius than conductor B.
A
B
2) Compare the potential at the surface of conductor A
with the potential at the surface of conductor B.
a) VA > VB
b) VA = VB
c) VA < VB
Electrical Potential
Two ways to find V at any point in space:
r
r r
• Use electric field:
V = − ∫ E • dl
∞
• Sum or Integrate over charges:
q1
q2 r2
P
r1
r3
q3
dq
r
1
qi
V =∑
i 4πε 0 ri
P
V=
dq
4πε 0 ∫ r
1
Examples of integrating over a distribution of charge:
• line of charge (review this one)
• ring of charge
• disk of charge
You should be able to do these.
Infinite line charge or conducting cylinder.
r
Linear charge density λ
λ
E=
2πε 0 r
b
Va − Vb = ∫ Edr = ∫
a
Suppose we set rb
to infinity,
potential is infinite
rb
λ
λ
=
ln( )
2πε 0 r 2πε 0
ra
Instead, set ra=r and rb=r0 at
some fixed radius r0.
Cover homework hints
Potential from a charged sphere
1 q
V (r) =
4πε0 r
(where V (∞) ≡ 0 )
Er
Equipotential
•
•
•
•
The electric field of the charged sphere has spherical symmetry.
The potential depends only on the distance from the center of the
sphere, as is expected from spherical symmetry.
Therefore, the potential is constant on a sphere which is
concentric with the charged sphere. These surfaces are called
equipotentials.
Notice that the electric field is perpendicular to the equipotential
surface at all points.
Equipotentials
Defined as: The locus of points with the same potential.
•
Example: for a point charge, the equipotentials are spheres centered on
the charge.
The electric field is always perpendicular
to an equipotential surface!
r r
VB − VA = − ∫ E • dl
B
Why??
A
Along the surface, there is NO change in V (it’s an equipotential!)
Therefore,
r r
− ∫ E • dl = ∆V = 0
B
A
r r
We can conclude then, that E • dl is zero.
If the dot product of the field vector and the displacement vector is zero,
then these two vectors are perpendicular, or the electric field is always
perpendicular to the equipotential surface.
EXAMPLES of Equipotential Lines
Conductors
+
+
+
+
+
+
+
+
+
+
+
+
+
+
• Claim
The surface of a conductor is always an
equipotential surface (in fact, the entire
conductor is an equipotential).
• Why??
If surface were not equipotential, there would be
an electric field component parallel to the surface
and the charges would move!!
Preflight 6:
B
A
3) The two conductors are now connected by a wire. How do the
potentials at the conductor surfaces compare now ?
a) VA > VB
b) VA = VB
c) VA < VB
4) What happens to the charge on conductor A after it is
connected to conductor B ?
a) QA increases
b) QA decreases
c) QA doesn’t change
Charge on Conductors?
• How is charge distributed on the surface of a conductor?
– KEY: Must produce E=0 inside the conductor and E normal to the
surface .
Spherical example (with little off-center charge):
+ +
+
- -- +
- +
+ -+q - +
+ - +
+ - +
+
+ + +
+
E=0 inside conducting shell.
charge density induced on
inner surface non-uniform.
charge density induced on
outer surface uniform
E outside has spherical
symmetry centered on spherical
conducting shell.
Equipotential Example
• Field lines more closely
spaced near end with most
curvature – higher E-field
• Field lines ⊥ to surface
near the surface (since
surface is equipotential).
• Near the surface,
equipotentials have similar
shape as surface.
• Equipotentials will look
more circular (spherical) at
large r.
Conservation of Energy
• The Coulomb force is a CONSERVATIVE force
(i.e., the work done by it on a particle which
moves around a closed path returning to its
initial position is ZERO.)
• Therefore, a particle moving under the influence
of the Coulomb force is said to have an
electric potential energy defined by:
U = qV
this “q” is the “test charge”
in other examples...
• The total energy (kinetic + electric potential) is then
conserved for a charged particle moving under the
influence of the Coulomb force.
Lecture 6, ACT 3
3A
Two test charges are brought
separately to the vicinity of a
positive charge Q.
Q
r
q
– charge +q is brought to pt A, a
distance r from Q.
– charge +2q is brought to pt B, a
Q
distance 2r from Q.
– Compare the potential energy of q
(UA) to that of 2q (UB):
(b) UA = UB
(a) UA < UB
3B
A
B
2r
2q
(c) UA > UB
• Suppose charge 2q has mass m and is released from
rest from the above position (a distance 2r from Q).
What is its velocity vf as it approaches r = ∞?
(a)
vf =
1
4 πε
0
Qq
mr
(b)
v
f
=
1
2 πε
0
Qq
mr
(c) v
f
= 0
Clicker Problem
3A
• Two test charges are brought
separately to the vicinity of positive
charge Q.
– charge +q is brought to pt A, a
distance r from Q.
– charge +2q is brought to pt B, a
distance 2r from Q.
Q
r
q
A
Q
2r
– Compare the potential energy of q (UA)
to that of 2q (UB):
(a) UA < UB
(b) UA = UB
(c) UA > UB
• The potential energy of q is proportional to Qq/r.
• The potential energy of 2q is proportional to Q(2q)/(2r).
• Therefore, the potential energies UA and UB are EQUAL!!!
2q
B
Clicker Problem
3B
• Suppose charge 2q has mass m and is released from
rest from the above position (a distance 2r from Q).
What is its velocity vf as it approaches r = ∞?
(a)
vf =
1
Qq
4πε 0 mr
(b)
vf =
1
Qq
2πε 0 mr
(c) v
= 0
f
• What we have here is a little combination of 170 and 272.
• The principle at work here is CONSERVATION OF ENERGY.
• Initially:
• The charge has no kinetic energy since it is at rest.
• The charge does have potential energy (electric) = UB.
• Finally:
• The charge has no potential energy (U ∝ 1/R)
• The charge does have kinetic energy = KE
UB = KE
1
4 πε
0
Q( 2q ) 1
= mv
2r
2
2
f
v 2f =
1
2 πε
0
Qq
mr
How to obtain vector E field from V
r r
V = − ∫ E • dl
r
• Determine V from E:
∞
Example: V due to spherical charge distribution.
Determining E from V:
r r b
∆V = Vb − Va = − ∫ E • dl = ∫ dV
b
a
For an infinitesimal step:
r r
dV = − E • dl = − E dl cos θ
a
E
θ
dl
Cases:
• θ = 0:
• θ = 90o:
• θ = 180o:
directional derivative
dV = E dl (maximum)
dV = 0
dV depends on direction
dV = -E dl
Can write:
r r
dV = − E • dl = −( E x iˆ + E y ˆj + E z kˆ) • (dxiˆ + dy ˆj + dz kˆ)
= −( E x dx + E y dy + E z dz )
Potential Gradient
Take step in x direction: (dy = dz = 0)
dV = − ( E x dx + E y dy + E z dz ) = − E x dx
dV
∂V
Ex = −
=−
partial derivative
dx y , z const .
∂x
Similarly:
Ey = −
And:
∂V
∂y
Ez = −
∂V
∂z
r
r
∂V ˆ ∂V ˆ ∂V ˆ
E = −(
i+
j+
k ) = −∇V
∂x
∂y
∂z
r
∂ ˆ ∂ ˆ ∂ ˆ
gradient operator
∇ =( i +
j + k)
∂x
∂y
∂z
Gradient of V points in the direction that V increases the fastest
with respect to a change in x, y, and z.
E points in the direction that V decreases the fastest.
E perpendicular to equilpotential lines.
Potential Gradient
Example: charge in uniform E field
E
U = qEy
q
V = U/q = Ey
where V is taken as 0 at y = 0.
r
r
∂ ˆ ∂ ˆ ∂ ˆ
E = −∇V = − ( i +
j + k ) Ey
∂x
∂y
∂z
= −(0 iˆ + E ˆj + 0kˆ) = − E ˆj
Given E or V in some region of space,
can find the other.
y
o
Cylindrical and spherical symmetry cases: Example: E of point charge:
∂V
∂
q
For E radial case and r is distance
Er = −
=− (
)
from point (spherical) or axis
∂r
∂r 4πε 0 r
(cylindridal):
∂V
Er = −
∂r
= −(
q
4πε 0
)(
q
−1
)
=
r2
4πε 0 r 2
r
r
∂
∂ ˆ ∂ ˆ
E = −∇V = − ( iˆ +
j + k ) { A( x 2 − 3 y 2 + z 2 ) )}
∂z
∂x
∂y
= −(2 Ax iˆ − 6 Ay ˆj + 2 Azkˆ)
= −2 A( x iˆ − 3 yˆj + zkˆ)
UI7PF7: Clicker Problem
This graph shows the electric
potential at various points
along the x-axis.
2) At which point(s) is the electric field zero?
A
B
C
D
UI7ACT1: Clicker Problem
1
The electric potential in a region of space is given by
V (x ) = 3x 2 − x 3
The x-component of the electric field Ex at x = 2 is
(a)
Ex = 0
(b)
Ex > 0
(c)
Ex < 0
UI7ACT1: Clicker Problem
1
The electric potential in a region of space is given by
V ( x ) = 3x 2 − x 3
The x-component of the electric field Ex at x = 2 is
(a)
Ex = 0
(b)
(c)
Ex > 0
We know V(x) “everywhere”
To obtain Ex “everywhere”, use
Ex < 0
r
r
E = −∇ V
∂V
Ex = −
∂x
E x = −6 x + 3x 2
E x (2 ) = −12 + 12 = 0
The Bottom Line/Take Home Message
If we know the electric field E everywhere,
W
V B − V A ≡ AB
q0
⇒
r
r
VB − V A = − ∫ E • dl
B
A
allows us to calculate the potential function V everywhere (keep
in mind, we often define VA = 0 at some convenient place)
If we know the potential function V everywhere,
r
r
E = −∇ V
allows us to calculate the electric field E everywhere
• Units for Potential! 1 Joule/Coul = 1 VOLT
(but remember we measure potential differences with a voltmeter)