What You Already Know
ÎCoulomb’s
ÎElectric
ÎGauss’
law
fields
law
ÎElectric
fields for several configurations
‹ Point
‹ Line
‹ Plane
(nonconducting)
‹ Sheet (conducting)
‹ Ring (along axis)
‹ Disk (along axis)
‹ Sphere
‹ Cylinder
‹ Dipole (along || and ⊥ axes)
PHY2049: Chapter 24
1
Chapter 24: Electric Potential
ÎElectric
Potential Energy
ÎElectric
Potential
ÎEquipotential
Surfaces
ÎPotential
of Point Charge
ÎPotential
of Charge Distribution
ÎCalculating
the Field from the Potential
ÎPotential
Energy from a System of Charges
ÎPotential
of Isolated Charged Conductors
PHY2049: Chapter 24
2
Reading Quiz: Chapter 24
ÎAn
equipotential surface is:
‹ a)
‹ b)
‹ c)
‹ d)
‹ e)
a surface where the electric field is constant
always parallel to the electric field
a surface where the potential is zero
always perpendicular to the electric field
a surface where the electric field is zero
PHY2049: Chapter 24
3
Reading Quiz: Chapter 24
ÎThe
volt is a unit of:
‹ a)
potential energy
‹ b) electric field
‹ c) potential
‹ d) force
PHY2049: Chapter 24
4
Reading Quiz: Chapter 24
ÎElectric
potential is:
‹ a)
a scalar quantity
‹ b) a vector quantity
‹ c) can be either scalar or vector
PHY2049: Chapter 24
5
Electric Work and Potential Energy
ÎPoint
charges q1, q2: Work moving charge q2 from A → B
B
WAB = ∫ F ⋅ ds = ∫
rˆ ⋅ ds
( rˆ ⋅ ds = dr )
r2
B kq1q2
kq1q2 kq1q2
=∫
dr =
−
2
A r
rA
rB
A
WAB
B kq1q2
ÎPath
A
• Depends only on endpoints
• Path independent
• Like gravitation
independence: “conservative force”
‹ Define
potential energy of two point charge
kq1q2
U B − U A ≡ −WAB ⇒ U =
r
PHY2049: Chapter 24
6
Electric Force is Conservative
ÎHolds
in all electrostatic situations (not just point charge)
‹ Proof:
ÎWork
integrate over any charge distribution
done by electric field moving charge q from i to f
‹ Calculate
from difference of potential energies
i
Charges
Work:
q
f
Welec = −ΔU fi = U i − U f
PHY2049: Chapter 24
7
Problem: Electric Potential Energy
ÎTwo
identical +12 mC point charges are initially spaced 5
cm from each other. If they are released at the same
instant from rest, how fast will they be moving when they
are very far from each other? Assume m1 = m2 = 1.0 g.
Ki + Ui = K f + U f
kq 2
=2
0+
di
(
1 mv 2
f
2
(9 ×10 ) (0.012 )
−3
10
( ) ( 0.05)
9
vf =
)
kq 2
+0 ⇒ vf =
md i
2
= 1.6 × 105 m/s
PHY2049: Chapter 24
8
Gravitational & Electric Potential Energy
Gravity
Electric
A
h
B
WG = mgh = U A − U B
-
+
+
+
+
+
+
+
A
d
B
WE = qEd = U A − U B
Point B at lower potential energy than point A (q>0)
PHY2049: Chapter 24
9
Electric Potential
Î Potential
= PE per unit charge
V =U /q
Î Potential
-------------------
b
difference: general E field
b
Vb − Va = − ∫ E ⋅ ds
a
Î Potential
difference: constant E
d
E
+Q
Va − Vb = Ed
Î Potential
+q: Moves from higher to lower V
‹ −q: Moves from lower to higher V
‹
PHY2049: Chapter 24
a
+++++++++++
higher at + charges and
“falls” to lower value at − charges
10
Units for V and E
ÎUnits
‹V
of potential: “volt”
= U/q ⎯→ Volt = Joule / Coulomb
ÎUnits
of electric field
= Eq ⎯→ E = F/q → Newton / Coulomb
‹ V = Ed ⎯→ E = V/d → Volt / Meter
‹F
PHY2049: Chapter 24
11
Example of Potential of Point Charge
ÎPoint
charge q (using V = 0 at r = ∞)
kq
V=
r
ÎExample:
(
Potential at surface of proton (r = 10-15 m)
)(
9 × 10 1.6 × 10
kq
V=
=
−15
r
10
9
−19
) = 1.44 ×106 = 1.44 MV
PHY2049: Chapter 24
12
Energy Units: Electron Volts
Î1
eV = energy of charge e accelerated through 1 Volt
(
)
1eV = 1.6 × 10−19 C i1V
= 1.6 × 10−19 J
Î Let
q = 4e and V = 2000 V
K = 4 × 2000 = 8000eV = 8keV
(
)
K = 8000 × 1.6 × 10−19 = 1.28 × 10−15 J
PHY2049: Chapter 24
13
ConcepTest: Electric Energy
ÎA
proton and an electron are each accelerated across a
region of constant E field. Which has larger acceleration?
‹ (a)
proton
‹ (b) electron
‹ (c) both have equal acceleration
‹ (d) neither one accelerates
F = Ee
a = F/m = Ee/m
me mp Electron is much lighter than proton
PHY2049: Chapter 24
14
ConcepTest: Electric Energy
ÎWhich
has the biggest increase in KE?
‹ (a)
proton
‹ (b) electron
‹ (c) both have the same increase in KE
‹ (d) KE = 0 for both
K = Fd = Eed
Ve > Vp
PHY2049: Chapter 24
15
Equipotential Surfaces
ÎEquipotentials:
‹ No
ÎE
Contours of constant potential
work to move charge along contour: W = -qΔV = 0
⊥ equipotential surface
E⎥⎥ ≠ 0, would need work to move charge along surface
‹ See http://www.falstad.com/emstatic/
‹ If
PHY2049: Chapter 24
16
Equipotential: Constant E Field
Constant E
Example: Capacitor
PHY2049: Chapter 24
17
Equipotential: Point Charge
Equipotentials
PHY2049: Chapter 24
18
Equipotential: Dipole
PHY2049: Chapter 24
19
Topographic Map:
Equal Altitude Contours
Contour: Line of constant gravitational potential
PHY2049: Chapter 24
20
Calculating E From Electric Potential V
ÎElectric
field in terms of potential
dU = − F ⋅ ds = − Fx dx − Fy dy − Fz dz
Divide by q
dV = − E ⋅ ds = − Ex dx − E y dy − Ez dz
∂V
Ex = −
∂x
∂V
Ey = −
∂y
∂V
Ez = −
∂z
PHY2049: Chapter 24
21
Example: Electric Field of Point Charge
ÎGet
(
kq
V=
r
E by differentiating potential
r= x +y +z
2
2
2
)
1/ 2
kq
∂
Ex = −
∂x x 2 + y 2 + z 2
(
)
1/ 2
=
kqx
(
x +y +z
2
2
2
)
3/ 2
=
kqx
r
3
E y , Ez etc.
kq ⎛ x y z ⎞ kq
E = 2 ⎜ , , ⎟ ≡ 2 rˆ
r ⎝r r r⎠ r
Coulomb’s law
PHY2049: Chapter 24
22