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PHY2049: Chapter 24 1
What You Already KnowCoulomb’s law
Electric fields
Gauss’ law
Electric fields for several configurationsPointLinePlane (nonconducting)Sheet (conducting)Ring (along axis)Disk (along axis)SphereCylinderDipole (along || and ⊥ axes)
PHY2049: Chapter 24 2
Chapter 24: Electric Potential
Electric Potential Energy
Electric Potential
Equipotential Surfaces
Potential of Point Charge
Potential of Charge Distribution
Calculating the Field from the Potential
Potential Energy from a System of Charges
Potential of Isolated Charged Conductors
PHY2049: Chapter 24 3
Reading Quiz: Chapter 24An equipotential surface is:
a) a surface where the electric field is constant b) always parallel to the electric fieldc) a surface where the potential is zerod) always perpendicular to the electric fielde) a surface where the electric field is zero
PHY2049: Chapter 24 4
Reading Quiz: Chapter 24The volt is a unit of:
a) potential energyb) electric fieldc) potentiald) force
PHY2049: Chapter 24 5
Reading Quiz: Chapter 24Electric potential is:
a) a scalar quantityb) a vector quantityc) can be either scalar or vector
PHY2049: Chapter 24 6
Electric Work and Potential EnergyPoint charges q1, q2: Work moving charge q2 from A → B
Path independence: “conservative force”Define potential energy of two point charge
( )1 22
1 2 1 2 1 22
ˆ ˆB B
AB A A
BAB A A B
kq qW F ds r ds r ds drr
kq q kq q kq qW drr rr
= ⋅ = ⋅ ⋅ =
= = −
∫ ∫
∫• Depends only on endpoints• Path independent• Like gravitation
1 2B A AB
kq qU U W Ur
− ≡ − ⇒ =
PHY2049: Chapter 24 7
Electric Force is ConservativeHolds in all electrostatic situations (not just point charge)
Proof: integrate over any charge distribution
Work done by electric field moving charge q from i to fCalculate from difference of potential energies
Chargesi f
elec fi i fW U U U= −Δ = −Work:
q
PHY2049: Chapter 24 8
Problem: Electric Potential Energy Two 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.
i i f fK U K U+ = +
( )2 2
2120 2 0f f
i i
kq kqmv vd md
+ = + ⇒ =
( )( )
( )( )
295
3
9 10 0.0121.6 10 m/s
10 0.05fv
−
×= = ×
PHY2049: Chapter 24 9
Gravitational & Electric Potential Energy Gravity
G A BhW mg U U= = − E A BdW qE U U= = −
Electric
d
+++++++
-------
A
B A B
Point B at lower potential energy than point A (q>0)
h
PHY2049: Chapter 24 10
Electric Potential
Potential = PE per unit charge
Potential difference: general E field
Potential difference: constant E
Potential higher at + charges and “falls” to lower value at − charges
+q: Moves from higher to lower V−q: Moves from lower to higher V
/V U q=
a bV V Ed− =
bb a a
V V E ds− = − ⋅∫
b
a +++++++++++
-------------------
+QEd
PHY2049: Chapter 24 11
Units for V and EUnits of potential: “volt”
V = U/q ⎯→ Volt = Joule / Coulomb
Units of electric fieldF = Eq ⎯→ E = F/q → Newton / CoulombV = Ed ⎯→ E = V/d → Volt / Meter
PHY2049: Chapter 24 12
Example of Potential of Point ChargePoint charge q (using V = 0 at r = ∞)
Example: Potential at surface of proton (r = 10-15 m)
( )( )9 196
15
9 10 1.6 101.44 10 1.44 MV
10kqVr
−
−
× ×= = = × =
kqVr
=
PHY2049: Chapter 24 13
Energy Units: Electron Volts
1 eV = energy of charge e accelerated through 1 Volt
Let q = 4e and V = 2000 V
( )19
19
1eV 1.6 10 C 1V
1.6 10 J
−
−
= ×
= ×
i
( )19 15
4 2000 8000eV 8keV
8000 1.6 10 1.28 10 J
K
K − −
= × = =
= × × = ×
PHY2049: Chapter 24 14
ConcepTest: Electric Energy A 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 = Eea = F/m = Ee/mme mp Electron is much lighter than proton
PHY2049: Chapter 24 15
ConcepTest: Electric EnergyWhich 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 = EedVe > Vp
PHY2049: Chapter 24 16
Equipotential SurfacesEquipotentials: Contours of constant potential
No work to move charge along contour: W = -qΔV = 0
E ⊥ equipotential surfaceIf E⎥⎥ ≠ 0, would need work to move charge along surfaceSee http://www.falstad.com/emstatic/
PHY2049: Chapter 24 17
Equipotential: Constant E Field
Example: CapacitorConstant E
PHY2049: Chapter 24 18
Equipotential: Point Charge
Equipotentials
PHY2049: Chapter 24 19
Equipotential: Dipole
PHY2049: Chapter 24 20
Topographic Map:Equal Altitude Contours
Contour: Line of constant gravitational potential
PHY2049: Chapter 24 21
Calculating E From Electric Potential VElectric field in terms of potential
x y z
x y z
x y z
dU F ds F dx F dy F dz
dV E ds E dx E dy E dz
V V VE E Ex y z
= − ⋅ = − − −
= − ⋅ = − − −
∂ ∂ ∂= − = − = −
∂ ∂ ∂
Divide by q
PHY2049: Chapter 24 22
Example: Electric Field of Point ChargeGet E by differentiating potential
2 2ˆ, ,kq x y z kqE r
r r rr r⎛ ⎞= ≡⎜ ⎟⎝ ⎠
( ) ( )1/ 2 3/ 2 32 2 2 2 2 2
, etc.
x
y z
kq kqx kqxEx rx y z x y z
E E
∂= − = =
∂ + + + +
( )1/ 22 2 2r x y z= + +
Coulomb’s law
kqVr
=