Physics1302.300Spring2017
Class13—February14,2017
MainTopics:
•Electricpotentialenergy
•Electricpotential
NoclassFriday.
Review:PotentialandPotentialEnergy
• Potentialisascalarassociatedwithapointinspace.Itisthe
energyrequired(+forrepulsion;negativeforattraction)to
bringa1.00Ctestchargefrominfinitytoapointx.
• Potentialenergyisascalarassociatedwithacharge.Itisthe
energyrequiredtobringthatchargefrominfinitytoits
currentposition.
• Fieldtopotential(orforcetopotentialenergy)requires
integration.
• Potentialtofield(orpotentialenergytoforce)requires
differentiation.
Physics1302.300Spring2017
2
Problem
• A proton is released from rest at the surface of a 1.0 cm
diameter sphere that has been charged to 1000 V. What is
the proton’s speed when it is 1.0 cm from the sphere’s
surface?
–
–
–
–
–
(a) 2.11 x 105 m/s
(b) 2.79 x 105 m/s
(c) 3.13 x 105 m/s
(d) 3.57 x 105 m/s
(e) 4.01 x 105 m/s
Physics1302.300Spring2017
Problem
• A proton is released from rest at the surface of a 1.0 cm
diameter sphere that has been charged to 1000 V. What is
the proton’s speed when it is 1.0 cm from the sphere’s
surface?
Proton moves from r=0.5 cm to r=1.5 cm
1 ⎞
⎛ 1
ΔV = kQ ⎜
−
⎝ 0.5 1.5 ⎟⎠
ΔK = ΔU = eΔV
2K
v=
= 3.57 x 10 5 m / s
m
Physics1302.300Spring2017
Problem
• A proton is released from rest at the surface of a 1.0 cm
diameter sphere that has been charged to 1000 V. What is
the proton’s speed when it is 1.0 cm from the sphere’s
surface?
Proton moves from r=0.5 cm to r=1.5 cm
1 ⎞
⎛ 1
ΔV = kQ ⎜
−
⎝ 0.5 1.5 ⎟⎠
ΔK = ΔU = eΔV
2K
v=
= 3.57 x 10 5 m / s
m
Physics1302.300Spring2017
Problem
• A proton is released from rest at the surface of a 1.0 cm
diameter sphere that has been charged to 1000 V. What is
the proton’s speed when it is 1.0 cm from the sphere’s
surface?
Proton moves from r=0.5 cm to r=1.5 cm
1 ⎞
⎛ 1
ΔV = kQ ⎜
−
⎝ 0.5 1.5 ⎟⎠
ΔK = ΔU = eΔV
2K
v=
= 3.57 x 10 5 m / s
m
Physics1302.300Spring2017
Potential of a Ring of Charge
V=k
Q
R2 + z 2
Potential of a Disk of Charge
V=
∫
R
k
1
2π rσ dr =
r +z
2kQ R rdr
2kQ
V= 2 ∫
= 2
2
2
0
R
R
r +z
0
2
2
∫
R
0
(
1
Q
k
2π r
dr
2
2
2
πR
r +z
R2 + z 2 − z
)
PotentialofaLineofCharge
Potential for a Parallel Plate Capacitor
• V=Ed=4πkds = 4πkdQ/A
• E=V/d
Physics1302.300Spring2017
Equipotentials
• Equipotentials arecurvesthatconnectpointsatthesame
potential.
• Equipotentials arealwaysperpendiculartofieldlines.
• Contourlinesontopographicmapsareexamplesof
gravitationalequipotentials.
Physics1302.300Spring2017
11
Equipotentials
• Equipotentials arecurvesthatconnectpointsatthesame
potential.
• Equipotentials arealwaysperpendiculartofieldlines.
Physics1302.300Spring2017
12
GravitationalPotential
• Contourlinesontopographicmapsaregravitational
equipotentials.
• Waterflowsareperpendiculartothecontourlines,inthe
directionofthenegativegradient
Physics1302.300Spring2017
13
Question
• Potentialistheintegraloffield.Positivecharges
movetowardsnegativepotential.Whichpotential
energygraphdescribesthisfield?
–
–
–
–
–
(a)
(b)
(c)
(d)
(e)
Physics1302.300Spring2017
14
Question
• Whichsetofequipotentialsurfacesmatchesthisfield?
–
–
–
–
–
(a)
(b)
(c)
(d)
(e)
Physics1302.300Spring2017
15
PotentialEnergyandPotential
• Potentialenergyisassignedto
anobject;potentialisassigned
toapointinspace
• Aforceisassignedtoan
object;afieldisassignedtoa
pointinspace
Physics1302.300Spring2017
h
Question
• Threemetalspheresareconnectedbyathinwire.Whichof
thefollowingstatementsiscorrect?
–
–
–
–
–
(a)
(b)
(c)
(d)
(e)
1.
2.
3.
4.
5.
Physics1302.300Spring2017
V1 = V2 = V3 and E1 = E2 = E3
V1 = V2 = V3 and E1 > E2 > E3
V1 > V2 > V3 and E1 = E2 = E3
V1 > V2 > V3 and E1 > E2 > E3
V3 > V2 > V1 and E1 = E2 = E3
17