physics 201 - galileogalileo.phys.virginia.edu/~pqh/202_3n.pdf · electric potential energy an...
TRANSCRIPT
Electric Potential and Energy
Summary of last lecture
Force on a point charge q0 in the presence ofan electric field:
~F = q0~E(r)
~E(r) can be calculated using eitherCoulomb’s law or Gauss’s law.
From last semester, a Force does Work.Furthermore, Conservative Force (like e.g.1/r2 force) ⇒ Concept of Potential Energyand Concept of Potential.
Physics 201 – p. 2/19
Electric Potential and Energy
Summary of last lecture
Force on a point charge q0 in the presence ofan electric field:
~F = q0~E(r)
~E(r) can be calculated using eitherCoulomb’s law or Gauss’s law.
From last semester, a Force does Work.Furthermore, Conservative Force (like e.g.1/r2 force) ⇒ Concept of Potential Energyand Concept of Potential.
Physics 201 – p. 2/19
Electric Potential and Energy
Summary of last lecture
Force on a point charge q0 in the presence ofan electric field:
~F = q0~E(r)
~E(r) can be calculated using eitherCoulomb’s law or Gauss’s law.
From last semester, a Force does Work.Furthermore, Conservative Force (like e.g.1/r2 force) ⇒ Concept of Potential Energyand Concept of Potential.
Physics 201 – p. 2/19
Electric Potential and Energy
Electric Potential Energy
Recall that the gravitational force is
Fg = −Gm1 m2
r2
and it is a conservative force ⇒ A potentialenergy is associated with it.
The work done by gravity: from point A topoint B is (positive direction pointing upward):
WAB = KEB − KEA = UAg − UB
g =mghA − mghB
Physics 201 – p. 3/19
Electric Potential and Energy
Electric Potential Energy
Recall that the gravitational force is
Fg = −Gm1 m2
r2
and it is a conservative force ⇒ A potentialenergy is associated with it.
The work done by gravity: from point A topoint B is (positive direction pointing upward):
WAB = KEB − KEA = UAg − UB
g =mghA − mghB
Physics 201 – p. 3/19
Electric Potential and Energy
Electric Potential EnergyAn electric force does work WAB = +5.0 × 10−5 Jin moving a charge q0 = +2.0µC from A to B. 1)Find the potential energy difference; 2) Find thepotential difference between B and A.
How does one goes from the work done byan electric force to the potential energydifference?
What is the potential difference?
Physics 201 – p. 4/19
Electric Potential and Energy
Electric Potential EnergyAn electric force does work WAB = +5.0 × 10−5 Jin moving a charge q0 = +2.0µC from A to B. 1)Find the potential energy difference; 2) Find thepotential difference between B and A.
How does one goes from the work done byan electric force to the potential energydifference?
What is the potential difference?
Physics 201 – p. 4/19
Electric Potential and Energy
Electric Potential Energy Ue
Force on +q: ~F = q ~E . Assume a uniformelectric field ~E pointing vertically downward.
Work done by this force in moving the particlefrom A to B with hA > hB:
WAB = q E hA − q E hB = KEB − KEA
Conservative electric force ⇒
WAB = q E hA − q E hB = KEB − KEA =UA
e − UBe
Physics 201 – p. 5/19
Electric Potential and Energy
Electric Potential Energy Ue
Force on +q: ~F = q ~E . Assume a uniformelectric field ~E pointing vertically downward.
Work done by this force in moving the particlefrom A to B with hA > hB:
WAB = q E hA − q E hB = KEB − KEA
Conservative electric force ⇒
WAB = q E hA − q E hB = KEB − KEA =UA
e − UBe
Physics 201 – p. 5/19
Electric Potential and Energy
Electric Potential Energy Ue
Force on +q: ~F = q ~E . Assume a uniformelectric field ~E pointing vertically downward.
Work done by this force in moving the particlefrom A to B with hA > hB:
WAB = q E hA − q E hB = KEB − KEA
Conservative electric force ⇒
WAB = q E hA − q E hB = KEB − KEA =UA
e − UBe
Physics 201 – p. 5/19
Electric Potential and Energy
Electric Potential
Electric Potential:
V = Ue
q
Unit: V = joule/coulomb=Volt
Notice that V is not a vector. It is a scalar
potential difference between B and A is:
VB−VA = UBe
q−
UAe
q= −
WAB
q= −E(hA−hB) < 0
Only valid for a uniform electric field!
hA > hB ⇒ VB < VA ⇒ The electric fieldpoints in te direction of decreasing potential.
Physics 201 – p. 7/19
Electric Potential and Energy
Electric Potential
Electric Potential:
V = Ue
q
Unit: V = joule/coulomb=Volt
Notice that V is not a vector. It is a scalar
potential difference between B and A is:
VB−VA = UBe
q−
UAe
q= −
WAB
q= −E(hA−hB) < 0
Only valid for a uniform electric field!
hA > hB ⇒ VB < VA ⇒ The electric fieldpoints in te direction of decreasing potential.
Physics 201 – p. 7/19
Electric Potential and Energy
Electric Potential
Electric Potential:
V = Ue
q
Unit: V = joule/coulomb=Volt
Notice that V is not a vector. It is a scalar
potential difference between B and A is:
VB−VA = UBe
q−
UAe
q= −
WAB
q= −E(hA−hB) < 0
Only valid for a uniform electric field!
hA > hB ⇒ VB < VA ⇒ The electric fieldpoints in te direction of decreasing potential.
Physics 201 – p. 7/19
Electric Potential and Energy
Electric PotentialMore generally:
VB − VA = − ~E.~s
where ~s is the displacement vector pointing fromA to B.
The positive charge which moves in thedirection of the electric field, goes from highto low potential. How about negative charge?
IMPORTANT POINT: As with the gravitationalcase, only the potential energy and potentialdifference makes physical sense.
Physics 201 – p. 8/19
Electric Potential and Energy
Electric PotentialMore generally:
VB − VA = − ~E.~s
where ~s is the displacement vector pointing fromA to B.
The positive charge which moves in thedirection of the electric field, goes from highto low potential. How about negative charge?
IMPORTANT POINT: As with the gravitationalcase, only the potential energy and potentialdifference makes physical sense.
Physics 201 – p. 8/19
Electric Potential and Energy
Electric PotentialAn electric force does work WAB = +5.0 × 10−5 Jin moving a charge q0 = +2.0µC from A to B. 1)Find the potential energy difference; 2) Find thepotential difference between B and A.
UBe − UA
e = −WAB = −5.0 × 10−5 J < 0. Thepotential energy is higher at A than at B.
V B− V A = −WAB/q = −
5.0×10−5 J+2.0µC
= −25 V .So VB < VA.
Physics 201 – p. 10/19
Electric Potential and Energy
Electric PotentialAn electric force does work WAB = +5.0 × 10−5 Jin moving a charge q0 = +2.0µC from A to B. 1)Find the potential energy difference; 2) Find thepotential difference between B and A.
UBe − UA
e = −WAB = −5.0 × 10−5 J < 0. Thepotential energy is higher at A than at B.
V B− V A = −WAB/q = −
5.0×10−5 J+2.0µC
= −25 V .So VB < VA.
Physics 201 – p. 10/19
Electric Potential and Energy
Electric PotentialA 12-V battery powers a 60.0 W headlight for onehour. How many electrons have passed throughthe terminals of the battery during that time?
Energy consumed by the headlight in onehour:Energy = Power x time⇒ E = 60.0W × 3600s = 2.2 × 105 J .
Physics 201 – p. 11/19
Electric Potential and Energy
Electric Potential
Equal to the change in potential energy of thetotal number of electrons which pass throughthe terminals:E = Q∆V = Q12V = 2.2 × 105 J⇒ Q = 1.8 × 104 C.
Each electron carries a charge of magnitude1.6 × 10−19 C. The total number of electrons isN = 1.8×104 C
1.6×10−19 C= 1.1 × 1023.
Physics 201 – p. 12/19
Electric Potential and Energy
Electric Potential
Equal to the change in potential energy of thetotal number of electrons which pass throughthe terminals:E = Q∆V = Q12V = 2.2 × 105 J⇒ Q = 1.8 × 104 C.
Each electron carries a charge of magnitude1.6 × 10−19 C. The total number of electrons isN = 1.8×104 C
1.6×10−19 C= 1.1 × 1023.
Physics 201 – p. 12/19
Electric Potential and Energy
Conservation of EnergyPoint B has an electric potential that is 25 Vgreater than that of point A. A particle of mass1.8 × 10−5 kg and a charge whose magnitude is3.0 × 10−5 C. We will neglect gravity and friction.(a) If the particle has a positive charge and isreleased from rest at B, what speed vA does theparticle have when it arrives at A? (b) If theparticle has a negative charge and is releasedfrom rest at A, what speed vB does the particlehave when it arrives at B? (c) What if thenegatively charged particle is released from restfrom B?Key point: Energy is conserved. ThereforeKEA + UA
e = KEB + UBe .
Physics 201 – p. 13/19
Electric Potential and Energy
Conservation of Energy
(a) 12mv2
A + qVA = 12mv2
B + qVB.⇒
12mv2
A = q(VB − VA)
⇒ vA =√
2q(VB−VA)m
= 9.1 m/s.
(b) A negatively charged particle acceleratingfrom rest from A is the same as a positivelycharged particle accelerating from rest from B⇒ vB = 9.1m/s.
(c) It will never reach B.
Physics 201 – p. 14/19
Electric Potential and Energy
Conservation of Energy
(a) 12mv2
A + qVA = 12mv2
B + qVB.⇒
12mv2
A = q(VB − VA)
⇒ vA =√
2q(VB−VA)m
= 9.1 m/s.
(b) A negatively charged particle acceleratingfrom rest from A is the same as a positivelycharged particle accelerating from rest from B⇒ vB = 9.1m/s.
(c) It will never reach B.
Physics 201 – p. 14/19
Electric Potential and Energy
Conservation of Energy
(a) 12mv2
A + qVA = 12mv2
B + qVB.⇒
12mv2
A = q(VB − VA)
⇒ vA =√
2q(VB−VA)m
= 9.1 m/s.
(b) A negatively charged particle acceleratingfrom rest from A is the same as a positivelycharged particle accelerating from rest from B⇒ vB = 9.1m/s.
(c) It will never reach B.
Physics 201 – p. 14/19
Electric Potential and Energy
Conservation of Energy
Another unit of energy: eV or electronvolt.One eV is the change in potential energy ofan electron moving a potential difference ofone volt.
1 eV = 1.6 × 10−19 J .
Physics 201 – p. 15/19
Electric Potential and Energy
Conservation of Energy
Another unit of energy: eV or electronvolt.One eV is the change in potential energy ofan electron moving a potential difference ofone volt.
1 eV = 1.6 × 10−19 J .
Physics 201 – p. 15/19
Electric Potential and Energy
Electric Potential of a point chargeA charge q1 = 2.00µC is located at the origin anda charge q2 = −6.00µC is located at (0, 3.00 m).Find the total electric potential at point P situatedat (4.00m, 0). See the figure drawn in class.
The electric field of a point charge is notconstant. We cannot use the formula derivedabove.
So what’s the electric potential of a pointcharge then?
Physics 201 – p. 16/19
Electric Potential and Energy
Electric Potential of a point chargeA charge q1 = 2.00µC is located at the origin anda charge q2 = −6.00µC is located at (0, 3.00 m).Find the total electric potential at point P situatedat (4.00m, 0). See the figure drawn in class.
The electric field of a point charge is notconstant. We cannot use the formula derivedabove.
So what’s the electric potential of a pointcharge then?
Physics 201 – p. 16/19
Electric Potential and Energy
Electric Potential of a point charge
~E(r) = k qr2 r̂ (NOT CONSTANT) ⇒
VB − VA = −
∫ B
A~E.d~s = k q
rB
− k qrA
Boundary condition: VA = 0 when rA isinfinite:
⇒
V = k qr
Physics 201 – p. 17/19
Electric Potential and Energy
Electric Potential of a point charge
~E(r) = k qr2 r̂ (NOT CONSTANT) ⇒
VB − VA = −
∫ B
A~E.d~s = k q
rB
− k qrA
Boundary condition: VA = 0 when rA isinfinite:
⇒
V = k qr
Physics 201 – p. 17/19
Electric Potential and Energy
Electric Potential of a point charge
~E(r) = k qr2 r̂ (NOT CONSTANT) ⇒
VB − VA = −
∫ B
A~E.d~s = k q
rB
− k qrA
Boundary condition: VA = 0 when rA isinfinite:
⇒
V = k qr
Physics 201 – p. 17/19
Electric Potential and Energy
Electric Potential of a point chargeA charge q1 = 2.00µC is located at the origin anda charge q2 = −6.00µC is located at (0, 3.00 m).Find the total electric potential at point P situatedat (4.00m, 0). See the figure drawn in class.
The individual potentials add like scalars
The distance between q1 and P is 4.00 m. Thedistance between q2 and P is 5.00 m.
VP = k(q1
r1
+ q2
r2
)⇒
VP = 8.99 × 109 N.m2
C(2.00×10−6C
4.00m + −6.00×10−6C5.00m ) =
−6.29 × 103V .
Physics 201 – p. 19/19
Electric Potential and Energy
Electric Potential of a point chargeA charge q1 = 2.00µC is located at the origin anda charge q2 = −6.00µC is located at (0, 3.00 m).Find the total electric potential at point P situatedat (4.00m, 0). See the figure drawn in class.
The individual potentials add like scalars
The distance between q1 and P is 4.00 m. Thedistance between q2 and P is 5.00 m.
VP = k(q1
r1
+ q2
r2
)⇒
VP = 8.99 × 109 N.m2
C(2.00×10−6C
4.00m + −6.00×10−6C5.00m ) =
−6.29 × 103V .
Physics 201 – p. 19/19
Electric Potential and Energy
Electric Potential of a point chargeA charge q1 = 2.00µC is located at the origin anda charge q2 = −6.00µC is located at (0, 3.00 m).Find the total electric potential at point P situatedat (4.00m, 0). See the figure drawn in class.
The individual potentials add like scalars
The distance between q1 and P is 4.00 m. Thedistance between q2 and P is 5.00 m.
VP = k(q1
r1
+ q2
r2
)⇒
VP = 8.99 × 109 N.m2
C(2.00×10−6C
4.00m + −6.00×10−6C5.00m ) =
−6.29 × 103V . Physics 201 – p. 19/19