physics 201 - galileogalileo.phys.virginia.edu/~pqh/202_3n.pdf · electric potential energy an...

Physics 201Professor P. Q. Hung

311B, Physics Building

Physics 201 – p. 1/19

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

~F = q0~E(r)

Physics 201 – p. 2/19

~F = q0~E(r)

Physics 201 – p. 2/19

Electric Potential Energy

Recall that the gravitational force is

Fg = −Gm1 m2

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

Recall that the gravitational force is

Fg = −Gm1 m2

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 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 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 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

e − UBe

Physics 201 – p. 5/19

e − UBe

Physics 201 – p. 5/19

Physics 201 – p. 6/19

Electric Potential

Electric Potential:

V = Ue

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= −

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

Electric Potential:

V = Ue

VB−VA = UBe

q= −

q= −E(hA−hB) < 0

Physics 201 – p. 7/19

Electric Potential

Electric Potential:

V = Ue

VB−VA = UBe

q= −

q= −E(hA−hB) < 0

Physics 201 – p. 7/19

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 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

Physics 201 – p. 9/19

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 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 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

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

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

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

Conservation of Energy

(a) 12mv2

A + qVA = 12mv2

B + qVB.⇒

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

(a) 12mv2

A + qVA = 12mv2

B + qVB.⇒

A = q(VB − VA)

⇒ vA =√

2q(VB−VA)m

= 9.1 m/s.

Physics 201 – p. 14/19

(a) 12mv2

A + qVA = 12mv2

B + qVB.⇒

A = q(VB − VA)

⇒ vA =√

2q(VB−VA)m

= 9.1 m/s.

Physics 201 – p. 14/19

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

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 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

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 of a point charge

~E(r) = k qr2 r̂ (NOT CONSTANT) ⇒

VB − VA = −

A~E.d~s = k q

− k qrA

Boundary condition: VA = 0 when rA isinfinite:

V = k qr

Physics 201 – p. 17/19

VB − VA = −

A~E.d~s = k q

− k qrA

V = k qr

Physics 201 – p. 17/19

VB − VA = −

A~E.d~s = k q

− k qrA

V = k qr

Physics 201 – p. 17/19

Physics 201 – p. 18/19

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

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

VP = k(q1

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

VP = k(q1

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

physics 201 - galileogalileo.phys.virginia.edu/~pqh/202_3n.pdf · electric potential energy an...

Documents