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Electric Potential. A. GPE = mgΔh. GPE = mgh A – mgh B. F = mg. GPE = Work (W) required to raise or lower the book. - Where W = (F gravity )( Δh). B. h A. h B. Gravitational Potential Energy. + + + + + + + + +. ΔEPE = q o EΔd.

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Gravitational potential energy

A

GPE = mgΔh

GPE = mghA – mghB

F = mg

  • GPE = Work (W) required to raise or lower the book.

    • -Where W = (Fgravity)(Δh)

B

hA

hB

Gravitational Potential Energy


Electric potential energy

+ + + + + + + + +

ΔEPE = qoEΔd

ΔEPE = qoEdA – qoEdB

dA

-WE(AB) = qoEdA – qoEdB

A

B

+

+

-WE(AB) = FedA – FedB

dB

Fe = qoE

Fe = qoE

- - - - - - - - - -

Electric Potential Energy

  • Does a proton at rest at point A have more or less potential energy than it would at point B?

More


Electric potential energy of point charges and work

F

F

+q

-qo

F

F

+q

+qo

Electric Potential Energy of Point Charges and Work

  • Much like the book is attracted to the earth due to gravity, two unlike charges are attracted to one another.

  • Conversely, like charges repel.

  • It takes positive work to move unlike charges away from one another and like charges closer together.


Electric potential energy1

-qo

+q

Electric Potential Energy

  • What would happen if the charged particle q was fixed in place and then particle qo was suddenly released from rest?

  • It would accelerate away from q.

  • It would accelerate towards q.

  • It would stay where it is.

  • How would the potential energy of this

    system change?

  • It would increase.

  • It would decrease.

  • It would remain the same.


Electric potential1
Electric Potential

SI Units: joule/coulomb = 1 volt (V)

  • The Electric Potential Difference is equal to the Work required to move a test charge from infinity to a point in an electric field divided by the magnitude of the test charge.

  • The Electric Potential is the energy per unit of charge (J/C).


Example 1 electric potential
Example 1: Electric Potential

  • An object with 2.5C of charge requires 1.00x10-3 Joules of energy to move it through an electric field. What is the potential difference through which the charge is moved?


Characteristics of a capacitor

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qo

B

Uniform Electric Field

Two equal and oppositely charged plates

qo

C

qo

A

Characteristics of a Capacitor

E

  • Since the electric field is constant, the force acting on a charged particle will be the same everywhere between the plates.

  • Fe = qoE

FA = FB = FC


Electric potential and work in a capacitor

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B

F = qoE

dB

dA

qo

Electric Potential and Work in a Capacitor

D

WAB = F·dB - F·dA

A

qo

WAB = qoEd

F = qoE

(Ue) -WAB

qo qo

V =

=

qo

C

If WAB = qoEd, then what is WCD?

  • WCD = 0 Joules because the force acts perpendicular to the direction of motion.

    • Do you remember that W = F·d·cos?


Electric potential of a capacitor an alternative

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Electric Potential of a Capacitor – An alternative

  • From mechanics, W = Fd.

  • From the previous slide, W = qoEd

  • From the reference table, V = W/qo

Two equal and oppositely charged plates

A

B

qo

F = qoE

Uniform

Electric

Field

V = WAB/qo = Fd/qo = qoEd/qo= Ed


Example 2 parallel plates

d

Example 2:Parallel Plates

A spark plug in an automobile engine consists of two metal conductors that are separated by a distance of 0.50 mm. When an electric spark jumps between them, the magnitude of the electric field is 4.8 x 107 V/m. What is the magnitude of the potential difference V between the conductors?

V = Ed

V = (4.8 x 107 V/m)(5.0 x 10-4m)

V = 24,000V


Example 3 parallel plates
Example 3: Parallel Plates

A proton and an electron are released from rest from a similarly charged plate of a capacitor. The electric potential is 100,000 V and the distance between the two plates is 0.10 mm.

  • Which charge will have greater kinetic energy at the moment it reaches the opposite plate?

  • Determine the amount of work done on each particle.

  • Determine the speed of each particle at the moment it reaches the opposite plate.

  • Determine the magnitude of the force acting on each particle.

  • Determine the magnitude of the acceleration of each particle.


Example 3 parallel plates cont

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Example 3: Parallel Plates(cont.)

  • Begin by drawing a picture and listing what is known:

    • V = 100,000V

    • d = 0.10 mm = 1.0 x 10-4m

    • qe = qp = 1.6 x 10-19C (ignore the sign. We are only interested in magnitude.)

p+

e-


Example 3 parallel plates 1 2
Example 3: Parallel Plates(#1 & #2)

  • For #1, you could answer #2 first to verify.

    • The answer is that the kinetic energy of both particles will be the same

      • Why?

      • because of the formula needed in question #2 applies to both charges, and work = energy.

      • Hence: Wproton = Welectron

        qprotonV = qelectronV

        Wproton = Welectron = (1.6x10-19C)(100,000V)

        Wproton = Welectron = 1.6x10-14 J


Example 3 parallel plates 3
Example 3: Parallel Plates(#3)

  • Apply the work-energy theorem to determine the final speed of the electron and proton.

    W = KE

  • Since the initial kinetic energy is equal to 0J:

    W = KEf

    W = ½ mvf2

  • Proton:

  • Electron:


Example 3 parallel plates 4
Example 3: Parallel Plates(#4)

  • Since F = qE, it will be the same for both particles because their charges are the same and the electric field is uniform between two parallel plates.

  • We also know that W = Fd. Since we know the distance between the plates and the work done to move either charge from one plate to another, we can determine the force as follows:


Example 3 parallel plates 5
Example 3: Parallel Plates(#5)

  • Since we have the force acting on each particle, we can now calculate the acceleration of each particle using Newton’s 2nd Law.


Equipotential lines
Equipotential Lines

  • Equipotential lines denote where the electric potential is the same in an electric field.

  • The potential is the same anywhere on an equipotential surface a distance r from a point charge, or d from a plate.

  • No work is done to move a charge along an equipotential surface. Hence VB = VA (The electric potential difference does not depend on the path taken from A to B).

  • Electric field lines and equipotential lines cross at right angles and point in the direction of decreasing potential.


Equipotential lines1

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Lines of Equipotential

Note: Electric field lines and lines of equipotential intersect at right angles.

Equipotential Lines

  • Parallel Plate Capacitor

Electric Field Lines

Decreasing Electric Potential / Voltage


Equipotential lines2

Note: Electric field lines and lines of equipotential intersect at right angles.

Lines of Equipotential

+

Equipotential Lines

  • Point Charge

Electric Field Lines

Note: A charged surface is also an equipotential surface!

Decreasing Electric Potential / Voltage


Equipotential lines examples
Equipotential Lines (Examples) intersect at right angles.

  • http://www.cco.caltech.edu/~phys1/java/phys1/EField/EField.html


Key ideas
Key Ideas intersect at right angles.

  • Electric potential energy (U) is the work required to bring a positive unit charge from infinity to a point in an electric field.

  • Electric potential (V) is the change in energy per unit charge as the charge is brought from one point to another.

  • The electric field between two charged plates is constant meaning that the force is constant between them as well.

  • The electric potential between two points is not dependent on the path taken to get there.

  • Electric field lines and lines of equipotential intersect at right angles.


Electric potential energy and work in a uniform electric field

+ intersect at right angles.

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B

F = qoE

dB

dA

qo

Electric Potential Energy and Work in a Uniform Electric Field

A

qo

F = qoE

Note: The force acting on the charge is constant as it moves from one plate to another because the electric field is uniform.

WAB = EPEB – EPEA

WAB = FdB – FdA

WAB = qoEdB – qoEdA

WAB = qoE(dB – dA) = qoEd


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