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Electric Potential Energy and Electric Potential. Chapter 16. Potential Energy. Potential Energy is stored energy do to an object’s position in a force field. Work = F · d = Δ PE = Δ U When work is done to move an object into a force field, Potential Energy Increases. - PowerPoint PPT Presentation

### Transcript of Electric Potential Energy and Electric Potential

• Electric Potential Energy and Electric PotentialChapter 16

• Potential EnergyPotential Energy is stored energy do to an objects position in a force field.

Work = Fd = PE = U

When work is done to move an object into a force field, Potential Energy Increases

• Gravitational P.E. UgTo bring a mass in from infinity to near the earth, we would encounter the Earths gravitational field.Work = Fd = Ug

Ug = Uf Ui = Ur Uinf

Ug= (GMem/r2)r

Ug = -GMm/r

• Uniform FieldsNear Earths surface, gravitational field is uniform. g = 9.8 m/s2.

We calculate Work = U = mgh

This is a simplification of Ug = -GMm/r

Work is essentially Fd

• Electric Potential EnergyTo bring a charge, q, in from infinity to near +Q, we would encounter the charges electric field.Work = Fd = Ue

Ue = Uf Ui = Ur Uinf

Ue= (kQq/r2)r

Ue = kQq/r

• Uniform Electric FieldParallel Plates with equal charge provide a uniform electric field.

Recall E = Fe/q

Fe = Eq

• Potential Energy Uniform Field (plate separation d)U = Uf Ui = UB UA

U = Work = Fd

U = (Eq)d = Eqd

E is electric field between plates, d is plate separation.

• Uniform Gravitational and Electric Field ComparisonUg = mgh for constant g field

Ue = qEd for constant E field

U is measured in Joules

• Non-Uniform Fields / PEFg = GMm/r2 Ug = -GMm/r

Fe = kQq/r2 Ue = kQq/r

• Electric Potential, V Electric Potential is defined as potential energy per unit charge.

Think about adding charge, one charge at a time, to a conducting sphere.

• Electric Potential, VV = Ue/q measured in Joules/Coulomb

Joule/Coulomb = Volt!

Uniform Electric Field: V = qEd/q = Ed

V = Ed gives the voltage difference between two parallel plates

Non-Uniform E Field: V = (kQq/r)/q = kQ/r

• Summarize New FormulasFe = kq1q2/r2

Ue = F (distance)= kq1q2/r electric potential energy between two charges (non uniform fields)Ue = qEdelectric potential energy for parallel plates with electric field EV = Ue/q= kq/r electric potential due to charge q (non uniform field)V = Edelectric potential for parallel plates with electric field E

• Electric Potential Energy for Several ChargesTo find the potential energy of a system of charges, add the potential energy between each pair of charges.

Ue = U12 + U13 + U23

• ExamplesRead Examples 16.1, 16.2, 16.3, 16.4

Copy these into your notes if you feel that is helpful!

Try # 11 14, 17, 18, 20 22, 24, 25, 28 page 562 { prepare to turn this assignment in}

• Equi-potential Surfaces*Surfaces where potential energy is constant are known as equi-potential surfaces.For the electric field, we are concerned with electric potential, V, in addition to electric potential energy, U. Equi-potential surfaces are surfaces with the same potential energy and the same electric potential.

• Conservative FieldsA conservative force is a force in which work done does NOT depend on the path taken. Gravitational Force is a conservative force.Electric Force is a conservative forceFriction is non-conservative.In moving an object along an equi-potential surface, no work is done.

• Equi-potential Surfaces

• Uniform Electric FieldRecall E = F/q F = Eq

U = Fd = Eqd

V = U/q

For a uniform field, V = Ed = Ex

• UnitsE [Joule/Coulomb] or [Volt/meter]

These are equivalent since E = F/q = V/d

• Example:Normally the Earth is electrically charged. This creates a constant electric field pointing down near the Earths surface of 150 V/m.A) What are the shapes of the equipotential surfaces?B) How far apart are two equi-potential surfaces with 1000V difference between them?

• Electron Volt measure for EnergyAn Electron-Volt is a common unit for energy

It is the amount of Kinetic energy acquired by an electron if it is accelerated through a voltage of 1 Volt.

1eV = KE = -U = qV = 1.6 X 10 19 J

• CapacitanceA capacitor is a device that stores charge.A good capacitor has the ability to store charge without appreciably increasing the electric potential.Work is done by a battery to move charge from one parallel plate to another. Separation of charge creates an electric field.Capacitance is defined as the amount of stored charge per unit of potential difference.

• CapacitanceCapacitance = Charge stored / Voltage

C = Q/V or Q = CV

1 Farad is a huge amount of capacitance. It is most commonly measured in microfarads = 10-6 F

• CapacitanceCapacitance of a parallel plate arrangement depends on the area of and the distance between the plates, as well as the material between them. (dielectrics increase capacitance)If the material between the plates is air, then

C = 0A/d

Where 0 is the permittivity of free spaceAnd 0 = 8.85 X 10 12 C2/Nm2

• ExampleWhat would be the plate area of an air filled 1.0F parallel plate capacitor if the plate separation were 1.0 mm?

• Energy!Potential increases as charge is added.

Slope = V/ Q = 1/C

Potential energy stored in a capacitor = Work Done = QVav = Q(V/2)

Uc = QV = CV2 = Q2/2C

• ExampleDuring a heart attack the heart beats erratically. One way to get it back to normal is to shock it with electrical energy. About 300J of energy is required to produce the desired effect. A defibrillator stores this energy in a capacitor charged by 5000V. What capacitance is required? What is the charge on the plates?

• Homework!Read sections 16.4 and 16.5 pages 552 559.Do # 57, 59, 60, 61, 64, 65 67, 69- 72

• Dielectric MaterialsDielectric materials placed between parallel plates have several purposes: Keep plates from coming in contactAllow for flexible platesIncrease capacitance of the capacitorDielectric constant, >1 = C/C0

• Two dielectric situations:Either the voltage difference is removed once the plates are charged, then the dielectric material inserted between platesOr the dielectric material is inserted between the plates while the voltage is maintained.These are different situations, but both result in increased capacitance.

• Remove voltage then insert dielectric

Voltage applied, V0, separates charge, Q0 and sets up electric field E0

Dielectric inserted and becomes polarized. Electric field does work to polarize dielectric molecules, which set up a smaller electric field in the opposite direction.

Net electric field is reduced. Therefore voltage is reduced.

• Remove voltage then insert dielectricCharge, Q, doesnt change once dielectric is inserted.Induced electric field in dielectric reduces original electric field to E and original voltage difference to V.

= E0 / E = V0 /V

Then C = Q/V = Q0/(V0/ ) = (Q0/V0) = C0

Uc = Q2/2C = Q2/2(C0) = U0/

• Remove voltage then insert dielectricCapacitance increases by factor of Since C = Q/V and voltage decreases, this makes sense!Stored energy decreases by factor of Some stored energy goes into polarizing the molecules in the dielectric, so this makes sense!

• Insert dielectric and maintain constant voltageWhen capacitor is maintained at constant voltage, the battery continues to supply charge to compensate for the induced electric field. C = Q/V = Q0/V0 = C0Charge and capacitance increase by a factor of .Uc = CV2 = C0V02

Stored energy increased by

• Maintain constant voltage across the capacitor with dielectricCharge (and therefore capacitance) increase by a factor of

Stored energy ( in the capacitor) increases by a factor of at the expense of the battery, which does more work.

• Capacitor Jewelry

• CapacitanceWith dielectric, C = C0 where C0 is capacitance without dielectric.

C = (0)A/d

• ExampleConsider a capacitor with dielectric underneath a computer key. The capacitor is connected to 12V and has a normal (uncompressed) plate separation of 3.0 mm and plate area of 0.75 cm2. a) What is the required dielectric constant if the capacitance is 1.10 pF?b) How much charge is stored on the plates under normal conditions?c) How much charge flows onto the plates if they are compressed to a separation of 2.0 mm?

• Circuitry Capacitors in Series and ParallelCapacitors in series are connected one after another.

The voltage from the battery is shared between the series capacitors.

Want to find equivalent capacitance.

• Series and Parallel CapacitorsCapacitors in parallel are connected in branches parallel to one another.

Each capacitor in parallel receives the same voltage from the battery.

Want to find the equivalent capacitance.

• SeriesTo find equivalent capacitance, consider what is constant. For series capacitors, the charge on each capacitor must be constant. (why?)Voltages across each capacitor add to the total voltage supplied.V1 + V2 + V3 = Vtot

Q/C1 + Q/C2 + Q/C3 =Q/Ceq

1/Ceq = 1/C1 + 1/C2 + 1/C3

• ParallelIn parallel, each capacitor gets the same voltage. With different capacitances, the charge on each is different.Charge adds to total charge separated by battery

Qtot = Q1 + Q2 + Q3

CeffV = C1V + C2V + C3V

Ceff = C1 + C2 + C3

• ExampleGiven two capacitors, one with a capacitance of 2.5 F and the other of 5.0 F, what are the charge on each and the total charge stored if they are connected to a 12 V battery ina) seriesb) parallel

• ExampleThree capacitors are connected as shown on page 560. Find the voltage across each capacitor.