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### Transcript of Electric Potential Energy and Electric Potential Chapter 16.

• Slide 1
• Electric Potential Energy and Electric Potential Chapter 16
• Slide 2
• Potential Energy Potential 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
• Slide 3
• Gravitational P.E. U g To bring a mass in from infinity to near the earth, we would encounter the Earths gravitational field. Work = Fd = U g U g = U f U i = U r U inf U g = (GM e m/r 2 )r U g = -GMm/r
• Slide 4
• Uniform Fields Near Earths surface, gravitational field is uniform. g = 9.8 m/s 2. We calculate Work = U = mgh This is a simplification of U g = -GMm/r Work is essentially Fd
• Slide 5
• Electric Potential Energy To bring a charge, q, in from infinity to near +Q, we would encounter the charges electric field. Work = Fd = U e U e = U f U i = U r U inf U e = (kQq/r 2 )r U e = kQq/r
• Slide 6
• Uniform Electric Field Parallel Plates with equal charge provide a uniform electric field. Recall E = F e /q F e = Eq
• Slide 7
• Potential Energy Uniform Field (plate separation d) U = U f U i = U B U A U = Work = Fd U = (Eq)d = Eqd E is electric field between plates, d is plate separation.
• Slide 8
• Uniform Gravitational and Electric Field Comparison U g = mgh for constant g field U e = qEd for constant E field U is measured in Joules
• Slide 9
• Non-Uniform Fields / PE F g = GMm/r 2 U g = -GMm/r F e = kQq/r 2 U e = kQq/r
• Slide 10
• 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.
• Slide 11
• Electric Potential, V V = U e /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
• Slide 12
• Summarize New Formulas F e = kq 1 q 2 /r 2 U e = F (distance) = kq 1 q 2 /r electric potential energy between two charges (non uniform fields) U e = qEdelectric potential energy for parallel plates with electric field E V = U e /q = kq/r electric potential due to charge q (non uniform field) V = Edelectric potential for parallel plates with electric field E field E
• Slide 13
• Electric Potential Energy for Several Charges To find the potential energy of a system of charges, add the potential energy between each pair of charges. U e = U 12 + U 13 + U 23
• Slide 14
• Examples Read 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}
• Slide 15
• Equi-potential Surfaces15 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.
• Slide 16
• Conservative Fields A 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 force Friction is non-conservative. In moving an object along an equi-potential surface, no work is done.
• Slide 17
• Equi-potential Surfaces
• Slide 18
• Uniform Electric Field Recall E = F/q F = Eq F = Eq U = Fd = Eqd V = U/q For a uniform field, V = Ed = Ex
• Slide 19
• Units E [Joule/Coulomb] or [Volt/meter] These are equivalent since E = F/q = V/d
• Slide 20
• 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?
• Slide 21
• Electron Volt measure for Energy An 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
• Slide 22
• Capacitance A 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.
• Slide 23
• Capacitance Capacitance = Charge stored / Voltage C = Q/V or Q = CV [Coulombs/Volt] = [Farad] 1 Farad is a huge amount of capacitance. It is most commonly measured in microfarads = 10 -6 F
• Slide 24
• Capacitance Capacitance 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 = 0 A/d Where 0 is the permittivity of free space And 0 = 8.85 X 10 12 C 2 /Nm 2
• Slide 25
• Example What would be the plate area of an air filled 1.0F parallel plate capacitor if the plate separation were 1.0 mm?
• Slide 26
• Energy! Potential increases as charge is added. Slope = V/ Q = 1/C Potential energy stored in a capacitor = Work Done = QV av = Q(V/2) U c = QV = CV 2 = Q 2 /2C
• Slide 27
• Example During 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?
• Slide 28
• Homework! Read sections 16.4 and 16.5 pages 552 559. Do # 57, 59, 60, 61, 64, 65 67, 69- 72
• Slide 29
• Dielectric Materials Dielectric materials placed between parallel plates have several purposes: Keep plates from coming in contact Allow for flexible plates Increase capacitance of the capacitor Dielectric constant, >1 = C/C 0
• Slide 30
• Two dielectric situations: Either the voltage difference is removed once the plates are charged, then the dielectric material inserted between plates Or the dielectric material is inserted between the plates while the voltage is maintained. These are different situations, but both result in increased capacitance.
• Slide 31
• Remove voltage then insert dielectric Voltage applied, V 0, separates charge, Q 0 and sets up electric field E 0 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.
• Slide 32
• Remove voltage then insert dielectric Charge, Q, doesnt change once dielectric is inserted. Induced electric field in dielectric reduces original electric field to E and original voltage difference to V. = E 0 / E = V 0 /V Then C = Q/V = Q 0 /(V 0 / ) = (Q 0 /V 0 ) = C 0 = (Q 0 /V 0 ) = C 0 Uc = Q 2 /2C = Q 2 /2(C 0 ) = U 0 /
• Slide 33
• Remove voltage then insert dielectric Capacitance 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!
• Slide 34
• When capacitor is maintained at constant voltage, the battery continues to supply charge to compensate for the induced electric field. C = Q/V = Q 0 /V 0 = C 0 Charge and capacitance increase by a factor of . Uc = CV 2 = C 0 V 0 2 Stored energy increased by Insert dielectric and maintain constant voltage
• Slide 35
• Maintain constant voltage across the capacitor with dielectric Charge (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.
• Slide 36
• Capacitor Jewelry
• Slide 37
• Capacitance With dielectric, C = C 0 where C 0 is capacitance without dielectric. C = ( 0 )A/d
• Slide 38
• Example Consider 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 cm 2. 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?
• Slide 39
• Circuitry Capacitors in Series and Parallel Capacitors in series are connected one after another. The voltage from the battery is shared between the series capacitors. Want to find equivalent capacitance.
• Slide 40
• Series and Parallel Capacitors Capacitors 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.
• Slide 41
• Series To 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. V 1 + V 2 + V 3 = V tot Q/C 1 + Q/C 2 + Q/C 3 =Q/C eq 1/C eq = 1/C 1 + 1/C 2 + 1/C 3
• Slide 42
• Parallel In parallel, each capacitor gets the same voltage. With different capacitances, the charge on each is different. Charge adds to total charge separated by battery Q tot = Q 1 + Q 2 + Q 3 C eff V = C 1 V + C 2 V + C 3 V C eff = C 1 + C 2 + C 3
• Slide 43
• Example Given 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 in a) series b) parallel
• Slide 44
• Example Three capacitors are connected as shown on page 560. Find the voltage across each capacitor.