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Today 3/24

Today 3/24. Plates of charge Capacatance Lab: DC Circuits (read lab) HW: 3/24 “Plate Potential 2” Due Thursday 2/27. W AB = qED. F Hq. F E = qE. W AB = qED. What is the change in potential energy in going from A to B?. A. B.

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Today 3/24

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  1. Today 3/24 • Plates of charge • Capacatance • Lab: DC Circuits (read lab) • HW: 3/24 “Plate Potential 2” Due Thursday 2/27

  2. WAB = qED FHq FE = qE WAB = qED What is the change in potential energy in going from A to B? A B Only applies when the field is uniform over the distance. DVAB‘s sign depends on the direction of E. In this case it’s positive. PEAB = qED PEAB = qDVAB DVAB = EDAB

  3. Two conducting plates connected by a battery. V=1000V What is the charge density  on each plate? d=1cm

  4. Two conducting plates connected by a battery. V=1000V What happens to  when the plates are moved twice as far apart? The E-field and  would both be one half the size. A current flowed through the battery as the plates moved apart. d=2cm

  5. Two conducting plates connected by a battery. 1cm What happens to  when a 1cm thick conducting plate is inserted between the two plates? V=1000V The E-field must be zero inside the conductor so arrange the charges in the 1cm plate to do this. V from the left plate to the right plate is still +1000V but what is E in the air spaces? V = Ed but d has been reduced to1cm since there is no field in the 1cm plate. E was doubled when the plate was inserted. d=2cm

  6. How did  change when the plate was inserted? V=1000V d=2cm E used to be half as big so half the field lines. V=1000V What about the charge? Consider only the E-field do to the induced charge on the center plate in the air spaces. 1cm d=2cm

  7. How did  change when the plate was inserted? V=1000V d=2cm V=1000V This induced charge does not contribute to the field in the air spaces. The only way to increase the field as the slab is inserted is to have more charge flow onto the outside plates. 1cm d=2cm

  8. How did  change when the plate was inserted? V=1000V d=2cm There must have been less charge before. V=1000V This extra plate increases the capacity to store charge. 1cm Putting a dielectric between plates stores more charge with the same voltage. d=2cm

  9. How does  ( and E) change as the distance between the plates is reduced? V=1000V V=1000V d=2cm Current flows from the neg plate to the pos plate. V must stay 1000J/C E field increases and the charge density,  increases. PE for some q would be the same between the plates in both cases! d=1cm

  10. How does  ( and E) change when a conducting plate is inserted? V=1000V V=1000V 1cm d=2cm d=2cm Current flows from the neg plate to the pos plate. V must stay 1000J/C E field increases and the charge density,  increases. PE for some q would be the same between the plates in both cases!

  11. Capacitors Capacitance tells me how many coulombs of charge are stored in a capacitor when it has 1 Volt across it. A 25F capacitor will have 25C of charge on it when it has 1 volt across it. Think of capacitance as “coulombs per volt”. Units: coulombs per volt or “Farads”Equation: C = q/V

  12. Capacitors Think of storing propane in a tank. The amount of propane depends on the pressure just as the amount of charge depends on voltage. We will use parallel plate capacitors so that all ideas from parallel plates apply. V = Ed E = /2o  = Q/A

  13. 6F 6F Capacitors How much charge is on the capacitor? Think “6x10-6 Coulombs per Volt and you won’t need the equation! 72x10-6 C or 72C 12 V

  14. 6F 6F Capacitors What is the E-field inside the capacitor if the plates have an area of 1 m2? Note that capacitance depends on the area and separation distance Q = 72x10-6 C or 72C E = /2o = 8.2x106 N/C What is the distance between the plates? V = Ed d = 12/8.2x106 = 1.5x10-6 m 12 V

  15. 6F Capacitors and Energy How many Joules of energy are stored in the capacitor? Ask yourself, “How much work must be done to charge the plates?” 12 V

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