LECT6_1661610 - 2 Charged Planes Equipotential surfaces are...

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2 Charged Planes Equipotential surfaces are parallel to the planes and to the E-field lines
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Capactitors & Capacitance Capacitor: a device for storing electrical potential energy Can also be rapidly discharged to release a large amount of energy at once Applications: camera flashes, automobile ignition systems, computer memory, laser flash lamps, defibrillators
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Laser Fusion at the Nat’l Ignition Facility, Livermore, CA. 10 6 J released in μ s: Power ~ 10 12 W Credit:: LLNL
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A Capacitor Capacitance is defined as the ability to store separated charge. C = Q / Δ V Unit: FARAD = C/V Note: +Q plate is connected to positive terminal of battery; –Q plate connected to – terminal.
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Parallel Plate Capacitor Note E-field inside is pretty uniform. E-field outside is relatively negligible Charges like to accumulate at inner edges of plates + + + + + + + + + + + + + + + + + + + + + + + + + + + + + - - - - - - - - - - - - - - - - - - - - - - - - - - - -
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Parallel Plate Capacitor Capacitance depends on geometry: C = ε 0 A / d
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Variable capacitor: C depends on “overlapping” area
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E = σ / ε 0 = (q/A)/ ε 0 = Δ V/d single sheet of charge +Q: E = σ /(2 ε 0 ) +Q –Q Parallel plate capac.:
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A 2A: C = ε 0 A/d C 2C Double the area…
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Δ V Δ V (held constant) d d/2 What happens to C & E? If plates are HELD at a fixed potential difference Δ V which does not change as you decrease d: + - + -
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Δ V Δ V (held constant) d d/2 E = Δ V/d so E 2E E = σ / ε 0 = (Q/A)/ ε 0 so Q 2Q C = Q/ Δ V so C 2C If plates are HELD at a fixed potential difference Δ V which does not change as you decrease d: + - + -
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Capacitors with insulators (for small d) With insulating material filling the gap, charges cannot travel from one plate to the other, but E- fields can permeate.
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A parallel-plate capacitor with A = 4 cm 2 , d = 1 mm. Find its capacitance. C = ε o A/d = (8.85x10 -12 C 2 /Nm 2 )(4x10 -4 m 2 )/(10 -3 m) = 3.54x10 -12 F = 3.54pF If the capacitor is connected to a 9 Volt battery, how much charge is on each plate? C = Q/ Δ V Q = C Δ V = (3.54x10 -12 F)(9V) =3.2x10 -11 C Example:
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Calculate the magnitude of the E-field inside the capacitor.
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