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Fields in Parallel Plates- Capacitors Physics Mrs. Coyle
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Electric Fields Between Parallel Plates Uniform Electric Field of Strength: E = V / d d is the distance between the plates Units: V/m =N/C=J/(Cm)
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Force Experienced by a Test Charge Inside the Parallel Plate Field The field is uniform so at any point in the field a test charge would feel a force F = qE
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Work and Energy V=Ed V=W/q 0 W/q 0 = Ed W=q 0 Ed Unit for work: Joule or the electron volt 1 eV= q e 1V= 1.6 x 10 -19 C x 1V 1eV= 1.6 x 10 -19 J
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Problem 1 Find: a)The field inside the plates. b)The force felt by a 9nC charge that is situated in the field.
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Continued c)Draw the equipotential surfaces between the plates (hint: equally spaced and parallel) d)If there three equipotential surfaces,what would be the distance between each? e)The work to move the charge from one equipotential to the other.
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Continued f)The potential difference between each equipotential surface and the negative plate.
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Equipotentials and Field Lines http://hyperphysics.phy-astr.gsu.edu/hbase/electric/equipot.html#c2
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Equipotentials and Field Lines
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Note Zero work is done when a charge is moved along an equipotential surface.
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Milikan’s Oil Drop Experiment He found the charge of the electron e = 1.602 x 10-19 C by noting that the droplets always carried whole number multiples of that number.
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Capacitor
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A capacitor stores charge (stores energy).
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Capacitors
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Leyden Jar 1746, Dutch physicist Pieter Van Musschenbroek Lived in “Leyden” Used by Ben Franklin
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Capacitance Capacitance C = Q/V Unit: Farad (F)= coulomb/ volt Q charge stored in capacitor V voltage across the capacitor
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Problem What is the charge of a 2µC capacitor that is charged, so that the potential difference is 8V?
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Question If the charge of a capacitor is changed, does that affect the potential difference or the capacitance?
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