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April 7 AP Physics

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AP Practice Exam April 28 ACT day 1 pm Not optional

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In: Two 1 kg spheres each carry a charge of magnitude 1C. How does F E, the strength of the electric force between the spheres, compare to F G, the strength of their gravitational attraction? a)F E < F G b)F E =F G c)F E >F G d)If the charges on the spheres are of the same sign, then F E >F G ; but if the charges on the spheres are of the opposite sign, then F E <F G e)Cannot be determined without knowing the distance between the spheres.

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Objective: To learn about static electricity: –Millikan Oil Drop Experiment –Electric potential energy –Electric potential –Capacitance –Parallel plate capacitors

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Homework Check 18, 20, 24, 30

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Robert Millikan Oil Drop experiment Measured charge to mass ratio of electron Published in 1913 Nobel prize 1932

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Millikan oil-drop experiment qE = mg q/m=g/E q e = -1.6x10 -19

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Electrical potential energy Work

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Electric potential energy The work done to bring a charge from infinity to a particular distance = - potential energy the charge has at that point. U E =-Work Also, the change in PE from one point to another = -work done to change the position. ΔU E =-W

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In a field generated by a point charge: U E =kq 1 q 2 r In a uniform field: ΔU E =qEΔd

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Both Gravitational potential and Electrical potential energy depend on position

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Electric potential The energy per each individual charge. Measured in Volts V = U E q

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Electric potential: uniform field V = -Ed

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Electric potential: Point charge V = kq r

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Equipotential lines

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Equipotential surfaces

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No work is done moving around on an equipotential surface. Work is done moving from one equipotential surface to another.

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Capacitance A capacitor is a device that stores energy in the form of charges. C = Q ΔV

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Parallel Plate capacitor Stores energy by building up opposite charges on each plate. C = Є o A d

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Energy stored in a capacitor U c = ½ QV= ½ CV 2

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Equivalent Capacitance parallel circuits

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C=Q/V V 1 =V 2 =V e Q 1 +Q 2 =Q e so…C 1 +C 2 =C e

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Equivalent Capacitance Series circuits

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C=Q/V V 1 +V 2 =V e Q 1 =Q 2 =Qe so…1/C 1 +1/C 2 =1/C e

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Out: What is the equivalent capacitance of the circuit between a and b ?

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