# April 7 AP Physics. AP Practice Exam April 28 ACT day 1 pm Not optional.

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April 7 AP Physics

AP Practice Exam April 28 ACT day 1 pm Not optional

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.

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

Homework Check 18, 20, 24, 30

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

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

Electrical potential energy Work

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

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

Both Gravitational potential and Electrical potential energy depend on position

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

Electric potential: uniform field V = -Ed

Electric potential: Point charge V = kq r

Equipotential lines

Equipotential surfaces

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

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

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

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

Equivalent Capacitance parallel circuits

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

Equivalent Capacitance Series circuits

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

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

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