Devices that can store electric charge are called capacitors.

Capacitors consist of 2 conducting plates separated by a small distance containing an insulator.

Capacitors are used in electronics, computers, camera flashes, and as protectors of circuits for surges or memory (RAM) of binary code.

Capacitors become quickly charged when a voltage is applied to it in a circuit.

One plate will be negatively charged and the other plate will have an equal positive charge.

The charge, Q, depends on the potential difference, V, applied to it. Q = CV. C is a proportionality constant called capacitance of the capacitor and is measured in coulombs/volt or farad (F).

Common capacitance ranges are from 1pF (picoFarad) to 1µF (microFarad) (10-12 to 10-6).

The Capacitance, C, depends on the area of the capacitor plates and the distance of insulated (or air) separation.

Recall the permittivity of free space constant has a value of 8.85 x 10-12 C2 /N*m2

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AC 0

(a) Calculate the capacitance of a capacitor whose plates are 20cm x 3.0 cm and are separated by a 1.0mm air gap. (b) What is the charge on each plate if the capacitor is connected to a 12-V battery? (c) What is the electric field between the plates?

(a) The area A = 20 cm x 3.0 cm = 60.0 cm2 = 6.0 x 10-3 m2 so the capacitance,

(b) The charge on each plate is Q=CV

(c) For a uniform electric field b/t the plates, E=V/d

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12 43

A charged capacitor stores energy equal to the work done to charge it.

Batteries charge capacitors by removing charge from one plate and putting charge on the other plate. This takes time.

The more charge already on a plate, the more work required to add more charge of the same sign.

Initially when a capacitor is uncharged, it takes NO WORK to move the first bit of charge over.

The work needed to move a small amount of charge Δq, when a potential difference, V, is across the plates, is ΔW=VΔq.

Voltage increases during the charging process from 0 to its final value Vf .

Since the average voltage during this process will be Vf /2 then the total work, W to move all of the charge, Q at once is the energy stored, U, in the capacitor: U=energy=1/2 QV.

Since Q=CV, then U=1/2QV=1/2CV2 =1/2Q2/C.

A camera flash unit stores energy in a 150µF capacitor at 200V. How much electric energy can be stored?

U = energy = ½ CV2 = Notice how the units work out: recall

1F=1C/V and 1V=1J/C. If this energy could be released in

1/1000s the power output would be 3000W.

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1 26

Though energy is not a substance in a place, it is helpful to think of it as being stored in the electric field between the plates.

Recall the electric field b/t 2 parallel plates separated by a small distance is uniform and related to electric potential, V by V=Ed. Also recall capacitance, C

d

AC 0

Therefore, energy stored, U in terms of the electric field is found

The quantity Ad is the volume between the plates where the electric field E exists.

Dividing both sides by the volume, we get the energy / volume or energy density

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Energy stored is proportional to the square of electric field.

Please do Ch 17 Rev p 524 #s 30, 32, 33, 35, 36, 37, 43, 44, & 45