 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) ( to ).

 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 C 2 /N*m 2

 (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 cm 2 = 6.0 x m 2 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

 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 V f.  Since the average voltage during this process will be V f /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=1/2CV 2 =1/2Q 2 /C.

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

 U = energy = ½ CV 2 =  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.

 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

 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 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