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Published byVanessa Wilkins Modified over 9 years ago
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A capacitor is a device that stores electrical potential energy by building up a difference in charge on two pieces of metal.
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The ability of a capacitor to store charge is called the capacitance of the capacitor.
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Capacitance is the ratio of net charge on each plate to the potential difference between the plates. C = Q/ΔV ( or Q = VC)
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The unit is the farad (F), which is one coulomb per volt (C/V). A typical capacitor ranges from microfarads to picofarads. F to pF 1 F = 10 -6 F 1 pF = 10 -12 F
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Capacitance for a parallel plate capacitor is calculated from: C = ε 0 A/d ε 0 is the permittivity of a vacuum A is the area of the plates d is the distance between the plates
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The permittivity depends on the material between the plates. For a vacuum (or air), the value is 8.85 x 10 -12 C 2 /Nm 2. Permittivity is greater for other materials.
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The material between the plates is an insulator and is called a dielectric. The presence of a dielectric increases the capacitance.
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The molecules of a dielectric align themselves with the electric field, negative end toward the positive plate, and vice-versa.
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The dielectric effectively reduces the charge at each plate, so the plate can store more charge for a given voltage. More charge means more capacitance for same voltage. (Q = VC)
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C = ε 0 A/d As the equation indicates, the larger the plates of a capacitor, the higher the capacitance; and the further apart the plates are, the lower the capacitance.
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List four ways to increase the charge on a capacitor.
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Once a capacitor is charged, the source can be removed and the capacitor will retain the charge. If the two plates are connected by a conducting material, the capacitor will rapidly discharge. This discharge continues until the potential is zero.
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Devices that use capacitors include: camera flash devices, computer keyboards, defibrillators, tasers and automobile blinkers.
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A charged capacitor stores electrical potential energy. The amount is given by the following equation: PE electric = ½QΔV
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PE electric = ½QΔV Since Q = ΔVC, PE electric = ½C(ΔV)2 and: PE electric = Q 2 /2C
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Here are the two big capacitor equations: Q = VC PE electric = ½CV 2
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A 3.0 F capacitor is connected to a 12 V battery. What is the magnitude of the charge on each plate of the capacitor, and how much electrical potential energy is stored in the capacitor?
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