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Unformatted text preview: V Q C = The greater the capacitance C of a capacitor, the greater the magnitude of Q of charge on either conductor for a given potential difference V ab . Hence the greater energy is stored by the capacitor. V Q C = Since capacitance depends on voltage, and voltage is calculated at a point a way from a charged object, then for different shaped objects we will have different capacitances . The simplest capacitor is a parallel plate capacitor consisting of two parallel plate conductors of area A , separated by distance d that is small in comparison with A . ε σ = E ε ε σ A Q E = = d A Q ds A Q ds E ds E q U V B A B A B A ε ε = = = ⋅ = ∆ = ∆ ∫ ∫ ∫ Inside the capacitor the Electric field is constant and depends on the charge density. This can be rewritten in terms of the charge as: V Q C = d A Q V ab ε = Hence the magnitude of the voltage across two parallel plate conductors is: V Q C = d A Q V ab ε = With this voltage result for parallel plate conductors we can immediate determine the capacitance. d A d A Q Q C ε ε = = Capacitance depends only on the geometry of the capacitor. In this case it is directly proportional to the area A of each plate, and inversely proportional to the distance d between each plate. d A C ε = Size of a One Farad capacitor. A parallel plate capacitor has a capacitance of 1.0 F. If the plates are 1.0mm apart, what is the area of the plates . ε =Permittivity of empty space = 8.85 x 1012 C 2 /N.m 2 d A C ε = 2 8 12 3 10 1 . 1 / 10 85 . 8 . 1 ) 10 . 1 ( m x m F x F x C d A = = = ε Any object of high dialectic value (e.g. human body part) brought into proximity to the sensor element changes the capacitance of the sensor element. This change in capacitance is detected by the electronics. The electronics can be set to trigger when a certain capacitance level is reached. Capacitor as a sensor saves the hand from being squeezed. Properties of a Parallel Plate Capacitor The plates of a parallelplate capacitor in vacuum are 5.00mm apart and 2.00m 2 in area. A potential difference of 10,000 Volts is applied across the capacitor. Compute: a) The capacitance b) The charge on each plate c) The magnitude of the electric field in the space between them. ε = 8.85 x 1012 C 2 /N.m 2 d A C ε = A potential difference of 10,000 Volts is applied across the capacitor the plates of a parallelplate capacitor in vacuum are 5.00mm apart and 2.00m 2 in area. Compute: a) The capacitance: a) The capacitance is a constant and for parallel plate capacitors is given by: F F x m x m m m F x d A C μ ε 00354 . 10 54 . 3 10 . 5 00 . 2 ) / 10 85 . 8 ( 9 3 2 2 12 = = = = The charge across each plate conductor is CV = Q = (3.54x109 )(10,000) = 3.54x105 C b) Compute the charge on each plate c) To compute the Electric field between the plates we can use the Gaussian result: C N x m N C x m x A Q E / 10 . 2 ) / 10 85 . 8 )( . 2 ( 10 54 . 3 6 2 2 12 2 5 = ⋅ = = = ε ε σ C N x m x V x d V E ab / 10 . 2 10 . 5 10 00 . 1...
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 Summer '09
 W.CHRISTENSEN
 Physics, Capacitance, Charge, Energy, Electric charge

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