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11 - Capacitance II - Physics 212 Fall 2009 RAM Capacitor...

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Physics 212 - Fall 2009 Capacitance II 9/18/2009 RAM Capacitor Charged = 1 Capacitor Uncharged = 0
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Recap Capacitance q = CV The amount of charge stored (q) depends on C - Capacitance - units: Farads=Coulomb/Volt V - potential difference across the plates C = q V = " 0 EA Ed = " 0 A d Determine Capacitance Depends only on geometry Capacitors in parallel add C eq = C i i " 1 C eq = 1 C i i " Capacitors in series:
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Example Find the equivalent capacitance for the combination of capacitances shown in the figure across which the potential difference is applied. Use C 1 =12E-6F, C 2 =5.3E-6F, C 3 =4.5E-6 F. If V=12.5 V, what is the charge on C 1 ? Reduce Circuit in “steps” to find capacitance ---> <-----Work back to find charge and potential C 1 and C 2 are in parallel C 12 = C 1 + C 2 = 12 μ F + 5.3 μ F = 17.3 μ F 1 C 12 and C 3 are in series 2 1 C 123 = 1 C 12 + 1 C 3 = C 12 + C 3 C 12 C 3 " C 123 = C 12 C 3 C 12 + C 3 C 123 = (17.3 μ F )(4.5 μ F ) (17.3 μ F + 4.5 μ F ) = 3.57 μ F 3 q 123 = C 123 V = (3.57 E " 6 F )(12.5 V ) = 44.6 μ C 4 Charge is same in series q 123 = q 12 = q 3 V 12 = q 12 C 12 = 44.6 E " 6 C 17.3 E " 6 μ F = 2.58 V 5 Voltage is the same in parallel V 12 = V 1 = V 2 q 1 = C 1 V 1 = (12 E " 6 F )(2.58 V ) = 31.0 μ C 6
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Dielectric Material C air = " 0 A d Dielectric - Insulating material such as mineral oil or plastic Molecules have permanent dipole moment
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