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Published byAlexandra Montgomery Modified over 9 years ago
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Capacitors A device storing electrical energy
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Capacitor A potential across connected plates causes charge migration until equilibrium VV – + –q+q Charge stored q = C V C = capacitance Unit = C/V = farad = F
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Parallel Plate Capacitance Plate area A, separation d d A Capacitance = A 0 /d 0 = 8.85 10 –12 C2C2 N m 2
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Circuit Element Symbols Potential Source + – VV Conductor Capacitoror Resistor
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At Equilibrium VV C + – Capacitor charges to potential V Capacitor charge Q = C V + – VV
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Energy in a Capacitor C = Q/ V so V = Q/C VV Q Work to push charge Q W = V Q = (Q/C) Q slope = 1/C QQ area = W
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Energy in a Capacitor Work to charge to Q is area of triangle W = 1/2 Q(Q/C) = 1/2 Q 2 /C VV Q Q/CQ/C CVCV Work to charge to V W = 1/2 V (C V) = 1/2 C( V) 2
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Combining Capacitors and Parallel Series
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Parallel Components All have the same potential difference Capacitances add (conceptually add A’s)
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Series Capacitors All have the same charge separation Reciprocals are additive (conceptually add d’s)
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Gauss’s Law Electric flux through a closed shell is proportional to the charge it encloses. E = Q in / 0 0 = 8.85 10 –12 C2C2 N m 2
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Field Around Infinite Plate With uniform charge density = Q/A E = AA 00 00 1 2, so E = = E(2A)
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Infinite ||-Plate capacitor Individually –q 1/2 / 0 +q 1/2 / 0 –q+q /0/0 00 Together
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Charge of a Capacitor Parallel plates of opposite charge Charge density = Q/A – + Fields cancel outside d Potential V = d / 0 = d Q/(A 0 ) Capacitance C = Q/ V = 0 A/d /0/0
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Parallel Plate Capacitance Plate area A, plate separation d Field E = 00 = Q A0A0 Potential V = Ed = Qd A0A0 Capacitance Q/ V = Q A 0 Qd A0A0 d =
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Capacitor with a Dielectric If capacitance without dielectric is C, dielectric is C. = dielectric constant
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Dielectric Parameters Dielectric constant –Dielectric permittivity = 0 Breakdown voltage –Actually field V/m
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