Capacitors Phy 1161: PreLecture 06 Today’s lecture will cover Textbook Sections 20-5 – 20-6

Comparison: Electric Potential Energy vs. Electric Potential  V AB : the difference in electric potential between points B and A  U AB : the change in electric potential energy of a charge q when moved from A to B  U AB = q  V AB q AB

Electric Potential: Summary E field lines point from higher to lower potential For positive charges, going from higher to lower potential is “downhill” For a battery, the (+) terminal is at a higher potential than the (–) terminal Positive charges tend to go “downhill”, from + to - Negative charges go in the opposite direction, from - to +  U AB = q  V AB

Important Special Case Uniform Electric Field Two large parallel conducting plates of area A +Q on one plate -Q on other plate Then E is uniform between the two plates: E=4  kQ/A zero everywhere else This result is independent of plate separation This is call a parallel plate capacitor

Parallel Plate Capacitor Potential Difference A B E=E 0 d V = V A – V B = +E 0 d A B d E= V = V A – V B = +2E 0 d Potential difference is proportional to charge: Double Q  Double V E 0 = 4πkQ/A Charge Q on platesCharge 2Q on plates

Capacitance The ability to store separated charge Definition: Units: Farad (F) – named in honor of Michael Faraday –1 F = 1C/V From Faraday’s notebook

Capacitor Any pair of conductors separated by a small distance. (e.g. two metal plates) Capacitor stores separated charge –Positive Q on one conductor, negative Q on other –Net charge is zero E d Q = CV U = (½) Q V Stores Energy

Capacitance of Parallel Plate Capacitor V = Ed AND E = Q/(  0 A) (Between two large plates) So: V = Qd/ /(  0 A) Remember: C  Q/V So: Equation based on geometry of capacitor A d A E+ - V  0 = 8.85x C 2 /Nm 2 If there is adielectric (κ>1) between plates C = κ C 0

Dielectric Placing a dielectric between the plates increases the capacitance. C =  C 0 Capacitance with dielectric Dielectric constant (  > 1) Capacitance without dielectric

MaterialConstantMaterialConstant Vacuum1Germanium16 Polyvinyl chloride 3.18Strontium titanate 310 Mica3 - 6Water80.4 Mylar3.1Glycerin42.5 Neoprene6.70Benzene2.284 Plexiglass3.40Glass5 – 10 Polyethylene2.25Air (1 atm)1 Liquid ammonia (-78 o C) 25Titanium dioxide (rutile) 173 perp 86 para Dielectrics

Voltage in Circuits Elements are connected by wires. Any connected region of wire has the same potential. V wire 1 = 0 VV wire 2 = 5 VV wire 3 = 12 VV wire 4 = 15 V V C 1 = _____ VV C 3 = _____ VV C 2 = _____ V C1C1 C2C2 C3C3 The potential difference across an element is the element’s “voltage.”

Voltage in Circuits Elements are connected by wires. Any connected region of wire has the same potential. V wire 1 = 0 VV wire 2 = 5 VV wire 3 = 12 VV wire 4 = 15 V V C 1 = 5 V V C 3 = 3 VV C 2 = 7 V C1C1 C2C2 C3C3 The potential difference across an element is the element’s “voltage.”

Capacitors in Parallel Both ends connected together by wire C1C1 C2C2 C eq Share Charge: Q eq = Q 1 + Q 2 Total Cap: C eq = (Q 1 + Q 2 )/V = C 1 + C 2 = V eq Same voltage: V 1 = V 2

Capacitors in Parallel Both ends connected together by wire C1C1 C2C2 C eq 15 V 10 V 15 V 10 V 15 V 10 V Share Charge: Q eq = Q 1 + Q 2 Total Cap: C eq = (Q 1 + Q 2 )/V = C 1 + C 2 = V eq Same voltage: V 1 = V 2

Capacitors in Series Connected end-to-end with NO other exits Same Charge: Q 1 = Q 2 = Q eq C eq C1C1 C2C Q -Q +Q -Q +Q -Q Share Voltage: V 1 +V 2 =V eq

Electromotive Force Battery –Maintains potential difference V –Not constant power –Not constant current –Does NOT produce or supply charges, just “pushes” them. + -