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Physics 2113 Lecture: 06 WED 01 OCT Capacitance II Physics 2113 Jonathan Dowling

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Capacitors in Parallel: V=Constant An ISOLATED wire is an equipotential surface: V=Constant Capacitors in parallel have SAME potential difference but NOT ALWAYS same charge! V AB = V CD = V Q total = Q 1 + Q 2 C eq V = C 1 V + C 2 V C eq = C 1 + C 2 Equivalent parallel capacitance = sum of capacitances AB C D C1C1 C2C2 Q1Q1 Q2Q2 C eq Q total V = V AB = V A –V B V = V CD = V C –V D VAVA VBVB VCVC VDVD ΔV=V PAR-V (Parallel: V the Same)

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Capacitors in Series: Q=Constant Q 1 = Q 2 = Q = Constant V AC = V AB + V BC A BC C1C1 C2C2 Q1Q1 Q2Q2 C eq Q = Q 1 = Q 2 SERIES: Q is same for all capacitors Total potential difference = sum of V Isolated Wire: Q=Q 1 =Q 2 =Constant SERI-Q: Series Q the Same

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Capacitors in Parallel and in Series In series : 1/C ser = 1/C 1 + 1/C 2 V ser = V 1 + V 2 Q ser = Q 1 = Q 2 C1C1 C2C2 Q1Q1 Q2Q2 C1C1 C2C2 Q1Q1 Q2Q2 In parallel : C par = C 1 + C 2 V par = V 1 = V 2 Q par = Q 1 + Q 2 C eq Q eq

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Example: Parallel or Series? What is the charge on each capacitor? C 1 =10 F C 3 =30 F C 2 =20 F 120V Q i = C i V V = 120V on ALL Capacitors (PAR-V) Q 1 = (10 F)(120V) = 1200 C Q 2 = (20 F)(120V) = 2400 C Q 3 = (30 F)(120V) = 3600 C Note that: Total charge (7200 C) is shared between the 3 capacitors in the ratio C 1 :C 2 :C 3 — i.e. 1:2:3 Parallel: Circuit Splits Cleanly in Two (Constant V)

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Example: Parallel or Series What is the potential difference across each capacitor? C 1 =10 F C 3 =30 F C 2 = 20 F 120V Q = C ser V Q is same for all capacitors (SERI-Q) Combined C ser is given by: C eq = 5.46 F (solve above equation) Q = C eq V = (5.46 F)(120V) = 655 C V 1 = Q/C 1 = (655 C)/(10 F) = 65.5 V V 2 = Q/C 2 = (655 C)/(20 F) = V V 3 = Q/C 3 = (655 C)/(30 F) = 21.8 V Note: 120V is shared in the ratio of INVERSE capacitances i.e. (1):(1/2):(1/3) (largest C gets smallest V) Series: Isolated Islands (Constant Q)

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Example: Series or Parallel? In the circuit shown, what is the charge on the 10 F capacitor? 10 F 10V 10 F 5F5F 5F5F 10V The two 5 F capacitors are in parallel Replace by 10 F Then, we have two 10 F capacitors in series So, there is 5V across the 10 F capacitor of interest by symmetry Hence, Q = (10 F )(5V) = 50 C Neither: Circuit Compilation Needed!

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Energy U Stored in a Capacitor Start out with uncharged capacitor Transfer small amount of charge dq from one plate to the other until charge on each plate has magnitude Q How much work was needed? dq

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Energy Stored in Electric Field of Capacitor Energy stored in capacitor: U = Q 2 /(2C) = CV 2 /2 View the energy as stored in ELECTRIC FIELD For example, parallel plate capacitor: Energy DENSITY = energy/volume = u = volume = Ad General expression for any region with vacuum (or air)

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Dielectric Constant If the space between capacitor plates is filled by a dielectric, the capacitance INCREASES by a factor This is a useful, working definition for dielectric constant. Typical values of are 10–200 +Q –Q DIELECTRIC C = A/d The and the constant o are both called dielectric constants. The has no units (dimensionless). The and the constant o are both called dielectric constants. The has no units (dimensionless).

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Atomic View E mol E cap Molecules set up counter E field E mol that somewhat cancels out capacitor field E cap. This avoids sparking (dielectric breakdown) by keeping field inside dielectric small. Hence the bigger the dielectric constant the more charge you can store on the capacitor.

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Example Capacitor has charge Q, voltage V Battery remains connected while dielectric slab is inserted. Do the following increase, decrease or stay the same: –Potential difference? –Capacitance? –Charge? –Electric field? dielectric slab

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Example Initial values: capacitance = C; charge = Q; potential difference = V; electric field = E; Battery remains connected V is FIXED; V new = V (same) C new = C (increases) Q new = ( C)V = Q (increases). Since V new = V, E new = V/d=E (same) dielectric slab Energy stored? u= 0 E 2 /2 => u= 0 E 2 /2 = E 2 /2

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Summary Any two charged conductors form a capacitor. Capacitance : C= Q/V Simple Capacitors: Parallel plates: C = 0 A/d Spherical : C = 4 0 ab/(b-a) Cylindrical: C = 2 0 L/ln(b/a) Capacitors in series: same charge, not necessarily equal potential; equivalent capacitance 1/C eq =1/C 1 +1/C 2 +… Capacitors in parallel: same potential; not necessarily same charge; equivalent capacitance C eq =C 1 +C 2 +… Energy in a capacitor: U=Q 2 /2C=CV 2 /2; energy density u= 0 E 2 /2 Capacitor with a dielectric: capacitance increases C’=κC

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