Slide 1

Physics 2113 Lecture: 17 WED 25 FEB Capacitance II Physics 2113 Jonathan Dowling Slide 2 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 V=V PAR-V (Parallel: V the Same) Slide 3 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 Slide 4 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 Slide 5 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) Slide 6 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) = 32.75 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) Slide 7 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 Simplification Needed! Slide 8 Slide 9 Slide 10 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 Slide 11 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) Slide 12 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 10200 but it is always greater than 1! +Q Q DIELECTRIC C = A/d The and the constant o are both called dielectric constants. The has no units (dimensionless). Trick: Just substitute o for o in all the previous formulas! The and the constant o are both called dielectric constants. The has no units (dimensionless). Trick: Just substitute o for o in all the previous formulas! Slide 13 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. Slide 14 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 Example: Battery Connected Voltage V is Constant but Charge Q Changes Slide 15 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 increases Example: Battery Connected Voltage V is Constant but Charge Q Changes Slide 16 Capacitor has charge Q, voltage V Battery remains is disconnected then dielectric slab is inserted. Do the following increase, decrease or stay the same: Potential difference? Capacitance? Charge? Electric field? dielectric slab Example: Battery Disconnected Voltage V Changes but Charge Q is Constant Slide 17 Initial values: capacitance = C; charge = Q; potential difference = V; electric field = E; Battery remains disconnected Q is FIXED; Q new = Q (same) C new = C (increases) V new = Q/C new = Q/( C) (decreases). Since V new < V, E new = V new /d = E/ (decreases) dielectric slab Energy stored? Example: Battery Disconnected Voltage V Changes but Charge Q is Constant Slide 18 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=C