Physics 2102 Jonathan Dowling Physics 2102 Lecture 8 Capacitors II Capacitors in parallel and in series In parallel Ceq C1 C2 Veq V1 V2 Qeq Q1 Q2 In series 1 Ceq 1 C1 1 C2 Veq V1 V2 Qeq Q1 Q2 Q1 C1 Q2 C2 Qeq Q1 Q2 C1 C2 Ceq Example 1 What is the charge on each capacitor Q CV V 120 V Q1 10 F 120V 1200 C Q2 20 F 120V 2400 C Q3 30 F 120V 3600 C Note that Total charge 7200 C is shared between the 3 capacitors in the ratio C1 C2 C3 i e 1 2 3 10 F 20 F 30 F 120V Example 2 What is the potential difference across each capacitor Q CV Q is same for all capacitors 10 F Combined C is given by 1 1 1 1 Ceq 10 F 20 F 30 F 20 F 120V 30 F Ceq 5 46 F Q CV 5 46 F 120V 655 C Note 120V is shared in the V1 Q C1 655 C 10 F 65 5 V ratio of INVERSE V2 Q C2 655 C 20 F 32 75 V capacitances i e 1 1 2 V3 Q C3 655 C 30 F 21 8 V 1 3 largest C gets smallest V Example 3 10 F In the circuit shown what is the charge on the 10 F capacitor 5 F 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 Hence Q 10 F 5V 50 C 5 F 10V 10 F 10 F 10V Energy 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 Q Q dq 2 2 q Q CV U Vdq dq C 2C 2 0 0 Energy Stored in Electric Field Energy stored in capacitor U Q2 2C CV2 2 View the energy as stored in ELECTRIC FIELD For example parallel plate capacitor Energy DENSITY energy volume u 2 2 Q Q Q 0 Q 0 E 2 U 2 2CAd 2 0 A Ad 2 0 A 2 0 A 2 2 2 d volume Ad General expression for any region with vacuum or air Example 10 F capacitor is initially charged to 120V 20 F capacitor is initially uncharged Switch is closed equilibrium is reached How much energy is dissipated in the process 10 F C1 Initial charge on 10 F 10 F 120V 1200 C 20 F C2 After switch is closed let charges Q1 and Q2 Charge is conserved Q1 Q2 1200 C Q1 400 C Q Q Q2 Also Vfinal is same 1 2 Q1 Q2 800 C C1 C2 2 Vfinal Q1 C1 40 V Initial energy stored 1 2 C1Vinitial2 0 5 10 F 120 2 72mJ Final energy stored 1 2 C1 C2 Vfinal2 0 5 30 F 40 2 24mJ Energy lost dissipated 48mJ Dielectric Constant DIELECTRIC Q Q C A d 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 Example Capacitor has charge Q voltage V Battery remains connected while dielectric slab is inserted Do the following increase decrease dielectric slab or stay the same Potential difference Capacitance Charge Electric field Example Initial values capacitance C charge Q potential difference V electric field E Battery remains connected V is FIXED Vnew V same Cnew C increases Qnew C V Q increases Since Vnew V Enew E same Energy stored dielectric slab u 0E2 2 u 0E2 2 E2 2 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 Ceq 1 C1 1 C2 Capacitors in parallel same potential not necessarily same charge equivalent capacitance Ceq C1 C2 Energy in a capacitor U Q2 2C CV2 2 energy density u 0E2 2 Capacitor with a dielectric capacitance increases C C
View Full Document
Unlocking...