Chapter 26 Capacitance and Dielectrics

1. Chapter 26 Capacitance and Dielectrics PHYS 2326-11

2. Concepts to Know • Capacitors • Capacitance • Energy Density • Dielectric • Dielectric Constant • Permittivity • Series and Parallel Circuits

3. Capacitance • Capacitance is the ability for an object to hold a charge for a given potential energy C = Q/ΔV C is the capacitance in Farads that can hold a charge Q with an increase in potential of ΔV volts. 1 Farad = 1 coulomb of charge per Volt This is a huge value

4. Capacitors • You’re responsible for knowing • Parallel Plate Capacitors • Spherical Capacitors • Cylindrical Capacitors • Typical values range from a few pico farads to a few thousand micro farads

5. Parallel Plate Capacitor • most common type of component although most are rolled up or contain many plates C = εo A/d Capacitance is proportional to area /inverse proportional to separation

6. Energy Stored in Capacitor • Work done to charge a capacitor given by eqn 26.11 which is the integral of the dw required for each dq of charge added U = ½ * QV = ½ * C V2 = Q2 /2C • Eqn 26.13 is the concept of energy density of the electric field u = ½ εo E2

7. Dielectrics • When nonconducting material is inserted in a capacitor, in between the plates – the voltage drops. • That means since C=Q/V that C increases • C = κCo • ε = κεo – permittivity of free space • The equations for capacitors are the same except using the permittivity of the material

8. Dielectric Materials • Dielectrics have either polar or non polar molecules. In the electric field the non polar molecules tend to become polarized • The electric field is E = Eo/κ = σ/κεo between 2 plane sheets • E becomes the difference between the original Eo and E induced (opposite direction to Eo ) • Study examples 26.7 & 8

9. Capacitor Circuits • 2 capacitors in parallel – ΔV is the same and the total charge is the sum of each • C = C1 + C2+ …. • 2 capacitors in series – Q is the same and ΔV is the sum of the individual ΔVs. • 1/C = 1/C1 + 1/C2 +… Parallel Series

10. Solving for Equivalents • Reduce the number of real capacitors down to equivalent capacitors, combining simple parallel and series combinations C6 C1 C2 C5 a c a a C3 = C3 b = C4 b b a C4 = C4 C7 b

11. Example 1 • 4 capacitors arranged in previous slide • potential is 10V above ground for a, ground for b. Find a) voltage and charge on each capacitor b) find voltage on pt c wrt ground. • C1 = 12nF, C2 = 24nF, C3 = 22nF, C4=15nf

12. Example 1 • C1 and C2 in series – combine to a C5 1/C5 = 1/C1+ 1/C2 = 8 nF • C3 and C5 are parallel – combine to C6 C6 = C5 + C3 = 30nF • C6 and C4 are series – combine to C7 1/C7 = 1/C6 + 1/C5 = 10nF • V for C7 = 10V, Q=CV = (10nF)(10) = 100nC

13. Example 1 A) V7 = V Q7 = C7V7 Q7 = 100nC Q6 = Q7 Q6 = C6V6 V6 = 3.33V Q4=Q7 Q4=C4V4 V4=6.667V V3 = V6 Q3=C3V3 Q3 =73.33nC V5 = V6 Q5=C5V5 Q5=26.67nC Q2=Q5 Q2=C2V2 V2=1.111V Q1=Q5 Q1=C1V1 V1=2.22V b) Vc=V2+V4

14. Example 2 • A parallel plate capacitor made from 2 squares of metal, 2mm thick and 20cm on a side separated by 1mm with 1000V between them • Find: • a) capacitance b)charge per plate c) charge density d)electric field e) energy stored f) energy density

15. Example 2 • Given: • L = 20cm = 0.2m • d = 1mm = 1E-3m • V= 1E3 V • epsilon = 8.85E-12 F/m • Equations: • C=εo A/d, A=L2 , Q=CV, σ = Q/A, V=-Ed, U=1/2 QV, u= U/Ad =1/2 εo E2

16. Example 2 • a) C=εo A/d = (8.85E-12)(0.2)2/(1E-3) = 0.354nF • b) Q=CV = (0.354nF)(1E3V) = 0.354 μC • c) σ = Q/A = (0.354 μC)/(0.04 m2 ) = 8.85E-6 C/ m2 • d) V=Ed, E = V/d = 1000/0.001 = 1.0E6 V/m (this is a magnitude) • e) U= (1/2)(0.354E-6)(1000)=1.77E-4J • f) u = (1/2)(8.85E-12)(1.0E+6) 2 =4.425J/m3