Electromagnetism - Lecture 5. Capacitors & Electrostatic Energy

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1 Electromagnetism - Lecture 5 Capacitors & Electrostatic Energy Examples of Capacitors Calculations of Capacitance Electrostatic Energy Introduction of Dielectrics General Result for Electrostatic Energy Density 1

2 Capacitors A capacitor is formed from two conductors with equal and opposite surface charges +σ and σ separated by an insulating gap. Capacitance C is the ratio of the total charge Q on each conductor to the potential difference V across the gap: C = Q V = σa V The unit of capacitance is the Farad F = C/V Practical capacitors are between pf and µf 2

3 Parallel Plate Capacitor The electric field for infinite plates is obtained from Gauss s Law or by using the superposition of two uniform fields There is a uniform field in the gap of width d: E gap = σ ɛ 0 ˆd The field outside the plates is zero! The potential difference and capacitance are: V = Ed = Qd C = Q Aɛ 0 V = Aɛ 0 d Note that C is a purely geometric property of the plates! 3

4 Notes: Diagrams: 4

5 Cylindrical Capacitor Two concentric conducting cylinders of length l and radii a and b, carry line charges +λ and λ Use Gauss s Law assuming l a, b: E gap = There is no field for r < a or r > b! λ 2πɛ 0 rˆr The potential difference and capacitance are: V = b a E.dr = C = Q V = Again C is a purely geometric property Q 2πɛ 0 l ln(b/a) 2πɛ 0l ln(b/a) 5

6 Energy Stored in a Capacitor Work is done to assemble charges ±Q on capacitor plates Q W = V dq = C dq This work is stored as electrostatic energy U E = 1 2 For a parallel plate capacitor: Q 2 C = 1 2 QV = 1 2 CV 2 U E = ɛ 0AV 2 2d This can be written in terms of the electric field between the plates: U E = ɛ 0(Ad)E 2 2 6

7 Notes: Diagrams: 7

8 Electrostatic Energy Density Electrostatic Energy is stored in a capacitor through the creation of the Electric field in the gap The energy density of an electric field is proportional to the square of its amplitude: du E dτ = 1 2 ɛ 0 E 2 A useful exercise is to prove this gives the correct electrostatic energy for a cylindrical capacitor 8

9 Electrostatic Energy of Nucleus A Uranium nucleus has Z = 92 protons and N = 146 neutrons uniformly distributed over a radius R m Electric field of nucleus: E(r < R) = Zer Ze ˆr E(r > R) = 4πɛ 0 R3 4πɛ 0 r ˆr 2 Total electrostatic energy by integration over energy density: du E = 1 2 ɛ 0 E 2 dτ U E = R 0 ( ) 2 Zer 4πɛ 0 R 3 2πɛ 0 r 2 dr + R ( ) 2 Ze 4πɛ 0 r 2 2πɛ 0 r 2 dr U E = (Ze)2 40πɛ 0 R + (Ze)2 8πɛ 0 R = J = 730MeV 9

10 Energy of Nuclear Fission A symmetric U fission creates two Pd daughter nuclei Note that fission actually prefers to be asymmetric! Nuclear radii obey R A 1/3 where A = Z + N Electrostatic energy of daughter nuclei compared to 238 U: ( ) U E UE = 2 = 0.63U 4(0.5) 1/3 E = 460MeV Predicted release of electrostatic energy in fission is 270 MeV Observed release is about 200 MeV 10

11 Notes: Diagrams: 11

12 Dielectrics in Capacitors Capacitance depends on the insulating material in the gap C = ɛ r C 0 where C 0 is the result for a vacuum ɛ r 1 is the dielectric constant of the material Energy stored in capacitor is increased by dielectric material: U = 1 2 CV 2 = ɛ r U 0 Electrostatic energy density is proportional to ɛ r du E dτ = 1 2 ɛ rɛ 0 E 2 12

13 Simple Model for Dielectrics If an electric dipole is placed between the capacitor plates it aligns itself with the electric field in the gap p E Potential energy of dipole: U = p.e The energy stored between the plates is increased by this amount In dielectric materials the atoms or molecules become polarized with intrinsic electric dipole moments pointing in the direction of the external field. This will be explained in more detail in a later lecture 13

14 Proof of Electrostatic Energy Density A set of n 1 charges at positions r j gives a potential at r i : V (r i ) = 1 4πɛ 0 n 1 j=1 Q j r i r j Work done to bring up a charge Q i from infinity to point r i : W i = Q i V (r i ) = Q i 4πɛ 0 n 1 j=1 Q j r i r j The total energy stored in the system of n charges is: U E = 1 4πɛ 0 n i=1 j<i Q i Q j r i r j = 1 8πɛ 0 n i=1 n j=1 Generalize to a double integral over the charge density ρ: U E = 1 8πɛ 0 ρ(ri )ρ(r j ) r i r j dτ idτ j Q i Q j r i r j 14

15 Notes: Diagrams: 15

16 Perform one integral over dτ to get the potential V : U E = 1 2 ρ(r)v (r)dτ Then use Poisson s equation to eliminate the other ρ: U E = ɛ 0 2 V 2 V dτ Now the tricky bit - integrate this by parts: 0 V.( V )dτ = [V ( V )] 0 0 ( V )( V )dτ Apply the boundary conditions V ( ) = 0 and V (0) = 0: U E = ɛ 0 2 ( V ) 2 dτ = ɛ 0 2 This proof comes from Jackson P E 2 dτ 16

Chapter 7: Polarization

Chapter 7: Polarization Joaquín Bernal Méndez Group 4 1 Index Introduction Polarization Vector The Electric Displacement Vector Constitutive Laws: Linear Dielectrics Energy in Dielectric Systems Forces