Electromagnetism Pysics 15b Lecture #6 Conductors Capacitance Purcell 3.2 3.5 Wat We Did Last Time Defined curl of vector field by In Cartesian: curlf = F = ˆx F z y F y ŷ F x z z F z x ẑ F y x F x y Stokes Teorem: Studied (ideal) conductor in electrostatic equilibrium E = 0 inside, ϕ = const, no net carge density inside F ds curlf ˆn lim C a 0 a FdV = F ds surface loop 1
Today s Goals Continue on electrostatics wit conductors Wat is te electric field near a conductor surface? in a space surrounded by a conductor? in between multiple conductors? A few examples in wic te field is easily calculable Concentric speres Plane conductor and a point carge Define capacitance, if time allows Last 15 minutes: Prof. Mazur s researc on ow students learn electromagnetism A 7-minute exercise of puzzle solving Not a quiz It s voluntary and confidential Electric Field Near Surface Consider a conductor wit surface carge density σ Wat is te E field immediately outside? We know te surface is an equipotential E field must be perpendicular to it area = A Draw a cylinder alf buried into te surface Carge inside te cylinder is σa E = 0 at te bottom (inside) E is parallel to te side of te cylinder Gauss s Law: 4πσ A = E da = EA E = 4πσ ˆn top were ˆn is to te surface E 2
Hollow Conductor Electric field inside an empty space completely surrounded by a conductor is zero Te inner surface of te conductor must be an equipotential φ = φ 0 = const No carge inside φ satisfies Laplace s Equation 2 φ = 0 E = 0 φ = φ 0 everywere inside is a solution, wic by te Uniqueness Teorem is te only solution E = φ = 0 Conducting case is used to sield te content from te effects of external electric fields Small oles in te wall OK Metallic screen works Slowly canging external field OK Multiple Conductors Multiple pieces of conductors C 1, C 2,, C N are arranged inside a conducting boundary B B B could be infinitely large, i.e., does not exist E field inside B is uniquely determined C 1 if te potential φ 1, φ 2, φ N, and φ B are C given 2 C 3 Actually obtaining te solution for Laplace s Equation wit 3-d boundary conditions may not be so easy Specifying carges q 1, q 2, q N, and φ B is also sufficient Pysically reasonable, but sligtly arder to prove Let s consider a few simple systems first 3
Concentric Speres A solid spere inside a ollow spere We know E = 0 for r < a and b < r < c For a < r < b, E = Q 1 ˆr r 2 On te r = a surface: σ = E ˆr 4π = Q 1 4πa 2 (from Gauss) Makes sense: Q 1 distributed uniformly over 4πa 2 On te r = b surface: σ = E (ˆr) = Q 1 4π 4πb 2 Q 1 is distributed uniformly over 4πb 2 Te carge cancellation is expected Consider an imaginary spere at a radius between b and c It contains Q 1 Q 1 = 0 total carge, and surface field E = 0 Q 1 a b c Q 2 Concentric Speres E r Outer surface at r = c must old carge Q 1 Q 2 Field outside: E ˆr = 4πσ = Q Q 1 2 c 2 i.e., Coulomb field wit Q 1 Q 2 φ r r 4
How About Tis? Q 2 Q 1 Conductive Plane A point carge Q is placed above a large conductive plane It attracts negative carge distribution on te plane s surface Q Field lines from Q terminate on te induced negative carge distribution Need to solve Laplace s Equation above te plane, wit a singularity at Q and a flat equipotential surface Generally a ard problem, but tere is a trick for tis one 5
Image Carge Tecnique Consider two carges Q and Q separated by 2 Coulomb field x 2 Easy enoug to calculate E is vertical on te mid-plane Te mid-plane is terefore an equipotential Te field above te mid-plane satisfies te boundary condition of te conductive plane problem By te Uniqueness Teorem, it s te solution z Q Tis tecnique works in a problem wit one (infinite) conductive plane Te Q carge is called te image carge of te original Q Surface Carge E field on te mid-plane points down E z (z = 0) = 2 Q r 2 cosθ = 2Q 2 (r 2 2 ) 3 2 It s related to te surface carge density ρ(r) = E z 4π = Q 2π(r 2 2 ) 3 2 If we integrate over te entire surface, we get Qr ρ(r) 2πrdr = 0 dr = 0 (r 2 2 ) 3 2 = Q Q r 2 2 0 z Q θ r Q r 6
Capacitance A conductor wit carge Q as a potential φ 0 w.r.t. infinity φ 0 = surface E ds Q because E is proportional to Q We define capacitance by te ratio C Q φ 0 It is determined by te sape of te conductor Example: a sperical conductor wit radius a E = Q r ˆr (for r > a) φ 2 0 = Q C = a a Dimension of capacitance = (lengt) Unit of capacitance = cm Capacitance Capacitance can also be defined for a pair of conductors Tis is in fact muc more common Give two conductors Q and Q and measure te potential difference φ 1 φ 2 = Capacitance is defined by 1 2 E ds Q Q C = φ 1 φ 2 Most useful case: two plates wit a small gap in between Q Q area A s 7
Parallel-Plate Capacitor If A is large (s is small), ten tis is similar to te double infinite seets of carge we saw in Lecture 3 E is uniform between te plates E top E = 4πσ = 4πQ Q/A A Potential difference is φ Q/A 1 φ 2 = Es = 4πQs A Capacitance is C = Q φ 1 φ 2 = A 4πs E bottom Real World Capacitors Capacitors appear everywere in electronics Computer memory (dynamic RAM) is an array of tiny capacitors Will do more on tis later In electronics, we must use SI Carge is in coulomb (C), potential is in volt (V) coulomb (C) SI unit for capacitance is farad (F) = volt (V) Conversion from CGS: 1 farad = 9 x 10 11 cm Since 1 farad is suc a large unit, we often use 1 µf = 10 6 F 1 pf = 10 12 F Ex: a sperical conductor wit a 10 cm radius as a capacitance of 10 cm = 11 pf 8
Summary Studied electric field near and around conductors Just outside a conductor surface, E = 4πσ ˆn In an empty space inside a conductor, E = 0 Concentric speres Image carge tecnique for carge(s) near a conductive plane Defined capacitance Isolated conductor: C Q relative to infinity φ 0 z Q Q Pair of conductors: Parallel-plate capacitor: Unit: cm (CGS) or farad (SI) C = Q (φ 1 φ 2 ) C = A 4πs area A s 9