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PHY481: Electromagnetism Exam 1 Solutions Charge on a conductor Lecture 14 Carl Bromberg - Prof. of Physics Exam 1 1) Stokes’s theorem: ∫ S (∇ × F ) ⋅ n̂ dA = ∫C F ⋅ d z F(x) = î × (xî + yĵ + zk̂) 1) d = ĵdy, z = 0, 2) d = k̂dz, y = 1 4) d = k̂dz, y = 0 1 a) d 2 = k̂dz d 4 = k̂ dz 1 0 0 0 1 1 x 1 y d 1 = ĵdy ⎛ ∂Fz ∂Fy ⎞ ⎛ ∂Fx ∂Fz ⎞ ⎛ ∂Fy ∂Fx ⎞ ∇ × F = î ⎜ − + ĵ ⎜⎝ − − = î(2) + ĵ( 0 ) + k̂(0) ⎟⎠ + k̂ ⎜ ⎟ ⎟ ⎝ ∂y ⎝ ∂x ∂z ⎠ ∂z ∂x ∂y ⎠ c) n̂ = î; dA = dxdy ; ∫ (∇ × F ) ⋅ n̂ dA = 2 S Lecture 14 F1 = yk̂ − 0 ĵ 0 ∫3 F ⋅ d = −1∫ dy = 1; ∫4 F ⋅ d = −0 ∫ dz = 0 C F2 = k̂ − zĵ F4 = 0k̂ − zĵ ∫1F ⋅ d = −0 ∫ dy = 0 ; ∫2 F ⋅ d = 1∫ dz =1; b) ∫ F ⋅ d = 2 d 3 = ĵ dy 1 = yk̂ − zĵ 3) d = ĵdy, z = 1, F3 = yk̂ − ĵ ∴ ∫ (∇ × F ) ⋅ n̂ dA = ∫ F ⋅ d = 2 S Carl Bromberg - Prof. of Physics C 1 Exam 1 (cont’d) 2) a) integral to obtain scalar potential from charge distribution 1 V (x) = 4πε 0 3 ρ( x ′ )d x ′ ∫ x − x′ b) differential equation relating charge density and the electric field ∇ ⋅ E(x) = ρ(x) ε0 c) differential equation relating electric field and potential E(x) = −∇V (x) d) integral to obtain electric field from the charge distribution E(x) = Lecture 14 1 4πε 0 x − x′ ∫ x − x′ 3 ρ( x ′ )d x ′ 3 Carl Bromberg - Prof. of Physics 2 Exam 1 (cont’d) 3a) Gauss’s law: b) ∫ S E ⋅ dA = qencl ε0 ET (x) = E0k̂; E B (x) = − E0k̂ θ n̂T = k̂; n̂ B = −k̂ ∫ E ⋅ dA = ∫ E0k̂ ⋅k̂ dA + ∫ ( − E0k̂ ) ⋅( −k̂ dA) S T R E = E0k̂ n̂ = r̂ n̂ = k̂ σ σ B 2E0 A = qencl ε 0 = σ A ε 0 c) n̂ = − k̂ E0 = σ 2ε 0 E = − E0 k̂ 2 n̂ = r̂; dA = R sin θ dθ dφ ; qencl = σπ R 2 ⎡ π2 ⎤ π 2 ∫ E ⋅ dA = ∫ ( E0k̂ ) ⋅r̂ dA + ∫ ( − E0k̂ ) ⋅r̂ dA = 2π E0 R ⎢⎢ ∫ dθ cosθ sinθ − ∫π dθ cosθ sinθ ⎥⎥ S T B ⎣0 ⎦ 2 ⎡ 2 1⎤ 1 0 ⎡ ⎤ x ⎥ 2 2 2 = 2π E0 R ⎢ ∫ xdx − ∫ xdx ⎥ = 2π E0 R ⎢ 2 = 2π E0 R ⎢⎣ 2 0 ⎥⎦ ⎣0 ⎦ 1 2 2 2π E0 R = qencl ε 0 = σπ R / ε 0 ∴ E0 = σ 2ε 0 Lecture 14 Carl Bromberg - Prof. of Physics 3 Exam 1 (cont’d) 4a) 3 q = ∫ ρ( x ′ )d x ′ = ∫ λ d = b) E(x) = x − x′ 1 4πε 0 ∫ x − x′ π 2 ∫ Rdφ = π Rλ −π 2 3 ρ( x ′ )d x ′ 3 2 2 32 ( x − x ′ = zk̂ − R ( î cos φ + ĵsin φ ) = z + R ) 1 zk̂ − R ( î cos φ + ĵsin φ ) E(x) = λ Rdφ ∫ 3 2 4πε 0 ( z 2 + R2 ) c) 3 3 1 ⎛ = ⎜ 4πε 0 ⎝ ( = 1 4πε 0 ⎞ ( î cos φ + ĵsin φ ) dφ ⎟ π zk̂ + ∫ 3 2 3 2 2 2 ⎠ ( z 2 + R2 ) −π 2 z +R ) λR λR λR ( z2 + R ) d) Ez (x) = Lecture 14 2 32 1 4πε 0 (π zk̂ − 2Rî ) R z → 0 (i.e., z >> R) πλ Rz (z 2 π 2 2 +R ) 2 32 = q 4πε 0 z ( z 1+ ( R z) 3 ) 2 32 Ez (x) ≈ q 4πε 0 z 2 Carl Bromberg - Prof. of Physics 4 Exam 1 (cont’d) 5) Identity for ∇ ⋅ ( A × B ) using ∇ × A and ∇ × B C = A × B where Ci = ε ijk Aj Bk ∇ ⋅ (C) = ∂Aj ∂B ∂ = ε ijk Aj Bk = Bk ε ijk + Aj ε ijk k ∂xi ∂xi ∂xi ∂xi ∂Ci ( ) Make replacements ε kij = ε ijk and − ε jik = ε ijk ∇ ⋅ ( C ) = Bk ( ε kij ) ∂Aj ∂xi ( + Aj −ε jik ∂Bk ) ∂x = Bk ( ∇ × A )k − Aj ( ∇ × B ) j i = B ⋅ (∇ × A ) − A ⋅ (∇ × B ) Lecture 14 Carl Bromberg - Prof. of Physics 5 Exam 1 (cont’d) 6a) Dipole moment 3 p = ∫ rρ(x)d x r̂ = k̂ cosθ + sin θ ( î cos φ + jsin φ ) 2 dA = R sin θ dθ dφ p = 2π R 3 ∫ r̂ (σ 0 cosθ ) sinθ dθ π 2π ⎡π ⎤ 2 3 2 = 2π R σ 0 ⎢ ∫ k̂ cos θ sin θ dθ ⎥ + R σ 0 ∫ cosθ sin θ dθ ∫ dφ ( î cos φ + jsin φ ) ⎣0 ⎦ 0 0 3 −1 ⎡ x3 ⎤ 3 2 3 = 2π R σ 0k̂ ∫ −x dx = 2π R σ 0k̂ ⎢ − ⎥ ⎣ 3 ⎦1 1 −1 3 = 4π R σ 0 3 k̂ b) Top is positive, bottom is negative, the same as a dipole c) From far away, potential is that of a point dipole : V (r,θ ) = Lecture 14 Carl Bromberg - Prof. of Physics p ⋅ r̂ 4πε 0 r 2 6 Exam 1 (cont’d) 7a) Prove ∞ f (− b a) 3 (d x should be dx, answer OK) a ∫ δ (ax + b) f (x)dx = −∞ nd x ′ = ax + b; dx = dx ′ a ; 2 integral is f at x ′ = 0 ∞ ∞ 1 f (− b a) δ (ax + b) f (x)dx = δ ( x ) f ( x a − b a)d x = ′ ∫ ∫ ′ ′ a a −∞ −∞ ∞ b) Prove 3 −1 3 ∫ δ (Ax + b) f (x)d x = f (−A b) / det A Jacobian det A −∞ −1 −1 x = A x′ − A b 3 3 d x = d x ′ Det A ∞ 3 ∫ δ (Ax + b) f (x)d x = −∞ ∞ −1 −1 3 ∫ δ (x′) f (A x′ − A b) d x′ Det A −∞ −1 = f (−A b) Det A Lecture 14 Carl Bromberg - Prof. of Physics 7 Conductors in static equilibrium Why these statements are true ? – Electric field E inside a conductor is zero. – Potential V inside a conductor is a constant. – Inside a conductor the charge density ρ is zero. – Charge Q on a conductor resides only on the surface. – A net charge Qnet here is always paired with charge -Qnet elsewhere. – On a conductor either V or Qnet, but not both, can be specified. – At a conductor’s surface, field component Et = 0, component En = σ /ε0 , but in force calculations, En = σ /2ε0 due to only “distant” charges. – In an empty cavity in a conductor, E = 0 and surface σcavity = 0. – The potential V of a conductor can be set by a battery VB. – The earth is an “infinite” source of charge at a constant V . – V or Qnet of conductors (and Qexternal) determine V everywhere. Lecture 14 Carl Bromberg - Prof. of Physics 8 Boundary value problems: 1-parallel plates Parallel plates (find potential, field & surface charge) The ground is a constant potential, so let V1 = 0 . Battery draws electrons from plate 2 and puts them on plate 1. Process “continues” until V2 = VB . Potential between the plates Between the plates the potential depends only on x. Laplace’s equation: ∇ 2V ( x ) = 0 Boundary conditions: V (0) = 0; V (d) = VB General solution (ODE) Apply boundary conditions d ⎛ dV ⎞ dV V (0) = c2 = 0 ∇V= = c1 ⎜⎝ ⎟⎠ = 0; dx dx dx VB V (d) = c1d = VB ; c1 = V (x) = c1x + c2 d 2 Lecture 14 Potential x V (x) = VB d Carl Bromberg - Prof. of Physics 9 Electric field Parallel plates (cont’d) ⎛ E = −∇V = −∇ ⎜⎝ VB En -> Surface charges Electric field x ⎞ VB ⎟⎠ − î d d Surface 1r charge density ε 0VB σ 1r = − d σ 1r VB E1n = =− ε0 d Surface 2 charge density σ 2 VB E2n = = ε0 d σ 2 Surface charge Q2 = −Q1r = σ A = Lecture 14 ε 0VB A d ε 0VB =+ d Force on each plate F2 = Q2 (E2n ) non-q 2 2 σ2 1 = A 2 ε0 Carl Bromberg - Prof. of Physics 10 Parallel plates (cont’d) Capacitance Two ways to know the electric field 1 1 1r VB σ E= ; E= ε0 d 2 2r E Capacitance is a geometrical factor relating the charge on a conductor and its potential. ε0 A C= d ε0 A Q=σA= VB = CVB d d 0 V = VB V =0 – x VB + ground Energy storage 2 ε0 ε 0 VB 2 3 1 ε0 A 2 U = ∫ E d x′ = Ad = VB 2 2 2 d 2 d Lecture 14 2 1 2 U = CVB 2 Carl Bromberg - Prof. of Physics 11