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NAME _____________________ STUDENT NO. ___________________ PHY481 Exam 1 NO books, notes, calculators, cell phones Put your NAME & SN on the front Page Show all work on the question page One sheet of scrap paper discarded at the end of the exam. Notations and Formulas Coordinates: Unit vectors: Cartesian î , ĵ, k̂ Cylindrical r̂, k̂ Spherical r̂, θ̂, φ̂ Position vector x = xî + yĵ + zk̂ rr̂ + zk̂ rr̂ Einstein notation for Cartesian coordinates (repeated indices are summed over) F ⋅ G = FiGi ; F × G = ε ijk F j Gk ; ε ijk ε klm = δ ilδ jm − δ imδ jl (corrected) ∇⋅F = ∂Fi ∂xi ; ( ∇ × F )i = ε ijk ∂Fk ∂x j Divergence and curl in curvilinear coordinates 1 ⎛ ∂ ( h2 h3 F1 ) ∂ ( h1h3 F2 ) ∂ ( h1h2 F3 ) ⎞ ∇⋅F = + + h1h2 h3 ⎜⎝ ∂u1 ∂u2 ∂u3 ⎟⎠ ∂ hF ⎞ ⎛∂ h F ( ∇ × F )i = 1 ⎜ ( k k ) − ( j j ) ⎟ hh ∂u ∂u j k ⎝ j k ⎠ Integrals 1 NAME _____________________ STUDENT NO. ___________________ 1) [15 pts] Consider a vector function F(x) = î × x , and let S be the unit square in the y-z plane, as shown in the figure a) [5 pts] Calculate the line integrals ∫ F ⋅ d for each of the four segments of the unit square. b) [5 pts] Calculate the loop integral ∫ F ⋅ d z F3 = yk̂ − ĵ d 3 = ĵ dy 1 F2 = k̂ − zĵ F4 = 0k̂ − zĵ d 2 = k̂dz d 4 = k̂ dz C c) [5 pts] Verify that this loop integral agrees with Stokes’s theorem: ∫ (∇ × F ) ⋅ n̂ dA = ∫ F ⋅ d . S C F1 = yk̂ − 0 ĵ x d 1 = ĵdy ANS. F(x) = î × (xî + yĵ + zk̂) = yk̂ − zĵ 1) d = ĵdy, z = 0, 2) d = k̂dz, y = 1 3) d = ĵdy, z = 1, a) 4) d = k̂dz, y = 0 1 1 0 0 0 0 1 1 ∫1F ⋅ d = −0 ∫ dy = 0 ; ∫2 F ⋅ d = 1∫ dz =1; ∫3 F ⋅ d = −1∫ dy = 1; ∫4 F ⋅ d = 0 ∫ dz = 0 b) ∫ F ⋅ d = 2 C ⎛ ∂F ∂Fy ⎞ ⎛ ∂Fx ∂Fz ⎞ ⎛ ∂Fy ∂Fx ⎞ + ĵ ⎜⎝ − − c) ∇ × F = î ⎜ z − ⎟⎠ + k̂ ⎜ ⎟ ⎟ = î(2) + ĵ( 0 ) + k̂(0) ⎝ ∂y ⎝ ∂x ∂z ⎠ ∂z ∂x ∂y ⎠ n̂ = î; dA = dxdy ; ∫ (∇ × F ) ⋅ n̂ dA = 2 S ∴ ∫ (∇ × F ) ⋅ n̂ dA = ∫ F ⋅ d = 2 S C 2 1 y NAME _____________________ STUDENT NO. ___________________ 2) [12 pts] Consider electric field and potential determined by a charge density ρ(x) . Answer the following questions clearly identifying any vector functions or variables. a) [3 pts] The scalar potential V (x) is determined by what integral over the source position x ′ . 1 V (x) = 4πε 0 3 ρ( x ′ )d x ′ ∫ x − x′ b) [3 pts] A charge density ρ(x) and the resulting electric field E(x) are related by what partial differential equation? ∇ ⋅ E(x) = ρ(x) ε0 c) [3 pts] An electric field E(x) and the scalar potential V (x) are related by what differential equation? E(x) = −∇V (x) d) [3 pts] The electric field E(x) is determined by what integral over the source position x ′ E(x) = 1 4πε 0 x − x′ ∫ x − x′ 3 ρ(x′)d 3 3 x′ NAME _____________________ STUDENT NO. ___________________ 3) [13 pts] The integral form of Gauss’s law relates the electric flux through a closed surface S to the charge enclosed by that surface. a) [2 pts] What is the integral form of Gauss’s law? b) [3 pts] Use Gauss’s law and symmetry to determine the electric field E(x) above and below an infinite sheet of charge with a constant charge density σ . c) [8 pts] Using a spherical Gaussian surface, again determine the electric field E(x) of an infinite sheet of charge with constant charge density σ . a) ∫ E ⋅ dA = qencl ε0 S θ R E = E0k̂ n̂ = r̂ n̂ = k̂ σ σ n̂ = −k̂ E = −E0k̂ b) ET (x) = E0k̂; E B (x) = − E0k̂ and n̂T = k̂; n̂ B = −k̂ ∫ E ⋅ dA = ∫ E0k̂ ⋅k̂ dA + ∫ ( − E0k̂ ) ⋅( −k̂ dA) = qencl S T ε0 = σ A ε0 B 2E0 A = σ A ε 0 E0 = σ 2ε 0 c) 2 n̂ = r̂; dA = R sin θ dθ dφ ; qencl = σπ R 2 ∫ E ⋅ dA = ∫ E0k̂ ⋅r̂ dA + ∫ ( − E0k̂ ) ⋅(k̂ ⋅r̂ dA) S T B ⎡ ⎤ π 2 = 2π E0 R ⎢ ∫ dθ cosθ sin θ − ∫ dθ cosθ sin θ ⎥ ⎢0 ⎥ π ⎣ ⎦ 2 1 0 ⎤ 2⎡ = 2π E0 R ⎢ ∫ xdx − ∫ xdx ⎥ ⎣0 ⎦ 1 1 ⎡ 2 ⎤ x ⎥ 2 2 = 2π E0 R ⎢ 2 = 2π E0 R ⎢⎣ 2 0 ⎥⎦ σ 2 2 2π E0 R = qencl ε 0 = σπ R / ε 0 ∴ E0 = 2ε 0 π 2 4 NAME _____________________ STUDENT NO. ___________________ 4) [20 pts] Consider a semicircle of charge with a linear charge density λ . a) [4 pts] Determine the total charge on the semicircle. b) [6 pts] Display the integral that will determine all components of the electric field E(x) on the z axis, AND draw on the figure the corresponding vector quantities. c) [6 pts] Determine the E(x) (vector) on the z axis. d) [4 pts] For z >> R , show that the z-component of field approaches the field of a point charge with a value equal to the total charge on the semicircle. a) 3 q = ∫ ρ( x ′ )d x ′ = ∫ λ d = 1 4πε 0 x − x′ π 2 ∫ Rdφ = π Rλ −π 2 ∫ x − x′ 3 ρ(x′)d 3 b) E(x) = c) x − x ′ = zk̂ − R ( î cos φ + ĵsin φ ) = ( z + R 3 3 2 ) 2 32 zk̂ − R ( î cos φ + ĵsin φ ) λ Rdφ ∫ ( z 2 + R2 )3 2 2 π 2 ⎞ 1 ⎛ λR λR ( ) = π z k̂ + î cos φ + ĵsin φ d φ ⎜ ⎟ ∫ 2 32 4πε 0 ⎝ ( 2 ⎠ ( z 2 + R2 )3 2 −π 2 z +R ) 1 λR (π zk̂ − 2Rî ) = 2 32 4πε 0 ( 2 z +R ) 1 πλ Rz q z Ez (x) = = 3 2 2 2 32 4πε 0 ( 2 4πε 0 3 z +R ) z 1+ ( R z) q Ez (x) ≈ R z → 0 (i.e., z >> R) 2 4πε 0 z This is field is the same as that of the point charge q . E(x) = d) x′ 1 4πε 0 ( 5 ) NAME _____________________ STUDENT NO. ___________________ 5) [10 pts] Use the Levi-Chivita tensor to determine an identity for ∇ ⋅ ( A × B ) involving ∇ × A and ∇ × B . Let C = A × B where Ci = ε ijk Aj Bk ∇ ⋅ (C) = ∂Ci ∂xi = ∂Aj ∂B ∂ ε ijk Aj Bk = Bk ε ijk + Aj ε ijk k ∂xi ∂xi ∂xi ( ) Make the replacements ε kij = ε ijk ∂Aj ( ) ∂x ∇ ⋅ ( C ) = Bk ε kij i ( + Aj −ε jik and − ε jik = ε ijk ∂B ) ∂xk i = Bk ( ∇ × A )k − Aj ( ∇ × B ) j = B ⋅ ( ∇ × A ) − A ⋅ ( ∇ × B ) 6 NAME _____________________ STUDENT NO. ___________________ 6) [15 pts] Consider a spherical shell with a surface charge density σ (x) = σ 0 cosθ , where θ is the standard polar angle. a) [8pts] Determine the dipole moment p of this charge distribution b) [2pts] State why the answer to a) makes sense. Hint: the dipole moment is a vector. c) [5 pts] Display an expression that represents the dominate term in the potential V (r,θ ) of this charge distribution in terms of the dipole moment p at a large distance from the shell. a) r̂ = k̂ cosθ + sin θ ( î cos φ + jsin φ ) 3 p = ∫ rρ(x)d x ; p= R 3 ∫ r̂ (σ 0 cosθ ) sinθ dθdφ 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 3 ⎡ x3 ⎤ 4π R σ 0 = 2π R σ 0k̂ ∫ −x dx = 2π R σ 0k̂ ⎢ − ⎥ = k̂ ⎣ 3 ⎦1 3 1 3 b) −1 2 3 Positive charge is displaced in the +z direction and negative charge in the –z direction, yielding a dipole moment in the +z direction. 2 Extra: The total charge in the upper half sphere, qT = π R σ 0 . Writing p = qT ”d”, the effective charge spacing ”d” = 4R 3 . Instead, with a constant charge density σ 0 (upper) and −σ 0 (lower) the total 2 charge in the upper half sphere is qT = 2π R σ 0 and the dipole moment is 3 p = π R σ 0 . The effective charge spacing ”d” = R / 2 , less than half the previous case. c) From far away this will look like a dipole along the positive z axis. Therefore, the dominate term will be the dipole term: p ⋅ r̂ V (r,θ ) = 2 4πε 0 r 7 NAME _____________________ STUDENT NO. ___________________ 7. [15 pts] ∞ ∫ δ (ax + b) f (x)dx = a) [10 pts] Prove −∞ f (− b a) 3 (d x should be dx, answer OK) a nd x ′ = ax + b; dx = dx ′ a ; 2 integral is f at x ′ = 0 ∞ ∫ δ (ax + b) f (x)dx = −∞ ∞ b) [5 pts] Prove ∫δ 3 ∞ 1 f (− b a) δ ( x ′ ) f ( x ′ a − b a)dx ′ = ∫ a −∞ a −1 3 (Ax + b) f (x)d x = f (−A b) / det A , −∞ where det A is the Jacobian of the transformation x ′ = Ax + b . 3 3 d x ′ = Det A d x ∞ 3 ∫ δ (Ax + b) f (x)d x = −∞ ∞ ∫ δ (x′) f (A −1 −1 3 x ′ − A b) d x ′ Det A −∞ −1 = f (−A b) Det A 8