Midterm Exam 1 Solutions - Electricity and Magnetism I | PHY 481, Exams of Physics

Download Midterm Exam 1 Solutions - Electricity and Magnetism I | PHY 481 and more Exams Physics in PDF only on Docsity! 1 PHY481 - Midterm I (2009) Time allowed 50 minutes. Do all questions - to get full credit you must show your working. The general solutions to Laplace’s equation with two co-ordinates allowed to vary are: V (x, y) = (a+ bx)(c+ dy) + ∑ k[A(k)cos(kx) +B(k)sin(kx)][C(k)cosh(ky) +D(k)sinh(ky)] (Cartesian); V (s, φ) = (A+Bln(s)) + ∑∞ n=1(Ans n + Bnsn )(Cncos(nφ) +Dnsin(nφ)) (Cylindrical); V (r, θ) = ∑ l=0(Alr l + Bl rl+1 )Pl(cosθ) (Spherical polar). Problem 1. Write down Gauss’s law in integral form and derive the differential form of Gauss’s law from it. Write down Faraday’s law in integral form and derive the differential form from it. Solution Gauss’s law in integral form is, ∮ ~E · d~a = q 0 (1) Using the divergence theorem and writing, q = ∫ ρ(~r)d~r, we find,∫ (~∇ · ~E)d~r = 1 0 ∫ ρ(~r)d~r (2) This is satisfied if ~∇ · ~E = ρ/0, the differential form of Gauss’s law. Faraday’s law is,∮ ~E · d~l = −∂φB ∂t (3) Using Stokes theorem and writing φB = ∫ ~B · d~a, we find,∫ ~∇∧ ~E · d~a = − ∂ ∂t ∫ ~B · d~a (4) This is satisfied if ~∇∧ ~E = −∂ ~B∂t , the differential form of Faraday’s law. Problem 2. A charge, q at position ~r′ = (0, 0, d) is above a conducting half space that lies in the region z < 0 so that its surface is at z = 0. Using the image charge method in Cartesian co-ordinates, write down the potential in the region z > 0 and show that the electric field is normal to the surface of the conductor. Find an expression for the induced charge at the surface of the conductor (do not integrate it) and find the work required to bring the charge from ∞ to the position ~r′. Solution The solution is constructed by noting that we can satisfy the boundary condition V (x, y, 0) = 0 with the function, V (~r) = k[ q (x2 + y2 + (z − d)2)1/2 − q (x2 + y2 + (z + d)2)1/2 ]. (5) Using ~E = −~∇V , we find, Ex = − ∂V ∂x = kq[ x (x2 + y2 + (z − d)2)3/2 − x (x2 + y2 + (z + d)2)3/2 ] (6) This reduces to zero at z = 0 as required. A similar expression is found for Ey, while the electric field in the z-direction is, Ez = − ∂V ∂z = kq[ z − d (x2 + y2 + (z − d)2)3/2 − z + d (x2 + y2 + (z + d)2)3/2 ] (7) In the limit z → 0, this gives Ez(z = 0) = − 2kqd(x2+y2+d2)3/2 . 2 The “screening” or induced charge density at the surface of the conductor is σ(x, y, 0) = 0Ez(x, y, 0) which yields, σ(x, y, 0) = −dq 2π(x2 + y2 + d2)3/2 (8) where k = 1/(4π0) was used. The force on the charge q, ~Fq = q ~E−q(x = 0, y = 0, d) = −kq2 4d2 ẑ. (9) This is equal to kqq′/(2d)2ẑ as it must. The work is W = − ∫ d ∞ −kq2 4z2 dz = −kq2 4d (10) This is one half the naive (incorrect) energy, −kq2/2d, because the electric field is finite only in the upper half plane. Problem 3. A grounded conducting cylinder of radius R has its axis along the ẑ direction and is in a uniform electric field ~E = E0x̂. Find an expression for the potential for s > R. Solution The electric field produces potential −E0x = −E0rcosφ. To satisfy this boundary condition, we need the n = 1 term in the solution to Laplace’s equation in cylindrical co-ordinates. We can therefore first see of only the n = 1 term is needed so that, V (s, φ) = (A1r + B1 r )cosφ (11) We can’t have the sinφ term because it is not consistent with the applied field. To satisfy the applied field boundary condition, we set A1 = −E0. To ensure that the potential is zero at the surface of the grounded cylinder, we set V (R,φ) = 0, which implies that B1 = R2E0. The solution is then, V (s, φ) = −E0cosφ(r − R2 r ) (12) We have to check that this has the property that E||(R,φ) = 0. We find E||(R,φ) = − 1 s ∂V ∂φ |s=R = −E0sinφ(R− R2 R ) = 0 (13) Problem 4. Using the multipole expansion, 1 |~r − ~r′| = ∑ l=0 (r′)l rl+1 Pl(cosθ) (14) Find an expression for the electrostatic potential of a dipole that has dipole moment ~p = qdẑ and is centered at the origin. From your expression for the potential, find the electric field of the dipole. Solution For a dipole charge configuration, we have, V (r, θ) = kq r ∑ l ( d 2r )lPl(cosθ)− kq r ∑ l=0 ( −d 2r )lPl(cosθ) (15) The even terms in the sum cancel, while the odd terms add so that, V (r, θ) = kq r ∑ l odd 2( d 2r )lP1(cosθ) = kq r [ d r cosθ + 2( d 2r )3P3(cosθ) + ...]. (16)