PHY481 Homework 4: Electric Potential & Field of Point Charges and Cavities - Prof. Philli, Assignments of Physics

Download PHY481 Homework 4: Electric Potential & Field of Point Charges and Cavities - Prof. Philli and more Assignments Physics in PDF only on Docsity! PHY481 - Solutions to Homework 4 Problem 4.1 - Solve using a Gaussian surface inside the metal. The electric field everywhere is zero on this surface is zero, therefore the total enclosed charge is zero. This means that the induced charge on the metal surface adjacent the cavity must equal to the charge inside the cavity. This also implies that if there is no charge inside the cavity, the net charge on the metal surface adjacent the cavity is zero. Problme 4.2 We solved a two slab problem in lectures using superposition - that method extends to this case. The detailed solution is on closed reserve in the BPS library Problem 4.9 The real charge q is at position (a, a), while the three image charges are at (−a, a), (a,−a), (−a,−a). The contribution of each of the charges to the potential at position (y, z) is found using cartesian co-ordinates, yielding, V (y, z) = kq[ 1 [(y − a)2 + (z − a)2]1/2 − 1 [(y + a)2 + (z − a)2]1/2 + 1 [(y + a)2 + (z + a)2]1/2 − 1 [(y − a)2 + (z + a)2]1/2 ] (1) Each term needs to be expanded for large r = (y2 + z2)1/2. For example the expansion for the first term is 1 [(y − a)2 + (z − a)2] = 1 r(1 − 2a(y+z) r2 + 2a 2 r2 )1/2 (2) Then use 1/(1 + δ)1/2 = 1 − δ/2 + 3δ2/8 + ...., with δ = −2a(y+z) r2 + 2a 2 r2 . Note that δ is different for each of the four terms in the potential V (y, z). When this expansion is carried out for the four terms in the potential function and the results are summed, the leading term in the final result is V (y, z) = 12kqa2yz/r5. After transforming to polar co-ordinates, this is the same as the answer given in PS. The detailed working of the problem is on closed reserve in the BPS library. Problem 4.11 again requires expansion of the electrostatic potential. In this case the potential is given by, V (r) = kq[ 1 r − r0 − 1 r + r0 + 2 [(r + r0/2)2 + 3r20/4] 1/2 − 2 [(r − r0/2)2 + 3r20/4] 1/2 ] (3) 1 In this case expand using, 1/(1 + δ)1/2 = 1 − δ/2 + 3δ2/8 − 15δ3/48 + ..., and 1/(1 + δ) = 1 − δ + δ2 − δ3 + .... In this problem the first three terms, monopole, dipole, quadrupole, are all zero. Full details are on closed reserve in the BPS library. Problem 4.14 is relatively straightforward and requires calculating the energy in the electric field and comparing it to the capacitor formula CV 2/2 in order to extract C for concentric spheres. The electric field is finite only between the two spheres, where its magnitude is kQ/r2 the energy in the electric field is given by, U = ∫ ud3r = 1 2 ǫ0 ∫ b a 4πr2( kQ r2 )2dr = 1 2 ǫ0k 2Q24π[ 1 a − 1 b ] (4) Equating this result to U = CV 2/2, where V is the voltage difference between the shells ie V = kQ[1/a − 1/b] enables us to extract C = 4πǫ0ab/(b − a). Problem 4.15 involves a point charge near a sphere and the formulas required are found as follows (see also Lecture 12). The potential is the sum of the contributions of the real charge and the image charge, V (r, θ) = kq r1 + kq r2 = k[ q (r2 + z20 − 2rz0cosθ) 1/2 + q′ (r2 + z′20 − 2rz ′ 0cosθ) 1/2 ] (5) where as usual we used the cosine rule to find expressions for r1 and r2. The conjugate relations are z′0 = R 2/z0, q ′ = −qR/z0. Subsituting these expressions in the above equation and simplifying yields, V (r, θ) = kq[ 1 (r2 + z20 − 2rz0cosθ) 1/2 − R (r2z20 + R 4 − 2rz0R2cosθ)1/2 ] (6) The electric field is given by, ~E = Err̂ + Eθθ̂ + Eφφ̂ = − ∂V ∂r r̂ − 1 r ∂V ∂θ θ̂ − 1 rsinθ ∂V ∂φ φ̂. (7) The component Eφ = 0 as there is no φ dependence in V . The other two components are, Er = −kq[− r − z0cosθ (r2 + z20 − 2rz0cosθ) 3/2 + rz20R − z0R 3cosθ (r2z20 + R 4 − 2rz0R2cosθ)3/2 ] (8) and Eθ = − kq r [ −rz0sinθ (r2 + z20 − 2rz0cosθ) 3/2 + rz0R 3sinθ (r2z20 + R 4 − 2rz0R2cosθ)3/2 ] (9) 2 = −q(a2 − b2) 2b(a2 + b2 − 2abcosθ)1/2 |π0 = −q(a2 − b2) 2b [± 1 a + b −± 1 a − b ] = ±q (24) The physical solution is −q, as is deduced using Gauss’s law following the procedure of problem 4.1. Problem 4.23 is analogous to the grounded cylinder problem. a) The electrostatic potential is found using one image line of charge density −λ, V (r, φ) = λ 2πǫ0 ln(c1/r1) − λ 2πǫ0 ln(c2/r2) + c3 (25) Using the cosine rule we find, V (r, φ) = λ 4πǫ0 [ln(r2 + x′20 − 2rx ′ 0cosφ) − ln(r 2 + x20 − 2rx0cosφ)] + c4 (26) Using the conjugate relation, x′0 = R 2/x0 and simplifying yields, V (r, φ) = λ 4πǫ0 [ln(d2r2 + R4 − 2rdR2cosφ) − ln(r2 + d2 − 2rdcosφ)] + c5 (27) b) The electric field in the φ direction Eφ(R, φ) = 0. The charge density on the cylinder is σ(φ) = ǫ0∂V/∂r where the sign is positive due to the fact that the normal to the cavity surface is in the −r̂ direction. From the potential, we find, ∂V ∂r = λ 4πǫ0 [ 2rd2 − 2R2dcosφ d2r2 + R4 − 2rdR2cosφ − 2r − 2dcosφ r2 + d2 − 2rdcosφ ]. (28) We then find, σ(φ) = ǫ0 ∂V ∂r |R = λ 2πR [ R2 − d2 R2 + d2 − 2dRcosφ ] (29) c) To find the total charge per unit length on the cavity wall, we can use Gauss’s law (easy) or integration (hard). Using integration, we have, λcav = ∫ 2π 0 Rdφσ(φ) = ∫ 2π 0 λ(R2 − d2)dφ 2π(R2 + d2 − 2dRcosφ) (30) To carry out the integral, use, ∫ 2π 0 dθ a + bcosθ = 2 (a2 − b2)1/2 Tan−1[ (a − b)tan(θ/2) (a2 − b2)1/2 ] (31) where a = R2 + d2, b = −2dR. We then find, λcav = λ(R2 − d2) 2π 2 R2 − d2 [Tan−1( (R − d)2 R2 − d2 tanπ) − Tan−1( (R − d)2 R2 − d2 tan0)] (32) 5 where we use the fact that (a2 − b2)1/2 = ((R2 + d2)2 − 4d2R2)1/2 = R2 − d2, and a − b = R2 + d2 − 2dR = (R − d)2. Both tanπ and tan0 are zero, and tan−10 has solutions 0,±π,±2π.... i.e. at all angles where sinx = 0. We can therefore find solutions λcav = ±nq, where n = 0,±1,±2..... The physical solution as deduced using Gauss’s law is λcav = −q. d) The force per unit length on the line charge is found by taking the electric field due to the image charge, evaluated at r = d, φ = 0 and noting that it acts in the negative r̂ direction, so that ~Fλ = −λE ′ r(d, 0)r̂ = −λ2 4πǫ0 [ 2d3 − 2R2d d4 + R4 − 2d2R2 r̂ = −(±) λ2d (2πǫ0(d2 − R2) r̂. (33) Notice again that in the surd we can choose either the positive or negative sign. The correct direction is the positive x̂ direction, which corresponds to the positive sign in the surd. Problem 4.24 can be solved directly by adding the electric fields due to the real charge and the image charge, or by using the formula in Lecture 12, to find Eθ = 0, Er = (kq/a 2)(1/16 − 2/25). Using the first method, we have that the magnitude of E is the sum of the electric field of the real charge and that of the image so that, ~E(0, 0,−2a) = [ kq (4a)2 + kq′ (2a + z′0) 2 ](−k̂) (34) we also have z′0 = R 2/z0 = a 2/2a = a/2, and q′ = −qR/z0 = −q/2, Substitution into the above equation gives, ~E(0, 0,−2a) = 7kq/(400a2)k̂. 6