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Omar Ashour Chapter 2 Review September 13, 2015 Problem 1 University Physics, problem 67 A region in space contains a total positive charge that is distributed spherically such that the volume charge density ρ (r) is given by for r ≤ R/2 ρ (r) = 3αr/ (2R) 2 ρ (r) = α[1 − (r/R) ] for R/2 ≤ r ≤ R for r ≥ R ρ (r) = 0 Here α is a positive constant having units of C/m3 a. Determine α in terms of Q and R b. Using Gausss law, derive an expression for the magnitude of the electric field as a function of r. Do this separately for all three regions. Express your answers in terms of the total charge. c. What fraction of the total charge is contained within the region R/2 ≤ r ≤ R ~ at r = R/2 d. What is the magnitude of E e. If an electron with charge q 0 = −e is released from rest at any point in any of the three regions, the resulting motion will be oscillatory but not simple harmonic. Why? Solution Part (a) ρ= dQ =⇒ Q = dV Z b ρ dV a dV = Adr = 4πr2 dr Z 3αr 6πα R/2 3 6πα 1 R 3 2 Q1 = 4πr dr = r dr = = παR3 2R R 0 R 4 16 32 0 Z R r 2 7 31 47 Q2 = α 1− 4πr2 dr = 4παR3 − = παR3 R 24 160 120 R/2 Z R/2 (1) (2) 1 Omar Ashour Chapter 2 Review September 13, 2015 From (1) and (2) Q = Q1 + Q2 = 47 3 233 παR3 + παR3 = παR3 120 32 480 α= 480Q 233πR3 Part (b) For the region where r ≤ R/2 Z ~ · dA ~ = Qen , the enclosed charge is a fraction of Q1 E 0 S r Z Qen = 0 r 3α 4πr2 dr = 2R Z r 3 0 480Q 233πR3 r 720Q 4πr2 dr = ·4 2R 233R4 E1 · 4πr2 = E1 = Z r r3 dr = 0 720Qr4 233R4 720Qr4 233R4 0 180Qr2 233π0 R4 For the region where R/2 ≤ r ≤ R Z ~ · dA ~ = Qen , the enclosed charge is the sum of Q1 and a fraction of Q2 E 0 S 0 2 ! 03 Z r 0 r r04 r r05 r 02 02 0 Qen = Q1 + α 1− 4πr dr = Q1 + 4πα r − 2 dr = Q1 + 4πα − R R 3 5R2 R/2 R/2 R/2 3 3 r R3 r5 R3 3 1 1 r5 1 3 3 1 r Qen = Q1 + 4πα − − + = παR + 4παR − − + 3 24 5R2 160 32 3 R3 24 5 R5 160 3 1 r 3 1 r 5 17 480Q 1 r 3 1 r 5 23 3 3 Qen = 4παR + − − = 4πR − − 128 3 R 5 R 480 233πR3 3 R 5 R 1920 3 1 Qen 4πR 480Q 1 r 3 1 r 5 23 E2 = = − − 2 2 3 4π0 r 4π0 r 233πR 3 R 5 R 1920 Z r 2 Omar Ashour Chapter 2 Review 480Q E2 = 233π0 r2 September 13, 2015 1 r 3 1 r 5 23 − − 3 R 5 R 1920 For the region where r ≥ R The enclosed charge is simply equal to the total charge Q. 1 Q E3 = · 2 4π0 r Part (c) The fraction of enclosed charge is simply the ratio between Q2 and Q. Q2 47 233 = παR3 ÷ παR3 = 0.807 Q 120 480 Part (d) Using either E2 or E3 for r = R/2 E= 180 Q 233 4π0 R2 Part (e) ~ = qE ~ F The condition for simple harmonic motion is to have F ∝ r which is not the case here as F ∝ 1 r2 3 Omar Ashour Chapter 2 Review September 13, 2015 Problem 2 University Physics, problem 48 (exam problem) A small conducting spherical shell with inner radius a and outer radius b is concentric with a larger conducting spherical shell with inner radius c and outer radius d. The inner shell has total charge +2q the outer shell has charge −2q. a. Calculate the electric field (magnitude and direction) in terms of q and the distance r from the common center of the two shells for i. r < a ii. a < r < b iii. b < r < c iv. c < r < d v. r > d b. What is the total charge on the i. inner surface of the small shell ii. outer surface of the small shell iii. inner surface of the large shell iv. outer surface of the large shell Solution Part (a) i. E1 = 0; charge enclosed is 0. ii. E2 = 0; inside a conductor. iii. E3 = 1 2q 1 q = 4π0 r2 2π0 r2 iv. E4 = 0; inside a conductor. v. E5 = 0; charge enclosed is −2q + 2q = 0. Part (b) To solve this part, we take a Gaussian surface in the regions where we want to find the charge. i. We take a Gaussian surface with radius r where a < r < b. Electric field is zero thus the charge enclosed is zero. ii. We take a Gaussian surface with radius r where b < r < c. Electric field in this region is the same as that of a point charge +2q, the charge on the surface is +2q iii. We take a Gaussian surface with radius r where c < r < d. Electric field in this region is zero, which means that the enclosed charge is zero. The enclose charge Qen = +2q + qinner surface = 0 =⇒ qinner surface = −2q iv. This spherical shell overall has a charge of −2q which is already present on the inner surface, thus the charge on the outer surface is zero. 4