ph1bA_08_q4s

ph1bA_08_q4s - Ph1b Analytic Quiz 4 Solution Chefung Chan,...

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Unformatted text preview: Ph1b Analytic Quiz 4 Solution Chefung Chan, Winter 2008 Problem 1 (10 points total) ¢¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¢ ¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡ ¢ Solution: We know E = − dV , therefore we have, dx φ24 − φ13 a−b φ13 − φ24 EII = b φ24 − φ13 EIII = a−b EI = 1 The diagram shows a cross-sectional view of 4 identical conducting plates of area A. Plates 1 and 3 are separated by a fixed distance a. Plates 2 and 4 are also separated by the fixed distance a. Plates 2 and 3 are separated by the smaller distance b. As the figure shows, plates 1 and 3 are connected by a wire and plates 2 and 4 are connected by a wire. The plates do have non-zero thickness, but it is so small that it can be ignored. The lateral dimensions of the plates are so big that all “edge effects” can be ignored. Plates 1 & 3 carry a total charge Q and have electric potential φ13 , while plates 2 & 4 carry a total charge −Q and are at potential φ24 . We define EI ,EII , and EIII to be the upward pointing electric fields in regions I , II , and III , respectively. By Gauss’s law, E is constant in each of the 3 regions. Furthermore, since the total charge on the system of plates is zero, the electric fields above and below the system are zero. (To see this, consider the face of each of the conductors as having some surface charge density σi , producing field 2πσi in opposite directions above and below the surface. The sum of all the charge is zero, so i σi = 0 and the resulting electric field above and below all the 4 plates must vanish.) (a) (3 points) Express the field in each of the three regions EI , EII , and EIII in terms of the potentials φ13 , φ24 , and a and b. ¦¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¦ ¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦¦ ¦¥ ¦¥ ¦¥ ¦¥ ¦¥ ¦¥ ¦¥ ¦¥ ¦¥ ¦¥ ¦¥ ¦¥ ¦¥ ¦¥ ¦¥ ¦¥ ¦¥ ¦¥ ¦¥ ¦¥ ¦¥ ¦¥ ¦¥ ¦¥ ¥ ¥¡¥¡¥¡¥¡¥¡¥¡¥¡¥¡¥¡¥¡¥¡¥¡¥¡¥¡¥¡¥¡¥¡¥¡¥¡¥¡¥¡¥¡¥¡¥¡ ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¥¦ b ¤¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¤ £ ¤£ ¤£ ¤£ ¤£ ¤£ ¤£ ¤£ ¤£ ¤£ ¤£ ¤£ ¤£ ¤£ ¤£ ¤£ ¤£ ¤£ ¤£ ¤£ ¤£ ¤£ ¤£ ¤£ ¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡¤¡£ £¡£¡£¡£¡£¡£¡£¡£¡£¡£¡£¡£¡£¡£¡£¡£¡£¡£¡£¡£¡£¡£¡£¡£¡ ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£¤ II a III a I 1 2 3 ¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡ ¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨ ¨§¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨ §¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§ ¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨¡¨ §¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡§¡ ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡§¨§ 4 (b) (2 points) What are the corresponding charge densities on the internal surfaces in the three regions [i.e., σ1 , σ2 (upper ), σ2 (lower ), σ3 (upper ), σ3 (lower ), and σ4 ] in terms of φ13 , φ24 , a and b? Solution: In the following parts, notation 2u, 3l, etc, is used to refer to plate 2(upper), plate 3(lower). Also, remind a fact that electric field inside conductor is zero. σ1 EI σ2u 1 σ1 EI 1 2 σ2u 2 Figure 1: Show what the gaussian surface looks like. Using part (a) results and by Gauss law, with gaussian surface including upper part of plate 2 but not including plate 1, EI A + 0 = 4πAσ2u φ24 − φ13 σ2u = 4π (a − b) Now consider gaussian surface including plate 1 and upper part of plate 2. Since the electric field inside conducting plate is zero, we have 0 = 4πA(σ1 + σ2u ) σ1 = −σ2u φ13 − φ24 = 4π (a − b) Similarly, φ24 − φ13 4πb φ13 − φ24 σ3u = 4πb φ13 − φ24 σ3l = 4π (a − b) φ24 − φ13 σ4 = 4π (a − b) σ2l = (c) (2 points) The sum of all the surface charges on plates 1 and 3 must equal Q. Use this fact to find the potential difference ∆φ between plates 1 and 2 in terms of Q, A, a and b. What is the capacitance C of the system? 2 Solution: From result in (b) and considering A(σ1 + σ3u + σ3l ) = Q, we have Q 1 φ13 − φ24 φ13 − φ24 φ13 − φ24 = + + A 4π a−b b a−b Letting φ13 − φ24 = ∆φ Q ∆φ 1 ∆φ ∆φ = + + A 4π a − b b a−b 2b + (a − b) 4πQ = ∆φ A (a − b)b 4πQ (a − b)b ∆φ = A a+b The capacitance C of the system : C = Q/∆φ A a+b = 4π (a − b)b Alternative solution: One can also work out the capacitance through considering 3 capacitors C12 , C23 , C34 connected in parallel. It’s because the potential across each capacitor is all ∆φ. Using C = A/(4πs), Csys = C12 + C23 + C34 A A A = + + 4π (a − b) 4πb 4π (a − b) A(a + b) = 4π (a − b)b (d) (2 points) Compute the potential energy stored on the system in terms of Q,A, a and b. Solution: The potential energy stored on the system : 1 1 U = C (∆φ)2 = Q∆φ 2 2 1 4πQ (a − b)b =Q 2 A a+b 2 2πQ (a − b)b = A a+b (e) (1 point) While a is kept fixed, there can be a force between plates 2 and 3 that will tend to either increase b (repulsive) or decrease b (attractive). Compute an expression for this force. For what values of a/b is the force attractive? 3 Solution: The force between plates 2 and 3 is, F =− dU (a is kept fixed) db 2πQ2 −(a − b)b a2 − b2 − (a + b)b + =− A (a + b)2 (a + b)2 2πQ2 −2ab − b2 + a2 =− A (a + b)2 2πQ2 (2ab + b2 + a2 ) − 2a2 = A (a + b)2 2πQ2 2a2 = 1− A (a + b)2 The force is attractive when F is negative, therefore we have, F <0 1− 2a2 <0 (a + b)2 (a + b)2 < 2a2 √ a + b < 2a 1 a/b > √ 2−1 √ > 2+1 Remark : If you consider the approach with F = E (Aσ ) instead of F = − dU , db notice you will have a extra factor of 2 which leads you to a wrong answer. 4 ...
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This note was uploaded on 04/28/2010 for the course PH 1b Anal taught by Professor Mckewn during the Winter '08 term at Caltech.

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