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Week 5 - Dielectrica, Resistance and Resistivity Further, the dignity of the science itself seems to require that every possible means be explored for the solution of a problem so elegant and so celebrated. Carl Friedrich Gauss Exercise 5.1: Simple dielectric capacitor Two conducting plates are placed in parallel to form a capacitor. Each plate has an area of 1 × 10−1 m. The distance between the plates is 2.00 cm. The potential difference between the plates is 5 kV. There is vacuum between the plates. a) What is the capacitance? Answer: A = 44.25 pF d (1) Q0 = C0 V0 = 0.22 µC (2) C 0 = ε0 b) What is the charge on each plate? Answer: 1 c) What is the electric field between the plates? Answer: E0 = V0 V = 2.5 × 105 d m (3) We disconnect the power supply and assume that all charge stays on the plates, while we insert a dielectric between the plates. The potential drops to 1 kV. d) What is the capacitance after the dielectric is inserted? Answer: C= Q = 221.2 pF V (4) e) What is the dielectric constant of the material? Answer: K= C0 4425 pF = = 5.0 C 221.2 pF (5) f) What is the permittivity of the dielectric? Answer: ε = Kε0 = 4.425 × 10−11 C2 /N · m2 (6) g) What is the induced charge on each face of the dielectric? Answer: Qi = σi A = σA(1 − 1 1 ) = Q(1 − ) = 0.177 µC K K (7) h) What is the electric field? Answer: E= E0 = 5 × 104 V/m K (8) Exercise 5.2: Audio Cables Copper and silver are the two materials which most often are used for Audio Cables. Silver has a slightly lower resistivity than copper. This means that for a given resistance, silver cables be made thinner than copper cables.1 Suppose that you have a copper wire of diameter 0.5 mm and length 5.0 m. 1 This has implications for audio quality. See ’Audio Wire’ Wikipedia. 2 a) What is the resistance of this wire? (Copper has ρC = 1.72 × 10−8 .) Answer: R = 0.44 Ω. b) What would be the diameter of a silver cable with the same length and resistance? (Silver has ρS = 1.47 × 10−8 .) Answer: d = 0.46 mm Exercise 5.3: Resistance and Resistivity Nuts a) The definition of resistivity (ρ = E/J) implies that an electric field exist inside a conductor. Yet in Chapter 21 we saw that there can be no electric fields inside a conductor. Is there a contradiction here? Answer: No. One key assumption in the derivation of this fact was that all the charges were static. They settle into their equilibrium position (where they feel no force) after a very brief period of time and thus there are no electric fields. However for a closed circuit with an applied field along the length of the wire, the charges are moving continuously. If you will there are no equilibrium. So the assumption is broken and therefore we have no contradiction. b) Suppose we have a closed loop of wire with electrons moving around because of an applied electric field. It’s a fact that the drift velocity of the electrons is on the order of v ∼ 10−4 m/s, but if there is an electric field inside the copper wire, how come the electrons does not accelerate? Answer: The value v ∼ 10−4 m/s is the average velocity of the electrons in the direction of the wire. They are continuously accelerating, but they are also continuously crashing into the atoms of the wire, losing their energy. And since there are so many of them, the net movement is as if they have a constant velocity of v. Exercise 5.4: Multiple Dielectrics Figure 1 3 A parallel-plate capacitor has the space between the plates filled with two slabs of dielectric; one with constant K1 and another with constant K2 . Each slab has thickness d/2, where d is the plate separation. Each plate has a charge density σ. See figure 1. a) Assume that the plates are very large compared to the distance d. Using Gauss law, find the electric field inside each dielectric. Answer: E1 = σ K 1 ε0 E2 = σ K 1 ε0 b) Calculate the potential between the plates. Answer: V = Qd K1 +K2 2ε0 A K1 K2 . c) Find the total capacitance. Answer: C= 2ε0 A K1 + K2 . d K1 K2 (9) Exercise 5.5: Leakage of a Dielectric There is no such thing as a perfect insulator (or conductor). Therefore any dielectric insulator will leak a little bit of current. Suppose then that you have a parallel plate capacitor with charge Q on each plate and you have a dielectric material with resistivity ρ and dielectric constant K placed between the two plates. a) Find the electric field using that the charge density may be expressed as σ = Q/A. Answer: E= Q σ . = Kε0 Kε0 A (10) b) What is the relation between current density and the electric field? Answer: J = E/ρ c) Show that the ’leakage’ current I carried by the dielectric is given by I= Q . Kε0 ρ (11) d) How does the capacitance C vary with time because of this leakage current? Explain. Exercise 5.6: Resistivity of a Conic conductor A material of resistivity ρ is formed into a solid, truncated cone of height h and radii r1 and r2 at either end (see figure 2). 4 Figure 2 a) Calculate the resistance of the cone between the two flat end faces. Hint: Imagine slicing the cone into very many thin disks, and calculate the resistance of one such disk. You might also need to use R = ρL/A. Answer: R = ρh πr1 r2 b) Check your result by showing it agrees with the resistance for a cylinder of length L and radius r, R = ρL/πr2 . c) The cone has a current density |J| = ar directed along the axis of the cone. Find the difference in current at the two ends of the cone. Would this conductor remain neutral over time? Answer: ∆I = aπ 4 r2 − r14 4 5 (12)