Week 5 - Dielectrica, Resistance and Resistivity

Transcription

1 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 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? A = pf d (1) Q0 = C0 V0 = 0.22 µc (2) C 0 = ε0 b) What is the charge on each plate? 1

2 c) What is the electric field between the plates? E 0 = V 0 d = V m () 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? C = Q V = pf (4) e) What is the dielectric constant of the material? K = C 0 C 4425 pf = = 5.0 (5) pf f) What is the permittivity of the dielectric? ε = Kε 0 = C 2 /N m 2 (6) g) What is the induced charge on each face of the dielectric? Q i = σ i A = σa(1 1 K ) = Q(1 1 ) = µc (7) K h) What is the electric field? E = E 0 K = V/m (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

3 a) What is the resistance of this wire? (Copper has ρ C = ) R = 0.44 Ω. The resistance of a cable of uniform resistivity is R = ρl A = 4ρL πd 2 (9) where d is the diameter of the cable. So for copper with ρ C = we get R = 4ρ CL πd 2 = π ( ) 2 Ω = 0.44 Ω. (10) b) What would be the diameter of a silver cable with the same length and resistance? (Silver has ρ S = ) d = 0.46 mm The diameter of a silver cable with the same resistance would be d = which is about a 15% shorter than for copper. 4ρL πr = 0.46 mm (11) Exercise 5.: 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? 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?

4 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 A parallel-plate capacitor has the space between the plates filled with two slabs of dielectric; one with constant K 1 and another with constant K 2. 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. E 1 = σ K 1ε 0 E 2 = σ K 1ε 0 b) Calculate the potential between the plates. V = Qd K 1+K 2 2ε 0A K 1K 2. In each slab of dielectric the electric field is decreased by the dielectric constant, and considering that it also is constant in each piece of dielectric the potential difference between the plates is V = ( σ K 1 ε 0 + σ K 2 ε 0 ) d 2 = Qd K 1 + K 2. (12) 2ε 0 A K 1 K 2 c) Find the total capacitance. C = 2ε 0A d K 1 K 2 (1) K 1 + K 2 4

5 C = 2ε 0A K 1 K 2 (14) d K 1 + K 2 Method 2: The situation would not change if we slided a little conductor in between the two dielectrics and then we would get two capacitors in series! Therefore we get 1 C = 1 C C 2 = 1 K 1ε 02A d with the capacitance equal to that of method K 2ε 02A d = d K 1 + K 2 (15) 2ε 0 A K 1 K 2 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. E = σ = Q Kε 0 Kε 0 A. (16) b) What is the relation between current density and the electric field? J = E/ρ c) Show that the leakage current I carried by the dielectric is given by I = Q Kε 0 ρ. (17) Since we have a dielectric with dielectric constant K the field is weakened by 1/K compared to vacuum. We have E = σ = Q Kε 0 Kε 0 A. (18) Now we know that J = E/ρ. Assuming a uniform current density throughout the dielectric (only a good approximation far from the edges) we get I = JA = Q Kε 0 ρ. (19) 5

6 d) How does the capacitance C vary with time because of this leakage current? Explain. It does not vary! That is because C is only dependent on the geometry (separation between plates, area etc). It is not dependent on the charge because V and Q are proportional quantities and C is their ratio. So when Q reduces, so does V and their ratio C remains constant! The point is that capacitance is not the measure of the charge stored, but rather the charge per unit potential. If you can place a high charge on your arrangement of conductors for a given potential, you have a high capacitance. Exercise 5.6: Resistivity of a Conic conductor Figure 2 A material of resistivity ρ is formed into a solid, truncated cone of height h and radii r 1 and r 2 at either end (see 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. R = ρh πr 1r 2 In the book we derived the equation R = ρl/a for a conductor with constant resistivity and constant area over the length L. For our cone conductor ρ is constant over the whole length (height h), but the area of a slice varies. Let s align our cone along the z-axis like shown in figure. If we make an infinitesimal slice of this cone then the area is constant over this slice and its resistance would be dr = ρdz πr 2. (20) But we need an equation relating z to r. This is found by considering realizing that we have two similar triangles in figure. Can you see them? Using this fact we find thats and therefore r r 1 z dz = = r 2 r 1 h (21) h r 2 r 1 dr. (22) 6

7 Figure The total resistance is then found by R = h 0 ρdz πr 2 = ρh r2 dr π(r 2 r 1 ) r 1 r 2 = ρh πr 1 r 2 (2) b) Check your result by showing it agrees with the resistance for a cylinder of length L and radius r, R = ρl/πr 2. Comparing our more general formula with that for a cylinder we have r 1 = r 2 = r which implies Agreement! R = ρh πr 1 r 2 = ρh πr 2. (24) 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? I = 2πa ( r 2 r 1) (25) The current is the flux of the current density trough the surface, i.e. I = J da (26) 7

8 and since this current density is parallel with the axis of the cone J da = JdA = Jrdrdθ. Thus we get I = J da = a 2π r 0 0 r 2 dθdr (27) = 2 πar (28) Therefore the difference in current between the upper and lower surface is I = 2πa ( r 2 r 1) (29) a non zero value. Since more charge is entering compared to what is coming out, or vice versa, some charge has to be building up within the cone. Therefore it will not remain neutral. 8