Physics 2 for Electrical Engineering

1. Ben Gurion University of the Negev www.bgu.ac.il/atomchip,www.bgu.ac.il/nanocenter Physics 2 for Electrical Engineering Lecturers: Daniel Rohrlich , Ron Folman Teaching Assistants: Ben Yellin, Yoav Etzioni Week 5. Dielectrics, currents and resistance – Energy stored in a capacitor • dielectrics•dielectrics and Gauss’s law•electrical current • resistance and Ohm’s law • resistance and temperature • electrical power Sources: Halliday, Resnick and Krane, 5th Edition, Chaps. 29-31.

2. Energy stored in a capacitor The flash on your camera uses a capacitor to store energy. Although the flash is charged by a battery, the battery cannot release the energy as quickly as the flash does. How much energy can a capacitor store?

3. Energy stored in a capacitor The flash on your camera uses a capacitor to store energy. Although the flash is charged by a battery, the battery cannot release the energy as quickly as the flash does. How much energy can a capacitor store? We have already learned that the potential energy U in a configuration of electrical charges does not depend on how the charges got to where they are. So let’s calculate U in a convenient way.

4. Energy stored in a capacitor Step 1: The capacitor is uncharged. What is the potential energy of the configuration of charges?

5. Energy stored in a capacitor Step 1: The capacitor is uncharged. What is the potential energy of the configuration of charges? U = 0

6. Energy stored in a capacitor Step 1: The capacitor is uncharged. What is the potential energy of the configuration of charges? U = 0 Step 2: The capacitor carries charges q' and –q' on its two parts. How much work dW does it take to move an infinitesimal amount of charge dq' from the negative part of the capacitor to the positive part? dq' q' –q'

7. Energy stored in a capacitor Step 1: The capacitor is uncharged. What is the potential energy of the configuration of charges? U = 0 Step 2: The capacitor carries charges q' and –q' on its two parts. How much work dW does it take to move an infinitesimal amount of charge dq' from the negative part of the capacitor to the positive part? dW = ΔV'dq' = (q'/C) dq' dq' q' –q'

8. Energy stored in a capacitor Step 1: The capacitor is uncharged. What is the potential energy of the configuration of charges? U = 0 Step 2: The capacitor carries charges q' and –q' on its two parts. How much work dW does it take to move an infinitesimal amount of charge dq' from the negative part of the capacitor to the positive part? dW = ΔV'dq' = (q'/C) dq‘ Step 3: What is the total work done in charging the capacitor to a charge q?

9. Energy stored in a capacitor Step 1: The capacitor is uncharged. What is the potential energy of the configuration of charges? U = 0 Step 2: The capacitor carries charges q' and –q' on its two parts. How much work dW does it take to move an infinitesimal amount of charge dq' from the negative part of the capacitor to the positive part? dW = ΔV'dq' = (q'/C) dq‘ Step 3: What is the total work done in charging the capacitor to a charge q? W = ∫q (q'/C) dq‘= q2 /2C

10. Energy stored in a capacitor Step 1: The capacitor is uncharged. What is the potential energy of the configuration of charges? U = 0 Step 2: The capacitor carries charges q' and –q' on its two parts. How much work dW does it take to move an infinitesimal amount of charge dq' from the negative part of the capacitor to the positive part? dW = ΔV'dq' = (q'/C) dq‘ Step 3: What is the total work done in charging the capacitor to a charge q? W = ∫q (q'/C) dq‘= q2 /2C Step 4: What is the potential energy stored in the capacitor?

11. Energy stored in a capacitor Step 1: The capacitor is uncharged. What is the potential energy of the configuration of charges? U = 0 Step 2: The capacitor carries charges q' and –q' on its two parts. How much work dW does it take to move an infinitesimal amount of charge dq' from the negative part of the capacitor to the positive part? dW = ΔV'dq' = (q'/C) dq‘ Step 3: What is the total work done in charging the capacitor to a charge q? W = ∫q (q'/C) dq‘= q2 /2C Step 4: What is the potential energy stored in the capacitor? U = q2 /2C = C(ΔV)2 /2

12. Energy stored in a capacitor Example 1: You have three identical capacitors and a battery. How should you combine them in one circuit so that the capacitors will store the most energy?

13. Energy stored in a capacitor Example 1: You have three identical capacitors and a battery. How should you combine them in one circuit so that the capacitors will store the most energy? Remember: Ceff = 2C/3 Ceff = 3C A B Ceff = C/3 Ceff = 3C/2 D C

14. C1 C2 Energy stored in a capacitor Example 2: A 3.55-μF capacitor C1 is charged to potential difference of 6.30 V. It is then connected with a 8.95-μF capacitor C2. (a) What is the final potential difference ΔV? (b) What is the initial potential energy Uin? (c) What is the final potential energy Ufin?

15. C1 C2 Energy stored in a capacitor Example 2: A 3.55-μF capacitor C1 is charged to potential difference of 6.30 V. It is then connected with a 8.95-μF capacitor C2. (a) What is the final potential difference ΔV? (b) What is the initial potential energy Uin? (c) What is the final potential energy Ufin? Answer: (a) Charge conservation!

16. C1 C2 Energy stored in a capacitor Example 2: A 3.55-μF capacitor C1 is charged to potential difference of 6.30 V. It is then connected with a 8.95-μF capacitor C2. (a) What is the final potential difference ΔV? (b) What is the initial potential energy Uin? (c) What is the final potential energy Ufin? Answer: (a) Charge conservation! q = (3.55 μF)(6.30 V) = (C1 + C2) ΔV q = 22.4 μC = (12.5 μF) ΔV ΔV = 1.79 V

17. C1 C2 Energy stored in a capacitor Example 2: A 3.55-μF capacitor C1 is charged to potential difference of 6.30 V. It is then connected with a 8.95-μF capacitor C2. (a) What is the final potential difference ΔV? (b) What is the initial potential energy Uin? (c) What is the final potential energy Ufin? Answer: (b) Uin = C1(6.30 V)2/2 = 70.4 μJ (c) Ufin = (C1 + C2) (ΔV)2/2 = 20.0 μJ

18. C1 C2 Energy stored in a capacitor Example 2: A 3.55-μF capacitor C1 is charged to potential difference of 6.30 V. It is then connected with a 8.95-μF capacitor C2. (a) What is the final potential difference ΔV? (b) What is the initial potential energy Uin? (c) What is the final potential energy Ufin? Answer: (b) Uin = C1(6.30 V)2/2 = 70.4 μJ (c) Ufin = (C1 + C2) (ΔV)2/2 = 20.0 μJ A puzzle: Isn’t energy conserved?

19. Another puzzle: A room with no windows contains three lamps. The switches to the lamps are outside the room, next to the door – which is closed. You can play with switches, turning the lamps on and off,but you can enter the room just once. How can you find out which switch controls which lamp, for all three lamps? (You have no special equipment.)

20. Energy stored in a capacitor Application: Where is the energy in a charged capacitor? Consider a parallel-plate capacitor: If we double the plate separation d, we double the energy. The energy, therefore, seems to live between the plates, where the electric field E = ΔV/d lives. If so, we can derive the energy density uE of the electric field:

21. C q q C' ΔV ΔV ' Dielectrics A simple experiment with a dielectric (non-conducting) material, e.g. glass: voltmeter

22. C q q C' ΔV ΔV ' Dielectrics With the dielectric filling the capacitor (right), ΔV decreases to ΔV ' = ΔV/κ where κ is called the dielectric constant of the dielectric. Hence the capacitance increases from C to C ' = κC. voltmeter

23. Dielectrics The advantages of a dielectric in a capacitor include 1. increased capacitance, 2. increased maximum potential ΔV, and 3. mechanical support between plates (which helps increase the capacitance even more since the plates can be very close).

25. Dielectrics Phenomenological explanation: Polar molecules (electric dipoles), which orient themselves randomly when E = 0, orient themselves in the direction of the electric field when E ≠ 0. The net effect is an induced charge inside the capacitor that decreasesE (and increases C)! E = 0 E ≠ 0

26. Dielectrics We can understand this explanation better by considering what happens when there is a conductor, instead of a dielectric, between the plates of a capacitor: The conductor screens the electric field via + and – charges on its surface, thus reducing the potential difference ΔV across the capacitor and increasing the capacitance C. E ≠ 0 - - - - - - - + + + + + + +

27. Dielectrics Example 1: At right is a parallel-plate capacitor of capacitance C. If the conductor inserted between its plates has the same cross-section but only 1/3 the width, what is the capacitance C ' of the capacitor with the conductor inside? E ≠ 0 - - - - - - - + + + + + + +

28. Dielectrics Example 1: At right is a parallel-plate capacitor of capacitance C. If the conductor inserted between its plates has the same cross-section but only 1/3 the width, what is the capacitance C ' of the capacitor with the conductor inside? Answer: The charge on the capacitor is unchanged. The potential difference is reduced by 2/3, i.e. ΔV ' = 2ΔV/3. Hence the new capacitance is C ' = q/ΔV ' = q/(2 ΔV/3) = 3(q/ΔV)/2 = 3C/2. E ≠ 0 - - - - - - - + + + + + + +

29. Dielectrics Example 2: At right is a parallel-plate capacitor of capacitance C. If the dielectric inserted between its plates has the same cross-section but only 1/3 the width, what is the capacitance C ' of the capacitor with the dielectric inside? E ≠ 0 - - - + + +

30. Dielectrics Example 2: At right is a parallel-plate capacitor of capacitance C. If the dielectric inserted between its plates has the same cross-section but only 1/3 the width, what is the capacitance C ' of the capacitor with the dielectric inside? Answer: The charge on the capacitor is unchanged. The potential difference ΔV is reduced to ΔV ' = (ΔV)(2/3 + 1/3κ). Hence the new capacitance is C ' = q/ΔV ' = 3 κC/(2κ + 1). E ≠ 0 - - - + + +

31. Dielectrics and Gauss’s law Let’s apply Gauss’s law to the surface shown: If there were no dielectric present, Gauss’s law would yield ε0EA = q, where q is the total charge on the inner surface of the capacitor and E is the field inside. But with a dielectric inside, the field inside is E ' and the total charge inside the surface is q – q ', where –q ' is the charge on the surface of the dielectric. Thus but since E ' = E/κ, we have - - - + + +

32. Dielectrics and Gauss’s law Although we obtained this version of Gauss’s law for the specific case of a dielectric in a parallel-plate capacitor, it is the correct generalization of Gauss’s law to include dielectrics:

33. Electrical current We now go beyond electrostatics! Definition: electrical current I is the flow of positive charge q through a given cross-sectional surface. The magnitude of I is and the unit of current is the ampere A = C/s. This definition holds equally for electrons in a conductor and protons in a beam. In a conductor, the electrons flow in the direction opposite to the current and I = nAevdrift, where n is the electron density in the conductor, A is the cross-sectional area of the wire, –e is the charge of the electron, and vdrift is the average speed of the electrons, which drift between collisions.

34. Electrical current Example: The density of copper is 8.95 g/cm3, and home copper wiring has a cross-sectional area A = 3.31 × 10–6 m2. For a current in the wire of 10.0 A, what is the drift speed vdrift? Note: each copper atom contributes one conduction electron.

35. Electrical current Example: The density of copper is 8.95 g/cm3, and home copper wiring has a cross-sectional area A = 3.31 × 10–6 m2. For a current in the wire of 10.0 A, what is the drift speed vdrift? Note: each copper atom contributes one conduction electron. Answer: The atomic weight of copper (see the Periodic Table) is 63.5 g/mol; using Avogadro’s number NA = 6.02214179 × 1023 /mol, we find that for copper, the density of atoms (which is also the density of conduction electrons) is n = (8.95 g/cm3) ×(6.02 × 1023 /mol)/ (63.5 g/mol)= 8.48 × 1022/cm3 electrons.

36. Electrical current Answer: The atomic weight of copper (see the Periodic Table) is 63.5 g/mol; using Avogadro’s number NA = 6.02214179 × 1023 /mol, we find that for copper, the density of atoms (which is also the density of conduction electrons) is n = (8.95 g/cm3) ×(6.02 × 1023 /mol)/ (63.5 g/mol)= 8.48 × 1022/cm3 electrons. Now vdrift = I/neA and from our data we calculate So why doesn’t it take hours, after you flick the light switch, for the light to go on?

37. Resistance and Ohm’s law Ohm’s law is not fundamental; some materials obey it, some don’t. Materials that obey Ohm’s law – including most metals – are called ohmic. Ohm’s law, as we will study it, connects two vector fields, the electric field E and the current density J, via the conductivity σ of an ohmic material: J = σE . (Here σis not surface charge density.) If we integrate J across any area element dA it crosses through, we get the current I, since charge does not build up in a conductor, I must be a constant along the wire.

38. Resistance and Ohm’s law Integrating E between two points on a wire, we obtain the potential difference ΔV between those two points: In the special case that E is constant along the wire and points in the same direction as dr, the integral reduces to El, where l is the length of the wire between the two points; then ΔV = El and E = ΔV / l. From Ohm’s law we obtain I = σEA, where A is the cross-sectional area of the wire, hence I = σ(ΔV)A / l. Now defining the resistance of the wire to be R = l / σA, we finally obtain ΔV = IR , also called Ohm’s law. The unit of R is the ohmΩ = V/A.

39. Resistance and Ohm’s law We also define the resistivityρ of an ohmic material to be the inverse ρ = 1/σ of the conductivity. So another expression for the resistance R of the wire is R = lρ/A. The units of ρ are Ω·m. Resistivities at 20°C.

40. Resistance and Ohm’s law Example 1: Calculate the resistance per unit length of gold wire having a radius of 0.321 mm, at 20°C.

41. Resistance and Ohm’s law Example 1: Calculate the resistance per unit length of gold wire having a radius of 0.321 mm, at 20°C. Answer: Resistance per unit length is R/l =ρ/A. From the table we find the resistivity of gold to be ρ = 2.44 × 10–8Ω·m. The cross-sectional area is A = π (0.321 mm)2 = 0.324 × 10–6 m2. Therefore R/l = ρ/A = (2.44 × 10–8Ω·m) / (0.324 × 10–6 m2) = 0.0753 Ω/m.

42. Resistance and Ohm’s law Example 2: a current I flows through an ohmic wire that thins, i.e. the cross-sectional area A gets smaller in the direction of I. How do the electric field E, the current density J and the drift velocity vdrift behave along the wire?

43. Resistance and Ohm’s law Example 2: a current I flows through an ohmic wire that thins, i.e. the cross-sectional area A gets smaller in the direction of I. How do the electric field E, the current density J and the drift velocity vdrift behave along the wire? Answer: If a current I flows through the wire and charge does not build up, then I is constant along the wire. But since A decreases, J must increase. Since J = σE, also E must increase. Finally, J is proportional to vdrift, so also vdrift must increase along the wire.

44. Resistance and temperature Over a limited range of temperature T, resistivity varies linearly with T – T0 , where T0 = 20°C: ρ = ρ0[1 + α(T – T0)] , where ρ0 is the resistivity at 20°C and α is a constant that characterizes the material. Example 1: When does a light bulb carry more current – when you just turn it on or when the glow is steady?

45. Resistance and temperature Over a limited range of temperature T, resistivity varies linearly with T – T0 , where T0 = 20°C: ρ = ρ0[1 + α(T – T0)] , where ρ0 is the resistivity at 20°C and α is a constant that characterizes the material. Example 2: The resistance of a platinum loop increases by 53.6% when it is dropped into melting indium. What is the melting point of indium?

46. Resistance and temperature Over a limited range of temperature T, resistivity varies linearly with T – T0 , where T0 = 20°C: ρ = ρ0[1 + α(T – T0)] , where ρ0 is the resistivity at 20°C and α is a constant that characterizes the material. Example 2: The resistance of a platinum loop increases by 53.6% when it is dropped into melting indium. What is the melting point of indium? Answer: From 1.536 = R/R0 = ρ/ρ0 = 1 + α(T – 20°C) and the value α = 3.92 × 10–3 /°C for platinum, we obtain the melting temperature T = [0.536 / (3.92 × 10–3) + 20]°C = 157°C.

47. C1 C2 Electrical power Remember the problem of charging one capacitor and then connecting it to a second, uncharged capacitor? We found that final potential energy was less than the initial potential energy. Where did the energy go?

48. R C Electrical power Remember the problem of charging one capacitor and then connecting it to a second, uncharged capacitor? We found that final potential energy was less than the initial potential energy. Where did the energy go? To answer this question, let’s replace one capacitor by a resistor. At time t = 0 we close the switch. The total potential around the circuit must vanish for all t: q/C + IR = 0. Hence dq/dt = I = –q/RC and we solve the differential equation to obtain q(t) = q(0) e–t/RC.

49. R C Electrical power At time t = 0 we close the switch. The total potential around the circuit must vanish for all t: q/C + IR = 0. Hence dq/dt = I = –q/RC and we solve the differential equation to obtain q(t) = q(0) e–t/RC. The potential energy in the capacitor at time t is U = [q(t)]2/2C = [q(0)]2 e–2t/RC/2C and the rate of loss of potential energy is dU/dt = –[q(0)]2 e–2t/RC/RC2 = –I2R. Conclusion: I2R is the rate of energy loss due to dissipation in the resistor! It equals the power delivered to the resistor.

50. Halliday, Resnick and Krane, 5th Ed., Chap. 31, Exercise 28: Part of an infinite array of identical 1-μΩ resistors is shown below. A battery connects two remote junctions. Show that the potential at any other junction is the average of the potentials at the four nearest junctions.