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Electrical Energies

Electrical Energies. Just as Newton’s Laws worked completely, but were difficult, so to, working with Electric Forces will be difficult.

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Electrical Energies

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  1. Electrical Energies Just as Newton’s Laws worked completely, but were difficult, so to, working with Electric Forces will be difficult. Just as with gravitation, in electricity we can solve many problems using the Conservation of Energy, a scalar equation that does not involve time or direction. This requires that we find an expression for the electric energy.

  2. Electric Potential Energy Since Coulomb’s Law has the same form as Newton’s Law of Gravity, we will get a very similar formula for electric potential energy: DPE = -  (kq1q2/r122) dr and choosing rstandard = infinity, gives: PEel = k q1 q2 / r12 Recall for gravity, PEgr = - G m1 m2 / r12 . Note that the PEelectric does NOT have a minus sign. This is because two like charges repel instead of attract as in gravity.

  3. Voltage Just like we did with forces on particles to get fields in space, (Edue to 1 = Fon 2/ q2) we can define an electric voltage in space (a scalar): Vdue to 1 = PEof 2 / q2 . We often use this definition this way: PEof 2 = q2 * Vdue to 1.

  4. Units The unit for voltage is, from the definition: Vdue to 1 = PEof 2 / q2 volt = Joule / Coulomb . Note that voltage, like field, exists in space, while energy, like force, is associated with a particle!

  5. Voltage due to a point charge Since the potential energy of one charge due to another charge is: PEel = k q1 q2 / r12 and since voltage is defined to be: Vdue to 1 = PEof 2 / q2 we can find a nice formula for the voltage in space due to a single charge: Vdue to 1 = k q1 / r12 .

  6. Voltages due to several point charges Since voltage, like energy, is a scalar, we can simply add the voltages created by individual point charges at any point in space to find the total voltage at that point in space: Vtotal = S k qi / r1i . For continuous distribution of charges, we can integrate: Vtotal = k dq / r .

  7. Voltages and Electric Fields Just like force and work, and hence potential energy, are related: D PE = - W = - F Ds, so too are electric field and voltage: D V = - E Ds . Note that voltage changes only in the direction of electric field. This also means that there is no electric field in directions in which the voltage is constant.

  8. Voltage and Field D V = - E Ds , or Ex = -DV / Dx . Note also the minus sign means that electric field goes from high voltage towards low voltage. Note also that this means that positive charges will tend to “fall” from high voltage to low voltage, but that negative charges will tend to “rise” from low voltage to high voltage!

  9. Voltage and Field D V = - E Ds , or Ex = -DV / Dx . Note that the units of electric field are (from its definition: E = F/q)Nt/Coul. But from the above relation, they are equivalently Volts/m. Hence: Nt/Coul = Volt/m.

  10. Voltage, Field and Energy The Computer Homework, Vol 3, #3, has an Introduction and problems concerning these ideas that relate voltage to field D V = - E Ds and voltage to energy PEof 2 = q2*Vat 1 for use with the Conservation of Energy Law.

  11. Review F1on2 = k q1 q2 / r122PE12 = k q1 q2 / r12 Fon2 = q2 Eat2 PEof2 = q2 Vat2 Eat2 = k q1 / r122Vat2 = k q1 / r12 use in use in S F = ma KEi + PEi = KEf +PEf +Elost VECTOR scalar Ex = -DV / Dx

  12. Energy example Through how many volts will a proton have to be accelerated if it is to reach a million miles per hour? DV = ? qproton = 1.6 x 10-19 Coul mproton = 1.67 x 10-27 kg vi = 0 m/s vf = 1 x 106 mph* (1 m/s / 2.24 mph) = 4.46x105 m/s .

  13. Example, cont. We can use the Conservation of Energy including the formulas for kinetic energy and potential energy: KEi + PEi = KEf + PEf + Elost , where KE = (1/2)mv2 and PE = qV: (1/2)mpvi2 + qpVi = (1/2)mpvf2 + qpVf + Elost Since vi=0 and Elost=0, and bringing qpVf to the left side, we have: qp(Vi-Vf) = (1/2)mpvf2 .

  14. Example, cont. qp(Vi-Vf) = (1/2)mpvf2 We note that (Vi-Vf) = -DV since the change is normally final minus initial. Thus, DV= -(1/2)mpvf2 / qp = -(1/2)(1.67x10-27kg)(4.46x105m/s)2 / 1.6x10-19 C= -1040 volts. We see that the proton must fall down (DV is negative) through 1040 volts to reach a million miles/hour.

  15. Voltages and Fields We’ve already seen that one way of calculating a voltage at a location is to add up all the voltages at that point due to all the point charges. However, another way is to use what we’ve already done with fields, and use the relation between voltage and field: Ex = -DV / Dx.

  16. Voltages from Fields Inside a hollow sphere of charge, we saw using Gauss’s Law that the Electric Field was zero. From Ex = -DV / Dx, with E=0 we see that DV = 0, and so the voltage is constant.

  17. Voltages and Fields Outside a hollow sphere of charge, we saw using Gauss’s Law that the Electric Field was the same as that due to a point charge. Thus, the voltage outside of a point charge is simply: V = kq/r (assuming V=0 at r=infinity).

  18. Voltages and Fields For a line of charge, or outside a cylinder of charge, the Electric Field was (from Gauss’s Law): Eline = 2 k l / r . Using Ex = -DV / Dx(or DV = -  E ds ) Vat 2 - Vat 1 = 2 k l ln(r1/r2) .

  19. Voltages and Fields Between parallel plates, E = s/e = constant, using Ex = -DV / Dx(or DV = -  E ds ) gives: E = -DV / Dx = -V / d.

  20. Change in Voltagesphere, cylinder and plate

  21. Voltage, Force and Energy The Computer Homework, Vol. 3, #4, on Electric Deflection, provides problems involving energy (PE = qV) and force (F = qE, where Ey = -DV/Dy ). This situation is what occurs in an old-style TV or computer monitor (CRT). There is a supplementary ppt slide set that helps set up this problem that is accessed from the course web page.

  22. Electric Circuits Now that we have the concept of voltage, we can use this concept to understand electric circuits. Just like we can use pipes to carry water, we can use wires to carry electricity. The flow of water through pipes is caused by pressure differences, and the flow is measured by volume of water per time.

  23. Electric Circuits In electricity, the concept of voltage will be like pressure. Water flows from high pressure to low pressure; electricity flows from high voltage to low voltage. But what flows in electricity? Charges! How do we measure this flow? By Current: current = I = Dq / Dt UNITS: Amp(ere) = Coulomb / second

  24. Circuit Elements In this second part of the course we will consider two of the common circuit elements: capacitor resistor The capacitor is an element that stores charge for use later (somewhat like a water tower). The resistor is an element that “resists” the flow of electricity.

  25. Capacitance A water tower holds water. A capacitor holds charge. The pressure at the base of the water tower depends on the height (and hence the amount) of the water. The voltage across a capacitor depends on the amount of charge held by the capacitor.

  26. Capacitance We define capacitance as the amount of charge stored per volt: C = Qstored / DV. UNITS: Farad = Coulomb / Volt Just as the capacity of a water tower depends on the size and shape, so the capacitance of a capacitor depends on its size and shape. Just as a big water tower can contain more water per foot (or per unit pressure), so a big capacitor can store more charge per volt.

  27. Parallel Plate Capacitor For a parallel plate capacitor, we can pull charge from one plate (leaving a -Q on that plate) and deposit it on the other plate (leaving a +Q on that plate). Because of the charge separation, we have a voltage difference between the plates, DV. The harder we pull (the more voltage across the two plates), the more charge we pull: C = Q /DV. Note that C is NOT CHANGED by either Q or DV; instead, C RELATES Q and DV!

  28. Parallel Plate Capacitor C = Q/V. For a parallel plate capacitor, the charge is related to the electric field, and the electric field is related to the voltage. We can then relate the charge to the voltage: E = s/eo where s=Q/A; and E = V/d. Recall also that k = 1/[4peo]. This leads to: V = Ed = Qd/eoA. Thus C = Q/V becomes C = Q/V = Q / (Qd/eoA) = eoA/d .

  29. Parallel Plate Capacitor For this parallel plate capacitor, the capacitance is related to charge and voltage (C = Q/V), but the actual capacitance depends on the size and shape: Cparallel plate = A / (4 p k d) where A is the area of each plate, d is the distance between the plates, and k is Coulomb’s constant (9 x 109 Nt-m2 / Coul2).

  30. Example What is the capacitance of a parallel plate capacitor where each plate is 5 cm by 5 cm in size, the two plates are separated by 2 mm with vacuum between the plates, and there is 20 volts difference between the plates?

  31. Examples: Parallel Plate Capacitor The capacitance depends on A, k and d: Cparallel plate = A / (4 p k d) where A = 5 cm x 5 cm = 25 cm2 = 25 x 10-4 m2, d = 2 mm = 2 x 10-3 m, and k = 9 x 109 Nt-m2/Coul2 , so C = (1) * (25 x 10-4 m2) / [4 * 3.14 * 9 x 109 Nt-m2/Coul2 * 2 x 10-3 m] = 1.10 x 10-11 F = 11 pF .

  32. Examples: Parallel Plate Capacitor Thus, one Farad would be a HUGE capacitance! If we have a DV = 20 volts, then to calculate the charge, Q, we can use: C = Q/V to get: Q = C*V = 11 x 10-12 F * 20 volts = 2.2 x 10-10 Coul = 0.22 nCoul = 220 pCoul. Note that we often drop the D in front of the V since we often are concerned by the change in voltage rather than the absolute value of the voltage - just as we do when we talk about height!

  33. Capacitance Note that if we doubled the voltage, we would not do anything to the capacitance. Instead, we would double the charge stored on the capacitor. However, if we try to overfill the capacitor by placing too much voltage across it, the positive and negative plates will attract each other so strongly, they will spark across the gap and destroy the capacitor. Thus capacitors have a maximum voltage!

  34. Another Example Consider a coaxial cable used as a capacitor. Charges are removed from the inner wire and deposited on the outer cylinder. What is the capacitance of this cable if the cable is 2 meters long, the wire has a radius of 2 mm and the cylinder has a larger radius of 1 cm (10 mm) ?

  35. Example #2 (cont.) Start with the definition of capacitance: C = Q/V . Then relate the voltage to the charge that has been moved. In the coaxial case, we know that the electric field between the wire and the cylinder is that due to the wire itself: E = 2kl/r, where l=Q/L . From this we can get the voltage, V = -Edr = -(2kl/r)dr = 2k(Q/L) ln(rcyl/rwire).

  36. Example #2 (cont.) Therefore, C = Q/V = C = Q / [2k(Q/L) ln(rcyl/rwire)] , or Ccoax = L / [2k ln(rcyl/rwire)]. Note that C does NOT depend on the Q or the V (it does relate them), but rather depends on how the capacitor is made (L, rcyl, and rwire).

  37. Example 2 (cont.) Ccoax = L / [2k ln(rcyl/rwire)]. In this case, we are given: L = 2 meters; rcyl = 1 cm = 10 mm; rwire = 2 mm. Thus, C = 2 m / [2 x 9 x 109 Nt-m2/Coul2 x ln(10 mm / 2 mm)] = 6.9 x 10-11 Coul2/Nt-m (UNITS: Nt-m = Joule, and a Joule/Coul = Volt, so units are Coul/Volt = Farad.C = 69 pF .

  38. Energy Storage If a capacitor stores charge and carries voltage, it also stores the energy it took to separate the charge. Note that we move the charge piece by piece from one plate to the next, and as each charge is moved, the voltage across the plates increases. Note that the whole charge is not moved across the whole voltage!

  39. Gravitational analogy This is like digging a hole: the first shovel of dirt (charge) is lifted (raised in voltage) only a slight amount. The next shovel of dirt is deeper and it must be lifted higher. Only the final shovel of dirt goes the whole distance between the bottom of the hole and the top of the hill.

  40. Energy Stored Starting from DPE = q DV, and adding each contribution of each charge, we have for the total stored energy: Estored = S qi Vi =  V dq =  (q/C) dq = (1/2)Q2/C (or using C = Q/V) Estored = (1/2)QV = (1/2)CV2.

  41. Energy Storage Note that previously we had: PE = q*V , and now for a capacitor we have: E = (1/2)*Q*V . Why the 1/2 factor for a capacitor?

  42. Energy Storage The reason is that in charging a capacitor, the first bit of charge is transferred while there is very little voltage on the capacitor (recall that the charge separation creates the voltage!). Only the last bit of charge is moved across the full voltage. Thus, on average, the full charge moves across only half the voltage!

  43. Effects of Materials Our derivation for the capacitance of parallel plates assumed we had vacuum between the plates. Would placing some material between the plates affect the capacitance? To answer that question, we need to consider what material is made of, and how it reacts to the presence of electric fields.

  44. Effects of Materials Materials are composed of atoms and molecules. Atoms and molecules are composed of charged parts (positive nucleus and negative electrons, positive and negative ions). When placed in an electric field, the charged particles are going to be pulled apart - the negative charges pulled towards the positive plate, the positive charges pulled towards the negative plate. +Q -q +q -Q

  45. Effects of Materials The net effect of this pulling is to create a separation of charges. This separation of charges will create an opposing electric field. This opposing electric field will tend to reduce the total field between the plates, and hence reduce the voltage between the plates (since E = -DV/Ds). This reduced voltage (for the same charge on the plates) will increase the capacitance (C=Q/V) ! +Q -q -q -q -q +q +q +q +q -Q

  46. Dielectric Constant Some materials are more “stretchable” than others, and so will cause the capacitance to increase more than others. Vacuum can not stretch, so it will provide the smallest possible capacitance. To describe this “stretchiness”, we define a dielectric constant, K, such that K = Cwith material / Cwith vacuum , or Cwith material = K Cwith vacuum . (Be sure to distinguish between k and K!)

  47. Capacitor terminals For a parallel plate capacitor with vacuum between the plates, it does not matter which plate is connected to the positive and which to the negative voltage. However, for large capacitors, a dielectric material is used that “stretches” better one way than another. This means that some capacitors will specify which terminal should be connected to the positive and which to the negative.

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