Electric Potential Energy and Voltage

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1 Slide 1 / 105 Electric Potential Energy and Voltage

2 Slide 2 / 105 How to Use this File Each topic is composed of brief direct instruction There are formative assessment questions after every topic denoted by black text and a number in the upper left. > Students work in groups to solve these problems but use student responders to enter their own answers. > Designed for SMART Response PE student response systems. > Use only as many questions as necessary for a sufficient number of students to learn a topic. Full information on how to teach with NJCTL courses can be found at njctl.org/courses/teaching methods

3 Slide 3 / 105 Electric Potential Energy and Voltage Electric Potential Energy Electric Potential (Voltage) Uniform Electric Field Capacitors and Capacitance

4 Slide 4 / 105 Electric Potential Energy

5 Slide 5 / 105 Electric Potential Energy Q+ q+ Start with two like charges initially at rest, with Q at the origin, and q at infinity. In order to move q towards Q, a force opposite to the Coulomb repulsive force (like charges repel) needs to be applied. Note that this force is constantly increasing as q gets closer to Q, since it depends on the distance between the charges, r, and r is decreasing.

6 Slide 6 / 105 Work and Potential Energy Q+ q+ Recall that Work is defined as: To calculate the work needed to bring q from infinity, until it is a distance r from Q, we need to use calculus, because of the non constant force. Then, use the relationship: Assume that the potential energy of the Q-q system is zero at infinity, and adding up the incremental force times the distance between the charges at each point, we find that the Electric Potential Energy, U E is:

7 Slide 7 / 105 Electric Potential Energy Again, just like in Gravitational Potential Energy, Electric Potential Energy requires a system - it is not a property of just one object. In this case, we have a system of two charges, Q and q. Another way to define the system is by assuming that the magnitude of Q is much greater than the magnitude of q, thus, the Electric Field generated by Q is also much greater than the field generated by q (which may be ignored). Now we have a field-charge system, and the Electric Potential energy is a measure of the interaction between the field and the charge, q.

8 Slide 8 / 105 Electric Potential Energy What is this Electric Potential Energy? It tells you how much energy is stored by work being done on the system, and is now available to return that energy in a different form, such as kinetic energy. Again, just like the case of Gravitational Potential Energy. If two positive charges are placed near each other, they are a system, and they have Electric Potential Energy. Once released, they will accelerate away from each other - turning potential energy into kinetic energy. These moving charges can now perform work on another system.

9 Slide 9 / 105 Electric Potential Energy Q+ q- If you have a positive charge and a negative charge near each other, you will have a negative potential energy. This means that it takes work by an external agent to keep them from getting closer together.

10 Slide 10 / 105 Electric Potential Energy Q+ q+ Q- q- If you have two positive charges or two negative charges, there will be a positive potential energy. This means that it takes work by an external agent to keep them from flying apart.

11 Slide 11 / Compute the potential energy of the two charges in the following configuration: +Q 1 +Q A positive charge, Q 1 = 5.00 mc is located at x 1 = m, and a positive charge Q 2 = 2.50 mc is located at x 2 = 3.00 m.

12 Slide 11 (Answer) / Compute the potential energy of the two charges in the following configuration: +Q 1 +Q A positive charge, Q 1 = 5.00 mc is located at x 1 = m, and a positive charge Q 2 = 2.50 mc is located at x 2 = 3.00 m. Answer [This object is a pull tab]

13 Slide 12 / Compute the potential energy of the two charges in the following configuration: -Q 1 +Q A negative charge, Q 1 = mc is located at x 1 = m, and a positive charge Q 2 = 4.50 mc is located at x 2 = 5.00 m.

14 Slide 12 (Answer) / 105

15 Slide 13 / Compute the potential energy of the two charges in the following configuration: -Q 1 -Q A negative charge, Q 1 = mc is located at x 1 = m, and a negative charge Q 2 = mc is located at x 2 = 7.00 m.

16 Slide 13 (Answer) / Compute the potential energy of the two charges in the following configuration: -Q 1 -Q Answer A negative charge, Q 1 = mc is located at x 1 = m, and a negative charge Q 2 = mc is located at x 2 = 7.00 m. [This object is a pull tab]

17 Slide 14 / 105 Electric Potential Energy of Multiple Charges To get the total energy for multiple charges, you must first find the energy due to each pair of charges. Then, you can add these energies together. Since energy is a scalar, there is no direction involved - but, there is a positive or negative sign associated with each energy pair. For example, if there are three charges, the total potential energy is: Where U xy is the Potential Energy of charges x and y.

18 Slide 15 / Compute the potential energy of the three charges in the following configuration: +Q 1 -Q 2 +Q A positive charge, Q 1 = 5.00 mc is located at x 1 = m, a negative charge Q 2 = mc is located at x 2 = m, and a positive charge Q 3 = 2.50 mc is located at x 3 = 3.00 m.

19 Slide 15 (Answer) / 105

20 Slide 16 / 105 Electric Potential (Voltage)

21 Slide 17 / 105 Electric Potential or Voltage Our study of electricity began with Coulomb's Law which calculated the electric force between two charges, Q and q. By assuming q was a small positive charge, and dividing F by q, the electric field E due to the charge Q was defined. The same process will be used to define the Electric Potential, or V, from the Electric Potential Energy, where V is a property of the space surrounding the charge Q: V is also called the voltage and is measured in volts.

22 Slide 18 / 105 Electric Potential or Voltage What we've done here is removed the system that was required to define Electric Potential Energy (needed two objects or a field and an object). Voltage is a property of the space surrounding a single, or multiple charges or a continuous charge distribution. It tells you how much potential energy is in each charg and if the charges are moving, how much work, per charge, they can do on another system.

23 Slide 19 / 105 Electric Potential or Voltage Voltage is the Electric Potential Energy per charge, which is expressed as Joules/Coulomb. Hence: To make this more understandable, a Volt is visualized as a battery adding 1 Joule of Energy to every Coulomb of Charge that goes through the battery.

24 Slide 20 / What is the Electric Potential (Voltage) 5.00 m away from a charge of 6.23x10-6 C?

25 Slide 20 (Answer) / What is the Electric Potential (Voltage) 5.00 m away from a charge of 6.23x10-6 C? Answer [This object is a pull tab]

26 Slide 21 / What is the Electric Potential (Voltage) 7.50 m away from a charge of -3.32x10-6 C?

27 Slide 21 (Answer) / 105

28 Slide 22 / 105 Electric Potential or Voltage Despite the different size of these two batteries, they both have the same Voltage (1.5 V). That means that every electron that leaves each battery has the same Electric Potential - the same ability to do work. The AA battery just has more electrons - so it will deliver more current and last longer than the AAA battery.

29 Slide 23 / 105 Electric Potential of Multiple Charges To get the total potential for multiple charges, you must first find the potential due to each charge. Then, you can add these potentials together. Since potential is a scalar, there is no direction involved - but, there is a positive or negative sign associated with each potential. For example, if there are three charges, the total potential is: Vnet = V 1 + V 2 + V 3 Where V net is the net potential and V 1, V 2, V 3... is individual potential.

30 Slide 24 / Compute the electric potential of three charges at the origin in the following configuration: +Q 1 +Q 2 -Q x(m) A positive charge, Q 1 = 5.00 nc is located at x 1 = m, a positive charge Q 2 = 3.00 nc is located at x 2 = m, and a negative charge Q 3 = nc is located at x 3 = 6.00 m.

31 Slide 24 (Answer) / Compute the electric potential of three charges at the origin in the following configuration: +Q 1 +Q 2 -Q x(m) A positive charge, Q 1 = 5.00 nc is located at x 1 = m, a positive charge Q 2 = 3.00 nc is located at x 2 = m, and a negative charge Q 3 = nc is located at x 3 = 6.00 m. Answer [This object is a pull tab]

32 Slide 25 / Two positive charges of magnitude +Q are placed at the corners A and B of equilateral triangle with a side r. Calculate the net electric potential at point C.

33 Slide 25 (Answer) / Two positive charges of magnitude +Q are placed at the corners A and B of equilateral triangle with a side r. Calculate the net electric potential at point C. Answer [This object is a pull tab]

34 Slide 26 / Four charges of equal magnitude are arranged in the corners of a square. Calculate the net electric potential at the center of the square due to all charges.

35 Slide 26 (Answer) / 105

36 Slide 27 / 105 Electric Potential or Voltage Another helpful equation can be found from by realizing that the work done on a positive charge by an external force (a force that is external to the force generated by the electric field) will increase the potential energy of the charge, so that: Note, that the work done on a negative charge will be negativ the sign of the charge counts!

37 Slide 28 / How much work must be done by an external force to bring a 1x10-6 C charge from infinity to the origin of the following configuration? +Q 1 +Q 2 -Q x(m) A positive charge, Q 1 = 5.00 nc is located at x 1 = m, a positive charge Q 2 = 3.00 nc is located at x 2 = m, and a negative charge Q 3 = nc is located at x 3 = 6.00 m.

38 Slide 28 (Answer) / 105

39 Slide 29 / Three charges of equal magnitude but different signs are placed in three corners of a square. In which of two arrangements more work is required to move a positive test charge q from infinity to the empty corner of the square? First, make a prediction and discuss with your classmates. Second, calculate work done on moving charge for each arrangement with the given numerical values: Q=9x10-9 C, q=1x10-12 C, s=1 m

40 Slide 29 (Answer) / 105

41 Slide 30 / 105 Electric Potential or Voltage Consider two parallel plates that are oppositely charged. This will generate a uniform Electric Field pointing from top to bottom (which will be described in the next section) A positive charge placed within the field will move from top to bottom. In this case, the Work done by the Electric Field is positive (the field is in the same direction as the charge's motion). The potential energy of the system will decreas this is directly analogous to the movement of a mass within a Gravitational Field.

42 Slide 31 / 105 Electric Potential or Voltage If there is no other force present, then the charge will accelerate to the bottom by Newton's Second Law. F External Force F Electric Field But, if we want the charge to move with a constant velocity, then an external force must act opposite to the Electric Field force. This external force is directed upwards. Since the charge is still moving down (but not accelerating), the Work done by the external force is negative.

43 Slide 32 / 105 Electric Potential or Voltage F External Force F Electric Field The Work done by the external force is negative. The Work done by the Electric Field is positive. The Net force, and hence, the Net Work, is zero. The Potential Energy of the system decreases.

44 Slide 33 / 105 Electric Potential or Voltage Now consider the case where we have a positive charge at the bottom, and we want to move it to the top. In order to move the charge to the top, an external force must act in the up direction to oppose the Electric Field force which is directed down. In this case, the Work done by the Electric Field is negative (the field is opposite the direction of the charge's motion). The potential energy of the system will increas again, this is directly analogous to the movement of a mass within a gravitational field.

45 Slide 34 / 105 Electric Potential or Voltage F External Force F Electric Field If the charge moves with a constant velocity, then the external force is equal to the Electric Field force. Since the charge is moving up (but not accelerating), the Work done by the external force is positive.

46 Slide 35 / 105 Electric Potential or Voltage F External Force F Electric Field The Work done by the external force is positive. The Work done by the Electric Field is negative. The Net force, and hence, the Net Work, is zero. The Potential Energy of the system increases.

47 Slide 36 / A positive charge is placed between two oppositely charged plates as shown below. Which way will the charge move? What happens to the potential energy of the charge/plate system? 㤴䉋䤰卓ㅉ㑇㥂㔱䝉䥅啕䴴 Q

48 Slide 36 (Answer) / A positive charge is placed between two oppositely charged plates as shown below. Which way will the charge move? What happens to the potential energy of the charge/plate system? 㤴䉋䤰卓ㅉ㑇㥂㔱䝉䥅啕䴴 Q Answer Down; decreases. [This object is a pull tab]

49 Slide 37 / A positive charge is placed between two oppositely charged plates. If the charge moves with a constant velocity (no acceleration) as shown below, what sign is the work done by the Electric field 䌴䭎メートル䬷卓 force? 䭕䉒ミクロン呏㐵䴳ㄶ What is the Lk sign of the work done by the external force? What is the total work done by the two forces?

50 Slide 37 (Answer) / A positive charge is placed between two oppositely charged plates. If the charge moves with a constant velocity (no acceleration) as shown below, what sign is the work done by the Electric field 䌴䭎メートル䬷卓 force? 䭕䉒ミクロン呏㐵䴳ㄶ What is the Lk sign of the work done by the external force? What is the total work done by the two forces? Answer Positive; negative, zero [This object is a pull tab]

51 Slide 38 / 105 Electric Potential or Voltage F Electric Field F External Force Similar logic works for a negative charge in the same Electric Field. But, the directions of the Electric Field force and the external force are reversed, which will change their signs, and the potential energy as summarized on the next slide.

52 Slide 39 / 105 Electric Potential or Voltage Work done by the external force is negative. Work done by the Electric Field is positive. Net force, and hence, the Net Work, is zero. Potential Energy of the system decreases F Electric Field - F External Force Work done by the external force is positive. Work done by the Electric Field is negative. Net force, and hence, the Net Work, is zero. Potential Energy of the system increases.

53 Slide 40 / A negative charge is placed between two oppositely charged plates as shown below. Which way will the charge move? What happens to the potential energy of the charge/plate system? 卓䭁卓䱒ㅆ䭑䅎㡖㐹䩅ヒ ル䴳 3k

54 Slide 40 (Answer) / A negative charge is placed between two oppositely charged plates as shown below. Which way will the charge move? What happens to the potential energy of the charge/plate system? 卓䭁卓䱒ㅆ䭑䅎㡖㐹䩅ヒ ル䴳 3k Answer Up; decreases [This object is a pull tab]

55 Slide 41 / A negative charge is placed between two oppositely charged plates, and due to an external force moves down with a constant velocity, as shown below. What sign is the work done by the 㥊卓䰰䴵㠰䭁㙈啎卓 external force? 䙖䩐卓 What Lk sign is the work done by the Electric field? What happens to the potential energy of the charge/plate system?

56 Slide 41 (Answer) / A negative charge is placed between two oppositely charged plates, and due to an external force moves down with a constant velocity, as shown below. What sign is the work done by the 㥊卓䰰䴵㠰䭁㙈啎卓 external force? 䙖䩐卓 What Lk sign is the work done by the Electric field? What happens to the potential energy of the charge/plate system? Answer Positive, negative, increases [This object is a pull tab]

57 Slide 42 / 105 Electric Potential or Voltage Like Electric Potential Energy, Voltage is NOT a vector, so multiple voltages can be added directly, taking into account the positive or negative sign. Like Gravitational Potential Energy, Voltage is not an absolute value - it is compared to a reference level. By assuming a reference level where V=0 (as we do when the distance from the charge generating the voltage is infinity), it is allowable to assign a specific value to V in calculations. The next slide will continue the gravitational analogy to help understand this concept.

58 Slide 43 / 105 Topographic Maps Each line represents the same height value.the area between lines represents the change between lines. A big space between lines indicates a slow change in height. A little space between lines means there is a very quick change in height. Where in this picture is the steepest incline?

59 Slide 44 / 105 Equipotential Lines 0 V 50 V 230 V 300 V 300 V 230 V 50 V 0 V These "topography" lines are called "Equipotential Lines" when we use them to represent the Electric Potential - they represent lines where the Electric Potential is the same. 0 V 50 V 230 V 300 V The closer the lines, the faster the change in voltage... the bigger the change in Voltage, the larger the Electric Field.

60 Slide 45 / 105 The direction of the Electric Field lines are always perpendicular to the Equipotential lines. Equipotential Lines The Electric Field lines are farther apart when the Equipotential lines are farther apart. + The Electric Field goes from high to low potential (just like a positive charge).

61 Slide 46 / 105 Equipotential Lines For a positive charge like this one the equipotential lines are positive, and decrease to zero at infinity. A negative charge would be surrounded by negative equipotential lines, which would also go to zero at infinity. + More interesting equipotential lines (like the topographic lines on a map) are generated by more complex charge configurations.

62 Slide 47 / 105 Equipotential Lines This configuration is created by a positive charge to the left of the +20 V line and a negative charge to the right of th20 V line. Note the signs of the Equipotential lines, and the directions Electric Field vectors (in red) which are perpendicular to the lines tangent to the Equipotential lines.

63 Slide 48 / At point A in the diagram, what is the direction of the Electric Field? A Up B Down C Left D Right +300 V +150 V 0 V -150 V -300 V B C E A D

64 Slide 48 (Answer) / At point A in the diagram, what is the direction of the Electric Field? A Up B Down C Left D Right +300 V +150 V 0 V -150 V -300 V B C E A Answer D D [This object is a pull tab]

65 Slide 49 / How much work is done by an external force on a +10 μc charge that moves from point C to B? +300 V +150 V 0 V -150 V -300 V B C E A D

66 Slide 49 (Answer) / 105

67 Slide 50 / How much work is done by an external force on a -10 μc charge that moves from point C to B? +300 V +150 V 0 V -150 V -300 V B C E A D

68 Slide 50 (Answer) / 105

69 Slide 51 / 105 Uniform Electric Field

70 Slide 52 / 105 Uniform Electric Field Let's begin by examining two infinite planes of charge that are separated by a small distance. The planes have equal amounts of charge, with one plate being charged positively, and the other, negatively. The above is a representation of two infinite planes (its rather hard to draw infinity).

71 Slide 53 / 105 Uniform Electric Field By applying Gauss's Law (a law that will be learned in AP Physics), it is found that the strength of the Electric Field will be uniform between the planes - it will have the same value everywhere between the plates. And, the Electric Field outside the two plates will equal zero.

72 Slide 54 / 105 Uniform Electric Field Point charges have a non-uniform field strength since the field weakens with distance. Only some equations we have learned will apply to uniform electric fields.

73 Slide 55 / 105 Uniform Electric Field Hill F N mg F N a The slope of the plane determines the acceleration and the net force on the object. mg Slope = 0 F net = 0 no acceleration!

74 Slide 56 / 105 Uniform Electric Field If we look at the energy of the block on the inclined plane... E 0 + W = E f where W = 0 2 mgδh = ½mv f If v f = v 0 + 2aΔx and v 0 = 0 then v f = 2aΔx mgδh = ½m(2aΔx) mgδh = maδx gδh = aδx a = gδh Δx

75 Slide 57 / 105 Uniform Electric Field A similar relationship exists with uniform electric fields and voltage. With the inclined plane, a difference in height was responsible for acceleration. Here, a difference in electric potential (voltage) is responsible for the electric field. V f V o

76 Slide 58 / 105 Uniform Electric Field The change in voltage is defined as the work done per unit charge against the electric field. Therefore energy is being put into the system when a positive charge moves in the opposite direction of the electric field (or when a negative charge moves in the same direction of the electric field). V f V o

77 Slide 59 / 105 Uniform Electric Field To see the exact relationship, look at the energy of the system. E 0 + W = E f where W = 0 2 qv 0 = qv f + ½mv f 2 qv 0 - qv f = ½mv f -qδv = ½mv f 2 where ΔV = V f - V 0 If v f 2 = v aΔx and v 0 = 0 then v f 2 = 2aΔx -qδv = ½m(2aΔx) -qδv = maδx If F = ma, and F = qe, then we can substitute ma = qe -qδv = qeδx -ΔV = EΔx E = -ΔV = -ΔV Δx d

78 Slide 60 / 105 Uniform Electric Field The equation only applies to uniform electric fields. 1 N C = V and a V = J m C E = _ 1 N C = (J/C) m and a J = N m 1 N C = (N m/c) m ΔV Δx ΔV = _ d It follows that the electric field can also be shown in terms of volts per meter (V/m) in addition to Newtons per Coulomb (N/C). This can be shown: 1 N C = 1 N C The units are equivalent.

79 Slide 61 / 105 Uniform Electric Field A more intuitive way to understand the negative sign in the relationship E = _ ΔV Δx is to consider that just like a mass falls down, from higher gravitational potential energy to lower, a positive charge "falls down" from higher electric potential (V) to lower. Since the electric field points in the direction of the force on a hypothetical positive test charge, it must also point from higher to lower potential. The negative sign just means that objects feel a force from locations with greater potential energy to locations with lower potential energy. This applies to all forms of potential energy.

80 Slide 62 / If the strength of the Electric field at point A is 5,000 N/C, what is the strength of the Electric field at point B? A B

81 Slide 62 (Answer) / If the strength of the Electric field at point A is 5,000 N/C, what is the strength of the Electric field at point B? A B Answer 5,000 N/C. [This object is a pull tab]

82 Slide 63 / If the strength of the Electric field at point A is 5,000 N/C, what is the strength of the Electric field at point B? A B

83 Slide 63 (Answer) / If the strength of the Electric field at point A is 5,000 N/C, what is the strength of the Electric field at point B? A B Answer 0 N/C. [This object is a pull tab]

84 Slide 64 / In order for a charged object to experience an electric force, there must be a: A large electric potential B small electric potential C the same electric potential everywhere D a difference in electric potential

85 Slide 64 (Answer) / In order for a charged object to experience an electric force, there must be a: A large electric potential B small electric potential C the same electric potential everywhere Answer D a difference in electric potential D [This object is a pull tab]

86 Slide 65 / How strong (in V/m) is the electric field between two metal plates 0.25 m apart if the potential difference between them is 100 V?

87 Slide 65 (Answer) / How strong (in V/m) is the electric field between two metal plates 0.25 m apart if the potential difference between them is 100 V? Answer [This object is a pull tab]

88 Slide 66 / An electric field of 3500 N/C is desired between two plates which are m apart; what Voltage should be applied?

89 Slide 66 (Answer) / An electric field of 3500 N/C is desired between two plates which are m apart; what Voltage should be applied? Answer [This object is a pull tab]

90 Slide 67 / 105 Uniform Electric Field For a field like this, potential energy or work can be calculated by using E electric field. Since the work done by the Electric Field is negative, and the force is constant on the positive charge, the Work-Energy Equation is used: U E = -W = -FΔx = -qeδx

91 Slide 68 / How much Work is done by a uniform 300 N/C Electric Field on a charge of 6.1 mc in accelerating it through a distance of 0.20 m?

92 Slide 68 (Answer) / How much Work is done by a uniform 300 N/C Electric Field on a charge of 6.1 mc in accelerating it through a distance of 0.20 m? Answer [This object is a pull tab]

93 Slide 69 / 105 F = kqq r 2 E = kq r 2 U E = kqq r V = kq r Use ONLY with point charges. Equations with the "k" are point charges ONLY. F = qe U E = qv E = - ΔV d U E = -qed Use in ANY situation. For point charges AND uniform electric fields ONLY for uniform electric fields

94 Slide 70 / 105 Parallel Plate Capacitors The simplest version of a capacitor is the parallel plate capacitor which consists of two metal plates that are parallel to one another and located a distance apart.

95 Slide 71 / 105 Parallel Plate Capacitors When a battery is connected to the plates, charge moves between them. Every electron that moves to the negative plate leaves a positive nucleus behind. The plates have equal magnitudes of charges, but one is positive, the other negative. V

96 Slide 72 / 105 Parallel Plate Capacitors Only unpaired protons and electrons are represented here. Most of the atoms are neutral since they have equal numbers of protons and electrons. V

97 Slide 73 / 105 Parallel Plate Capacitors Drawing the Electric Field from the positive to negative charges reveals that the Electric Field is uniform everywhere in a capacitor's gap. V Also, there is no field outside the gap.

98 Slide 74 / 105 Parallel Plate Capacitors ANY capacitor can store a certain amount of charge for a given voltage. That is called its capacitance, C. C = Q V This is just a DEFINITION and is true of all capacitors, not just parallel plate capacitors. V

99 Slide 75 / 105 Parallel Plate Capacitors C = Q V The unit of capacitance is the farad (F). A farad is a Coulomb per Volt. A farad is huge; so capacitance is given as picofarad (1pf = F), nanofarad (1nf = 10-9 F), microfarad (1µf = 10-6 F), millifarads (1mf = 10-3 F)

100 Slide 76 / What is the capacitance of a fully charged capacitor that has a charge of 25 μc and a potential difference of 50 V? A 0.5 μf B 2 μf C 0.4 μf D 0.8 μf

101 Slide 77 / A fully charged 50 F capacitor has a potential difference of 100 V across it's plates. How much charge is stored in the capacitor? A 6 mc B 4 mc C 5 mc D 9 mc

102 Slide 78 / 105 Parallel Plate Capacitors The Area of the capacitor is just the surface area of ONE PLATE, and is represented by the letter A. A The distance between the plates is represented by the letter d. d

103 Slide 79 / 105 Parallel Plate Capacitors For PARALLEL PLATE CAPACITORS, the capacity to store charge increases with the area of the plates and decreases as the plates get farther apart. V C # A C # 1/d

104 Slide 80 / 105 Parallel Plate Capacitors The constant of proportionality is called the Permitivity of Free Space and has the symbol # o. V # o = 8.85 x C 2 /N-m 2

105 Slide 81 / 105 Parallel Plate Capacitors So for PARALLEL PLATE CAPACITORS: C = # oa d The larger the Area, A, the higher the capacitance. The closer together the V plates get, the higher the capacitance.

106 Slide 82 / 105 Parallel Plate Capacitors Imagine you have a fully charged capacitor. If you disconnect the battery and change either the area or distance between the plates, what do you know about the charge on the capacitor? V The charge remains the same.

107 Slide 83 / 105 Parallel Plate Capacitors Imagine you have a fully charged capacitor. If you keep the battery connected and change either the area or distance between the plates, what do you know about the voltage across the plates? V The voltage remains the same.

108 Slide 84 / A parallel plate capacitor has a capacitance Co. If the area on the plates is double and the the distance between the plates drops by one half, what will be the new capacitance? A Co/4 B Co/2 C 4Co D 2Co

109 Slide 85 / A parallel plate capacitor is charged by connection to the battery and the battery is disconnected. What will happen to the charge on the capacitor and the voltage across it if the area of the plates decreases and the distance between them increases? A Both increase B Both decreas C The charge remains the same and voltage increases D The charge remains the same and voltage decreases

110 Slide 86 / A parallel-plate capacitor is charged by connection to a battery and remains connected. What will happen to the charge on the capacitor and the voltage across it if the area of the plates increases and the distance between them decreases? A Both increase B Both decrease C The voltage remains the same and charge increases D The voltage remains the same and charge decreases

111 Slide 87 / 105 Parallel Plate Capacitors After being charged the plates have equal and opposite voltage, V. There is a uniform electric field, E, between the plates. We learned earlier that with a UNIFORM E-FIELD that # V = -Ed; this is true in the case of the parallel plate capacitor. V +V/2 -V/2

112 Slide 88 / 105 Parallel Plate Capacitors The Electric Field is constant everywhere in the gap. The Voltage (also know as the Electric Potential) declines uniformly from +V to -V within the gap; it is zero at the location midway between the plates. +V/2 +V/4 0 -V/4 It is always perpendicular to the E- Field. -V/2

113 Slide 89 / 105 Parallel Plate Capacitors The energy stored in ANY capacitor is given by formulas most easily derived from the parallel plate capacitor. Consider how much work it would take to move a single electron between two initially uncharged plates.

114 Slide 90 / 105 That takes ZERO work since there is no difference in voltage. Parallel Plate Capacitors However, to move a second electron to the negative plate requires work to overcome the repulsion from the first one...and to overcome the attraction of the positive plate. V

115 Slide 91 / 105 To move the last electron from the positive plate to the negative plate requires carrying it through a voltage difference of V. The work required to do that is q# V... Parallel Plate Capacitors V +V/2 -V/2 Here the charge of a electron is -e, and the difference in potential is -V (-V/2 - V/2)... the work = ev.

116 Slide 92 / 105 Parallel Plate Capacitors If the work to move the first electron is zero. And the work to move the last electron is ev. V +V/2 The the AVERAGE work for ALL electrons is ev/2. -V/2

117 Slide 93 / 105 Parallel Plate Capacitors Then, the work needed to move a total charge Q from one plate to the other is given by W = QV/2 +V/2 That energy is V stored in the electric field within the capacitor. -V/2

118 Slide 94 / 105 Parallel Plate Capacitors So the energy stored in a capacitor is given by: U C = QV 2 Where Q is the charge on one plate and V is the voltage difference between the plates. V +V/2 -V/2

119 Slide 95 / 105 Parallel Plate Capacitors Using our equation for capacitance (C=Q/V) and our equation for electric potential energy in a capacitor, we can derive three different results. solve for V solve for Q U C = QV 2 substitute substitute U C = Q2 2C U C = 1/2 CV 2

120 Slide 96 / How much energy is stored in a fully charged capacitor that storing 15 nc of charge with 20 V voltage across its plates? A 10 μj B 15 μj C 30 μj D 60 μj

121 Slide 97 / How much energy is stored in a fully charge 3 mf parallel plate capacitor with 2 V voltage across its plates? A 6 mj B 3 mj C 5 mj D 12 mj

122 Slide 98 / How much energy is stored in a fully charged 12 pf capacitor that has 9 μc of charge? A 225 J B 375 J C 420 J D 580 J

123 Slide 99 / A parallel plate capacitor is connected to a battery. The capacitor becomes fully charged and stays connected to the battery. What will happen to the energy held in the capacitor if the area of the plates increases? A Remains the same B Increases C Decreases D Zero

124 Slide 100 / A paralle plate capacitor is conected to a battery. The capacitor becomes fully charged and disconnected for the battery. What will happen to the energy stored in the capacitor if the distance between the plates increases? A Remains the same B Increases C Decreases D Zero

125 Slide 101 / 105 Capacitance can be increased by inserting a dielectric (an insulator) into the gap. Parallel Plate Capacitors Before the plates are charges the atoms are unpolarized, the electron is bound to the nucleus and not oriented in any direction

126 Slide 102 / 105 Parallel Plate Capacitors When the plates charged, the atoms are polarized and line up to reduce the external E-field. This reduces the electric field, which lowers the voltage for a given charge (since V = Ed). Since C = Q/V, this increases the capacitance.

127 Slide 103 / 105 Parallel Plate Capacitors Every material has a dielectric constant, # (kappa), which is given in a table. For a vacuum, = 1; air is about 1. If a dielectric is present, then: V +V/ C = ## oa d -V/ The larger #, the larger C

128 Slide 104 / 105 Parallel Plate Capacitors These two equations are true for all capacitors. C = Q V U C = QV 2 This equation is true for Parallel Plate Capacitors. Unless indicated otherwise, # = 1. C = ## oa d Some combination of these can solve all problems related to the Voltage, Charge, Electric Field and Voltage of a Capacitor.

129 Slide 105 / 105

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