Slide 29 / 105 11 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
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 decrease - this is directly analogous to the movement of a mass within a Gravitational Field.
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.
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.
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 increase - again, this is directly analogous to the movement of a mass within a gravitational field.
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.
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.
Slide 36 / 105 12 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 + + + + + + + + - - - - - - -
Slide 37 / 105 13 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 䌴䭎㍍䬷 卓 䭕䉒 ㍈ 呏㐵䴳ㄶ Lk force? What is the sign of the work done by the external force? What is the total work done by the two forces? + + + + + + + + - - - - - - -
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.
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.
Slide 40 / 105 14 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 ㌱ 䴳 3k + + + + + + + - - - - - - - -
Slide 41 / 105 15 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 䙖䩐 卓 Lk external force? What sign is the work done by the Electric field? What happens to the potential energy of the charge/plate system? + + + + + + + - - - - - - - -
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.
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 l ittle space between lines means there is a very quick change in height. Where in this picture is the steepest incline?
Slide 44 / 105 Equipotential Lines These "topography" lines 300 V 300 V are called "Equipotential 230 V 230 V Lines" when we use them to represent the Electric 50 V 50 V Potential - they represent 0 V 0 V lines where the Electric Potential is the same. The closer the lines, the 300 V 230 V 0 V 50 V faster the change in voltage.... the bigger the change in Voltage, the larger the Electric Field.
Slide 45 / 105 Equipotential Lines The direction of the Electric Field lines are always perpendicular to the 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).
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.
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 the -20 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.
Slide 48 / 105 16 At point A in the diagram, what is the direction of the Electric Field? A Up +300 V +150 V -150 V -300 V 0 V C B B Down E A C Left D D Right
Slide 49 / 105 17 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 C B E A D
Slide 50 / 105 18 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 C B E A D
Slide 51 / 105 Uniform Electric Field
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).
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.
Slide 54 / 105 Uniform Electric Field + + + + + + + + - - - - - - - Point charges have a Only some equations we non-uniform field strength have learned will apply to since the field weakens uniform electric fields. with distance.
Slide 55 / 105 Uniform Electric Field Hill F N a The slope of the plane determines mg the acceleration and the net force on the object. F N Slope = 0 mg F net = 0 no acceleration!
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 mgΔh = ½mv f2 If v f2 = v 02 + 2aΔx and v 0 = 0 then v f2 = 2aΔx mgΔh = ½m(2aΔx) mgΔh = maΔx gΔh = aΔx gΔh a = Δ x
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. p + p + p + p + p + p + p + V f V o e - e - e - e - e - e - e -
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). p + p + p + p + p + p + p + V f p + V o e - e - e - e - e - e - e -
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 qV 0 = qV f + ½mv f2 qV 0 - qV f = ½mv f2 -qΔV = ½mv f2 where ΔV = V f - V 0 If v f2 = v 02 + 2aΔx and v 0 = 0 then v f2 = 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
Slide 60 / 105 Uniform Electric Field The equation only applies to uniform electric fields. ΔV ΔV = _ _ E = Δx 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: J 1 N V = and a V = C C m 1 N C = (J/C) and a J = N m m 1 N (N m/C) = C m 1 N C = 1 N The units are equivalent. C
Slide 61 / 105 Uniform Electric Field A more intuitive way to understand the negative sign in the relationship ΔV _ E = 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.
Slide 62 / 105 19 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 - - - - - - -
Slide 63 / 105 20 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
Slide 64 / 105 21 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
Slide 65 / 105 22 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?
Slide 66 / 105 23 An electric field of 3500 N/C is desired between two plates which are 0.0040 m apart; what Voltage should be applied?
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
Slide 68 / 105 24 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?
Slide 69 / 105 F = kQq F = qE Use in r 2 ANY Use situation. ONLY E = kQ U E = qV with point For point r 2 charges. charges AND uniform electric U E = kQq E = - ΔV Equations fields r d with the "k" are point charges V = kQ ONLY for U E = -qEd ONLY. r uniform electric fields
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.
Slide 71 / 105 Parallel Plate Capacitors When a battery is connected to the plates, charge moves between them. Every electron that moves p + to the negative plate leaves a positive V nucleus behind. The plates have e - equal magnitudes of charges, but one is positive, the other negative.
Slide 72 / 105 Parallel Plate Capacitors Only unpaired protons and electrons are represented here. p + p + p + p + p + p + p + Most of the atoms V are neutral since they have equal numbers of protons e - e - e - e - e - e - e - and electrons.
Slide 73 / 105 Parallel Plate Capacitors Drawing the Electric Field from the positive to negative p + charges reveals that p + p + p + p + p + p + the Electric Field is uniform everywhere V in a capacitor's gap. Also, there is no e - e - e - e - e - e - e - field outside the gap .
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. p + p + p + p + p + p + p + C = Q V V This is just a DEFINITION and is true of all capacitors, e - e - e - e - e - e - e - not just parallel plate capacitors.
Slide 75 / 105 Parallel Plate Capacitors C = Q V The unit of capacitance is the farad (F). A farad is a Coulomb per Volt. p + p + p + p + p + p + p + A farad is huge; so capacitance is given as picofarad (1pf = 10 -12 F), nanofarad (1nf = 10 -9 F), e - e - e - e - e - e - e - microfarad (1 µ f = 10 -6 F), millifarads (1mf = 10 -3 F)
Slide 76 / 105 25 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
Slide 77 / 105 26 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
Slide 78 / 105 Parallel Plate Capacitors The Area of the capacitor is just the surface area of A ONE PLATE, and is represented by the letter A. The distance between the plates is represented by d the letter d.
Slide 79 / 105 Parallel Plate Capacitors For PARALLEL PLATE CAPACITORS, the capacity to store charge increases with the area of the p + p + p + p + p + p + p + plates and decreases as the V plates get farther apart. e - e - e - e - e - e - e - C # A C # 1/d
Slide 80 / 105 Parallel Plate Capacitors The constant of proportionality is called the p + p + p + p + p + p + p + Permitivity of Free Space and has the symbol # o . V # o = 8.85 x 10 -12 C 2 /N-m 2 e - e - e - e - e - e - e -
Slide 81 / 105 Parallel Plate Capacitors So for PARALLEL PLATE CAPACITORS: C = # o A d The larger the Area, A, the higher the p + p + p + p + p + p + p + capacitance. The closer together the V plates get, the higher the capacitance. e - e - e - e - e - e - e -
Slide 82 / 105 Parallel Plate Capacitors Imagine you have a fully charged capacitor. If you disconnect the battery and change either the area or p + p + p + p + p + p + p + distance between the plates, what do you know about the charge V on the capacitor? e - The charge remains the e - e - e - e - e - e - same.
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 p + p + p + p + p + p + p + distance between the plates, what do you V know about the voltage across the plates? e - e - e - e - e - e - e - The voltage remains the same.
Slide 84 / 105 27 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
Slide 85 / 105 28 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
Slide 86 / 105 29 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
Slide 87 / 105 Parallel Plate Capacitors After being charged the plates have equal and opposite voltage, V. +V/2 There is a uniform p + p + p + p + p + p + p + electric field, E, between the plates. V We learned earlier that with a UNIFORM E-FIELD e - e - e - e - e - e - e - that # V = -Ed; this is true in -V/2 the case of the parallel plate capacitor.
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/2 p + p + p + p + p + p + p + +V to -V within the gap; it +V/4 is zero at the location midway between the 0 plates. -V/4 e - -V/2 e - It is always e - e - e - e - e - perpendicular to the E- Field.
Slide 89 / 105 Parallel Plate Capacitors The energy stored in ANY capacitor is given by formulas most easily derived from the parallel plate capacitor. p + Consider how much work it would take to move a single electron between two initially uncharged plates. e -
Slide 90 / 105 Parallel Plate Capacitors That takes ZERO work since there is no difference in voltage. However, to move a second electron to p + p + the negative plate requires work to overcome the V repulsion from the first one...and to overcome the e - e - attraction of the positive plate.
Slide 91 / 105 Parallel Plate Capacitors To move the last electron from the positive plate to the negative plate +V/2 requires carrying it p + p + p + p + p + p + through a voltage difference of V. V The work required to do that is q # V... e - e - e - e - e - e - -V/2 Here the charge of a electron is -e, and the difference in potential is -V (-V/2 - V/2)... the work = eV.
Slide 92 / 105 Parallel Plate Capacitors If the work to move the first electron is zero. +V/2 p + p + p + p + p + p + p + And the work to move the last electron is eV. V The the AVERAGE e - e - work for ALL e - e - e - e - e - -V/2 electrons is eV/2.
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 +V/2 p + p + p + p + p + p + p + W = QV/2 V That energy is stored in the electric field within the e - e - e - e - e - e - e - capacitor. -V/2
Slide 94 / 105 Parallel Plate Capacitors So the energy stored in a capacitor is given by: U C = QV 2 +V/2 p + p + p + p + p + p + p + Where Q is the charge on one plate and V is the voltage V difference between the plates. e - e - e - e - e - e - e - -V/2
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 Q solve for V substitute substitute U C = QV 2 U C = Q 2 U C = 1/2 CV 2 2C
Slide 96 / 105 30 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
Slide 97 / 105 31 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
Slide 98 / 105 32 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
Slide 99 / 105 33 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
Slide 100 / 105 34 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
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