1/7/2018. A model of the mechanism for electrostatic interactions. GRAVITATIONAL FORCE vs. ELECTROSTATCS FORCE OBJECT WITH MASS

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1 UNIT 3 Electrostatics: electric force, electric fiel, an electric potential. CHAPTER 15 THE ELECTRIC FIELD AP PHYSICS A moel of the mechanism for electrostatic interactions A moel for electric interactions, suggeste by Michael Faraay, involves some sort of electric isturbance in the region surrouning a charge object. Physicists call this electric isturbance an electric fiel. GRAVITATIONAL FORCE vs. ELECTROSTATCS FORCE OBJECT WITH MASS GRAVITATIONAL FORCE vs. ELECTROSTATCS FORCE OBJECT WITH MASS ELECTRIC FIELD Your iaper, your fiel!! Q Electric fiel is a property of a location in space that measures the force per unit charge that a charge object woul feel if place at that location. 1

2 ELECTRIC FIELD DUE TO A SINGLE POINTLIKE CHARGED OBJECT PHYSICAL QUANTITY: ELECTRIC FIELD F q = k Q q r Symbol Units Units symbol E Newton per Coulomb N/C E = k Q source r Type of PQ vector ELECTRIC FIELD DUE TO A SINGLE POINT LIKE CHARGED OBJECT We can interpret this fiel as follows: ELECTRIC FIELD DUE TO A SINGLE POINT LIKE CHARGED OBJECT 1. What is the strength an irection of the electric fiel 0.4 m to the right of a 5.0 C electric charge? E = 8150 N C The E fiel vector at any location points away from the object creating the fiel if Q is positive, an towar the object creating the fiel if Q is negative.. At what istance from a 8 C electric charge woul the electric fiel strength be N/C? = 0. 5 m 3. A, B, & C are ranom points aroun a C electric charge. Someboy else s iaper, someboy else s fiel! Fin the intensity of the electric fiel prouce by the electric charge at points A, B, & C. Assume each box to be = 0.1 m E A = N C C A C B E B = N C E c = 7000 N C

3 ELECTRIC FIELD FELT BY AN ELECTRIC CHARGE F q = E q F q = k Q q r Force = fiel property F q = E q E = F q q F G = g m ELECTRIC FIELD FELT BY AN ELECTRIC CHARGE INSIDE A UNIFORM EFIELD A 5 C electric charge is place insie a uniform electric fiel. a. If an electric force of 0.07 N is exerte on the electric charge, what is the magnitue of the electric fiel? ELECTRIC FIELD FELT BY AN ELECTRIC CHARGE INSIDE A UNIFORM EFIELD q A E b. What electric force woul be exerte if charge A is substitute by an electric charge of 4 C? c. What woul the magnitue of a new electric charge be, if an electric force of N is exerte on it? E = 14, 000 N C q A E F = N q = 4. 5μC ELECTRIC FIELD LINES (EFiel Lines) EFiel lines are a graphic representation of electric fiels use by physicists to stuy an analyze electric fiels. Experiment: Grass sees place near a charge object. Observation: Grass see aligne in a specific pattern of lines surrouning the charge object. Electric Fiel lines point in the irection of the electric fiel. Electric fiel lines o not exist but they are useful when analyzing electric fiels. 3

4 ELECTRIC FIELD LINES ELECTRIC FIELD LINES q q PROPERTY 1 Efiel lines start (leave) on positive charges. Efiel lines en (enter) on negative charges. PROPERTY The number of electric fiel lines is proportional to the magnitue of the charge. The bigger the magnitue of the electric charge the more the amount of Efiel lines. ELECTRIC FIELD LINES ELECTRIC FIELD LINES PROPERTY 3 EFiel lines will NOT cross each other Grass sees in an insulating liqui align with a similar electric fiel prouce by two oppositely charge objects ELECTRIC FIELD LINES E fiel lines Grass sees in an insulating liqui align with a similar electric fiel prouce by two objects with the same charge 4

5 PROPERTIES In which irection woul the electric fiel be at each point? A B C D E = 0 / / Q Q Q Q What is the magnitue an irection of the electric fiel at X? E net = E net = E net = K Q K Q k Q ( K Q ) K Q 4 4 K Q 4 E = 8kQ E net = 8 K Q USING THE SUPERPOSITION PRINCIPLE Draw E fiel lines for a large, uniformly charge plate of glass. Hint: place a ranom point in front of the plate. Ranom point in front of the plate 5

6 ELECTRIC FIELD USING THE SUPERPOSITION PRINCIPLE q q 1. Draw the irection of the electric fiel create by charge Q at point P. q x q x. Draw the irection of the electric fiel create by charge q at point P. q q 3. Draw the irection of the Net Electric fiel at point P. q x q x ELECTRIC FIELD USING THE SUPERPOSITION PRINCIPLE 1. Draw the irection of the Net Electric fiel at point P. ELECTRIC FIELD USING THE SUPERPOSITION PRINCIPLE E = k q. Fin the magnitue of the electric fiel at point P. 1 box = 0.1 m q = 4µC E = k q 1 E = k q 6

7 ELECTRIC FIELD USING THE SUPERPOSITION PRINCIPLE E net x = E net x = k q 1 cos θ E net y = 0 N k q C ELECTRIC FIELD USING THE SUPERPOSITION PRINCIPLE 1. Draw the irection of the Net Electric fiel at point P.. Fin the magnitue of the electric fiel at point P. E net x = 1.6 k q E net x = N C E net = N C, box = 0.1 m q A = 4µC q B = 4µC ELECTRIC FIELD USING THE SUPERPOSITION PRINCIPLE E = E = k q 1.3 sin θ k q E = 0.46 k q ELECTRIC FIELD USING THE SUPERPOSITION PRINCIPLE 1. Draw the irection of the Net Electric fiel at point P.. Write an expression for the magnitue of the electric fiel at point P in terms of k, q, x,. q x E = N C, 70 q ELECTRIC FIELD USING THE SUPERPOSITION PRINCIPLE q q E = x k q x cos θ 0 q q Charges Q an Q are locate on the x an yaxes, each at a istance from the origin, shown above. 1) What is the irection of the electric fiel at the origin? E = k q x x x E = k q x x x ) What is the magnitue of the electric fiel at the origin? 7

8 q E e p E = q 0 k q, 135 Pythagorean Theorem An electron e an a proton p are simultaneously release from rest in a uniform electric fiel E, as shown above. Assume that the only force exerte on the particles is from the electric fiel. At a later time when the particles are still in the fiel, the electron an the proton will have the same (A) irection of motion (B) magnitue of velocity (C) magnitue of momentum (D) kinetic energy (E) magnitue of acceleration Explain right an wrong answers E e p An electron an a proton have equal an opposite charges, so they will feel the same magnitue of force, but in opposite irections. The proton, because it is positive, will feel a force to the left, in the same irection as the electric fiel. The electron, because it is negative, will feel a force to the right, in the opposite irection as the electric fiel. Remember, the irection of electric fiel is the irection that a positive charge woul feel a force if place there. So that rules out (A) same irection of motion E F e p F Both feel the same magnitue of force but the proton is much more massive, so it will accelerate less. This rules out (B) magnitue of velocity an (E) magnitue of acceleration So if they are in the fiel for the same amount of time, will they have equal magnitue of momentum or equal kinetic energy? How can we reason about this? You Can t Escape Newtonian Mechanics!!! The energy of a system is change by work. The momentum of a system is change by an impulse. DKE = FDx Dp = FDt Imagine an infinitely large, uniform positively charge plate which prouces an electric fiel of magnitue E when by itself. You a another plate, negatively charge, at a istance from the first one an parallel to it. What is the magnitue of the E fiel to the left of the plates, between them, an to the right? The particles feel the same magnitue of force for the same amount of time, therefore they will en up with the same magnitue of momentum! The electron will actually have a greater isplacement uring this time, so the work one on it will be greater. The electron will en up with more KE than the proton! 8

9 CONSTANT ELECTRIC FIELDS Draw the path that a negatively charge E fiel lines give information about the force that the electric fiel exerts on a test charge at any location, an about the resulting acceleration of the test charge, but not the irection of motion of the test charge. particle will follow through the parallel charge plates. ELECTRIC FIELD DEFLECTS AN INK BALL Insie an inkjet printer a tiny ball of black ink of mass 1.1x10 11 kg with charge 6.7x10 1 C moves horizontally at 40 m/s. The ink ball enters an upwar pointing uniform E fiel of magnitue 1.0x10 4 N/C prouce by a negatively charge plate above an a positively charge plate below. The plates are use to eflect the ink ball so that it lans at a particular sport on a piece of paper. 9

10 ELECTRIC FIELD DEFLECTS AN INK BALL SOLUTION Draw a force iagram. Fin the magnitue of all forces exerte on the ink ball. Fin the time it takes the ink ball to go through the parallel plates. Fin the vertical acceleration of the ink ball. Fin the vertical isplacement the ink ball is eflecte so that it lans at a particular spot on a piece of paper. F q = E q F q = 6. 7x10 8 N xcomponent (constant velocity) Δt = ΔX v x Δt =. 5x10 4 s a = ΣF m a = m s ycomponent (constant acceleration) a Δt Δy = v y0 Δt Δy = 1. 9x10 4 m Electric Potential THE V FIELD Can we escribe electric fiels using the concepts of work an energy? To o so, we nee to escribe the electric fiel not as a forcerelate E an energyrelate fiel fiel, but as an energyrelate fiel. Electric potential ue to a single pointlike charge object Imagine an object with positive electric charge Q (we use capital Q to enote the object whose electric potential we are investigating). r Imagine three test objects: A, B an C, with charges q A = q, q B = q, an q C = 3q, that can be place at the location shown. r We wish to etermine the electric potential at a location that is a istance r from the object. q A = q q B = q q C = 3q Determine the electric potential energy that woul result from placing each test charge there. 10

11 q A = q q B = q q C = 3q r q A = q q B = q q C = 3q We place test objects A, B an C one at a time at the same location relative to Q. U Q an qa = k Qq A r æ = q A k Q ö ç è r ø U q = kq 1q r U Q an qb = k Qq B r U Q an qc = k Qq C r æ = q B k Q ö ç è r ø æ = q C k Q ö ç è r ø = U Q an A = 3U Q an A Although the potential energies iffer, they are proportional to the charges of the test objects. U Q an A = k Q r (q) Despite these ifferences, the ratio of resulting potential energy an the charge of the test object is ientical for all three objects: U Q an A q A U Q an B = k Q r (q) = U Q an B q B = U Q an C q C U Q an C = k Q r (3q) = kq r Since these ratios are all the same value, the ratio U Q an q is a mathematical q escription of the energy fiel cause by charge Q at a istance r from Q. r V = U Q an q test q test ELECTRIC POTENTIAL ( V ) V = U Q an q test q test r = kqq test rq test = kq r The electric potential at that location is inepenent of the test charge, an characterizes the energy per charge that woul result from placing a charge at that location. Q Electric potential is a property of a location in space that measures the energy per unit charge that a charge object woul feel if place at that location. 11

12 Symbol Units Units symbol Type of PQ PHYSICAL QUANTITY: ELECTRIC POTENTIAL V Joules per Coulomb (Volts) J/C [ v ] scalar V = U q q V = k Q r To etermine the V fiel at a specific location, place a test charge q test at that location an measure the electric potential energy of the system consisting of Q an q test. The V fiel at that location equals the ratio V = U Q an q test q test = kq r an is inepenent of the test charge. ELECTRIC POTENTIAL A, B, & C are ranom points aroun a C electric charge. Fin the electric potential prouce by the electric charge at points A, B, & C. Assume each box to be = 0.1 m A C C B V A = V V B = V V c = V If we know the electric potential V at a location, we can rearrange the efinition of the V fiel to etermine the electric potential energy of a system that inclues the source charge an a charge q at that location. U Q an q = qv Due to Q WB Insie an Xray machine, a hot filament ejects electrons. Imagine one of those electrons starts from rest an accelerates through a region where the V fiel increases by 40,000 V. The electron stops abruptly when it hits a piece of tungsten, proucing Xrays. How fast is the electron moving just before it reaches the tungsten? 1

13 Superposition Principle for Electric Potential U qi = U qf KE F m v V i q = V F q V i = 0 V V F = 40, 000 V m v = V F e v = V F e m v = 118, 599, 890 m s v = 0. 4c A B C D 9 V 18 V 7 V 36 V V = V 1 V V 3... = kq 1 r 1 kq r kq 3 r 3... The electric potential at a location in space is the scalar sum (arithmetic sum) of the iniviual contributions to the electric potential at that location prouce by each source charge. 0. m A 0. m Two 3µC charge point like objects are separate by 0. m. Write an expression for the electric potential at a points A an B. Fin the magnitue of the electric potential at points A an B. B V Anet = V Anet = k q k q 4 k q V Anet = 540, 000 V V Bnet = V Bnet = k q k q 3 k q V Bnet = 0, 500 V Suppose that the heart's ipole charges Q an Q are separate by a istance. Write an expression for the V fiel ue to both charges at point A, at a istance to the right of the Q charge. V net = k Q V net = k Q k q 0. m A 0. m One 3µC an one 3µC charge point like objects are separate by 0. m. Write an expression for the electric potential at a points A an B. Fin the magnitue of the electric potential at points A an B. B 13

14 V Anet = k q V Anet = 0 k q V Bnet = k q k q k q V Bnet = V Bnet = 67, 500 V Four objects with the same charge q are place at the corners of a square of sie. A. Determine the value of the Electric Fiel at a point that is locate in the center of the square (sketch). B. Determine the value of the Electric Potential at a point that is locate in the center of the square. y / / E net = 0 N C V net = 4 k q 4 k q V net = q q A C B x 1. Write an expression for the electric fiel at point A. Write an expression for the electric potential at point A SOLUTION 1. Write an expression for the electric fiel at point A E NET = kq kq E NET = kq. Write an expression for the electric potential at point A V NET = kq kq V NET = 0 y q q A C B x 1. Write an expression for the electric fiel at point B. Write an expression for the electric potential at point B 14

15 SOLUTION 1. Write an expression for the electric fiel at point B E NET = kq 3 kq E NET = 8kq 9. Write an expression for the electric potential at point B V NET = kq 3 kq V NET = kq 3 y q q A C B x 1. Write an expression for the electric fiel at point C. Write an expression for the electric potential at point C SOLUTION 1. Write an expression for the electric fiel at point C E NET = kq ( ) cosθ E NET = kq E NET = kq. Write an expression for the electric potential at point C V NET = kq kq V NET = 0 WB Can you think of locations relative to charge istributions where A) the V fiel at a particular location is zero but the E fiel is not? B ) the E fiel is zero but the V fiel is not? a) ur kq kq kq E = right right = right V = kq r r r r k(q) = 0 V r b) Look at the two situation below (point P is locate halfway between both charges): Fin Magnitue an irection of electric fiel at point P. Fin magnitue of electric potential at point P. Conclusions? ur kq kq E = right r r left = 0 N / C V = kq r kq r = kq r 15

16 THE VFIELD V net = E net = 0 4 k q V net = 0 E net = 8 k q E fiel is a vector quantity that escribes the force felt by a test charge at a location. V fiel is a scalar quantity that escribes the energy ae by placing a test charge at a location. Electric potential is a scalar quantity that can have a positive or negative value, epening on the FLAVOR of the source charge that is creating the V fiel. ISOLINES (EQUIPOTENCIAL LINES) Electric potential can be positive or negative, epening on the sign of the source charge creating the V fiel. 16

17 Equipotential surfaces represent the locations along which electric potential is constant. If a test charge is move along one of these surfaces, no work is neee (ΔV = 0; ΔU q = 0) ISOLINES an EFIELD LINES V point source = kq r The equipotentials prouce by a point source are spheres (they look like circles on a D crosssection) because V is the same at all locations equiistance from a point particle. Perpenicular to each other The electric potential of a point charge is given by V is a scalar quantity. It is not affecte by negative X V 1.5 V = kq (scalar) 1 X On your whiteboars, sketch the graph of V vs for a point charge. Assume that the charge is locate at the origin. The slope of the graph (ΔV/Δx) gets steeper as you get closer to the source charge This means that the electric potential increases more rapily as you get closer to the charge 17

18 In 3D V point source = kq r V point source = kq r The graph gets steeper approaching the source charge. This means that closer to the source, electric potential increases more quickly with respect to position. r 3D ANALOGY! Contour line a line on a map joining points of equal height above or below sea level. The contour lines are close together where the hill is the steepest. The gravitational potential increases in a short amount of istance. Contour maps: An analogy for equipotential surfaces 014 Pearson Eucation, Inc. 18

19 The equipotential lines show the same thing! As you get closer to the source, the equipotential lines get closer together. This means that closer to the source, the potential increases in a shorter amount of istance. Closer to the source charge, the equipotential lines get closer together. This means that the potential increases by the same amount in a shorter amount of istance. Negative Source Charge How about for a negative source charge? r V point source = kq r Similar iea, but now the potentials get more an more negative closer the source Negative Source Charge In 3D, an top view of contour lines (equipotentials) Similar iea, but now potential becomes more negative closer to the source. 19

20 In 3D V point source = kq r When release in a V fiel, a positively charge particle will accelerate towar lower electric potential. a When release in a V fiel, a positively charge particle will accelerate towar higher electric potential. When release in a V fiel, a negatively charge particle will accelerate towar higher electric potential. a a When release in a V fiel, a negatively charge particle will accelerate towar smaller electric potential. a 0

21 Deriving a relation between the E fiel an ΔV RELATING THE EFIELD AND THE VFIELD We attach a small object with charge q to the en of a very thin wooen stick an place the charge object an stick in the electric fiel prouce by the plate. Draw a Force Diagram Charge is the system Ignore force that Earth exerts on the charge. Charge is not accelerating 1

22 FORCE ANALYSIS WORK ENERGY ANALYSIS F SonO F QonO The stick moves away from the positively charge plate. System is charge an electric fiel. F SonO = F QonO F SonO = E q W = F SonO x cosθ Deriving a relation between the E fiel an ΔV W = F SonO x cosθ F SonO = E q W = E q x cos180 U q = E q x cos180 U q = E q x Divie everything by q Deriving a relation between the E fiel an ΔV U q q E = V x E q x = q V = E x Reorganize E = V x E = V r E = V Same quantity expresse in three ifferent ways! Ben brings a groune metal sphere with a wooen hanle near a Van De Graaff generator so that their potential is 450,000 v an the sphere oes not get charge. Ben pulls the sphere away. Preict the magnitue of the electric fiel when you see a spark. Vieo

23 E = V x E = E =, 647, 059 V m 450, Electric Fiel is so strong that it causes a DIELECTRIC breakown an we see a spark CONDUCTORS IN ELECTRIC FIELDS Electric fiel outsie a charge conuctor Charge Conuctor When a conucting sphere is given a net charge, it can be treate like a point charge at its center. Outsie the sphere: E = kq V = kq Where is the istance to the center of the sphere The EFiel within a conuctor is always zero. (vector aition) If it were not zero, the electrons woul continue to move. What is E & V insie the sphere??? The VFiel within a conuctor is always E = 0 N C V = kq r V = kq r (scalar aition) 3

24 Electric fiel insie/outsie a charge conuctor Electric fiel is a vector physical quantity. Electric fiel insie the charge conuctor cancel each other out. E=0 insie the conuctor Electric potential insie/outsie a charge conuctor Electric Potential is a scalar physical quantity. Electric potential insie the charge conuctor a up. V>0 insie the conuctor Grouning Grouning ischarges an object mae of conucting material by connecting it to Earth. Electrons will move between an within the spheres until the V fiel on the surfaces of an within both spheres achieves the same value. SHIELDING Negative charges insie the conuctor move to a higher potential 4

25 Negative charges insie the conuctor move to a higher potential An Electric Fiel E 1 is prouce insie the hollow conuctor E NET = E 1 E 0 E NET = 0 Electric Fiels cancel Key Concept The electric fiel insie a conuctor is always zero! This is why the metal frame of a car will protect you from lightning strikes! Dielectric materials in an electric fiel DIELECTRICS IN ELECTRIC FIELDS If an atom in a ielectric material resies in a region with an external electric fiel, the nucleus an the electrons are isplace slightly in opposite irections until the force that the fiel exerts on each of them is balance by the force they exert on each other. Electric Dipole 5

26 Polar water molecules in an external electric fiel Some molecules, such as water, are natural electric ipoles even when the external E fiel is zero. E fiel insie a ielectric A ielectric material cannot completely shiel its interior from an external electric fiel, but it oes ecrease the fiel. E fiel insie a ielectric Physicists use a physical quantity to characterize the ability of ielectrics to ecrease the E fiel: Dielectric constants for ifferent types of materials The ielectric constant (kappa) κ = E 0 External Efiel E i Inuce Efiel Electric force an ielectrics The force that object 1 exerts on object is reuce by compare with the force it woul exert in a vacuum. CAPACITORS F 1on = k q 1 q κ Insie the ielectric material, Coulomb's law is now written as: 6

27 A capacitor consists of two conucting surfaces separate by a nonconucting material. The simplest ones are parallel plate capacitors separate by air, rubber, paper or some other ielectric material. The purpose of a capacitor is to store electric potential energy. ParallelPlate Capacitor 9 V A battery provies a ifference in electric potential (ΔV) between its terminals. 9 V E = V x This potential ifference is the voltage of the battery 9 Volts = 9 Joules/Coulomb Each coulomb of particles that move between the terminals of the battery gain 9 J of energy. When the terminals are connecte through a circuit, an electric fiel propagates through the conucting wires, causing a flow of electrons opposite to the electric fiel. This is what causes an electric current. 7

28 CAPACITORS A negatively charge particle accelerates from regions of lower potential towar regions of higher potential (more negative electric potential). Capacitor in the process of charging Charge capacitor BATTERY Higher potential Charges will stop moving when the V of the battery is the same V of the capacitor. This process happens very quickly. BATTERY BATTERY BATTERY 8

29 An uncharge capacitor behaves like an ieal wire (zero resistance) for the initial moment that it is uncharge. A capacitor which no longer has charges separating behaves like an open switch (infinite resistance). Capacitors E = V x E = V x The magnitue of the electric fiel between the plates of the capacitor is equal to the magnitue of the potential ifference between the plates, ivie by the separation between the plates ( or x). If E V an E q plate, then q plate V E V In orer to create a stronger electric fiel, more separation of charge must occur between the plates. E q plate The greater the voltage across the plates, the more charge will separate between the plates of the capacitor. 9

30 q plate MATH CLASS MATH MODEL y = m x b PHYSICS q plate = C V V q plate Slope = C Capacitance V Capacitance The proportionality constant C in this equation is the capacitance of the capacitor. q plate = C V Capacitance is an inherent property of the capacitor itself, an epens on its geometry an type of ielectric material. q plate q plate = C V Capacitance is a property of the capacitor itself. Units: Faras 1 coulomb/volt = 1 fara QUANTITIES THAT AFFECT THE CAPACITANCE OF A CAPACITOR: PLATE AREA A capacitor with largersurfacearea plates shoul be able to maintain more charge separation because there is more room for the charge to sprea out V (1 C/V = 1 F) 30

31 QUANTITIES THAT AFFECT THE CAPACITANCE OF A CAPACITOR: DISTANCE SEPARATION A larger istance between the plates leas to a smallermagnitue E fiel between the plates. Because the magnitue of this E fiel is proportional to the amount of electric charge on the plates, a larger plate separation leas to a smallermagnitue electric charge on the plates. QUANTITIES THAT AFFECT THE CAPACITANCE OF A CAPACITOR: DIELECTRIC CONSTANT Material between the plates with a large ielectric constant becomes polarize by the electric fiel between the plates. Thus more charge moves onto capacitor plates that are separate by material of high ielectric constant. Capacitance of parallel plate capacitor The capacitance of a particular capacitor shoul increase if the surface area A of the plates increases, ecrease if the istance between them is increase, an increase if the ielectric C = κε 0A C = κa 4πk constant of the material between them increases: C = κε 0A C = κa 4πk κ = ielectric constant Tip Fin an expression for the area uner the graph. Determine the physical quantity base on its units. q plate V 31

32 C V A TRIANGLE = height base q ΔV A TRIANGLE = q V Units analysis: C J C J Units of energy C V A = A = q V f Units analysis: q V C q V i q V J C J Units of energy ELECTRIC POTENTIAL ENERGY STORED BY A CAPACITOR U C [ J ] A capacitor besies of storing electric charge, also stores electric potential energy. U C = q V CAPACITANCE C = q V ELECTRIC POTENTIAL ENERGY q V U C = Write a mathematical moel for U C in terms of C an V C V U C = Write a mathematical moel for U C in terms of C an q U C = q C CAPACITANCE CAPACITORS: MATHEMATICAL MODELS CHANGE IN ELECTRIC POTENTIAL ELECTRIC FIELD MATHEMATICAL MODELS Write a mathematical moel for E fiel in terms of q, k,, A. E = V V = q E = q x C C x U C = q V ELECTRIC POTENTIAL ENERGY U C = C V U C = q C C = κε 0A ε 0 = 1 4πk E = q xκε 0 A E = 4πkq κa E = q κε 0 A 3

33 E = 4πkq κa CAPACITORS GAUSS LAW E = q κε 0 A Estimate the capacitance of your physics textbook, assuming that the front an back covers (area A = m, separation = m) are mae of a conucting material. The ielectric constant of paper is approximately 6.0. Gauss Law is a relationship between the Electric fiel between the plates of the capacitor, the store charge an the Area of the plates. Determine what the potential ifference must be across the covers for the textbook to have a charge separation of 1x10 6 C (one plate has charge 1x10 6 C an the other has charge 1x10 6 C). C = C = κa 4πk π 9x C = pf V = V = q C 1x10 6 C 66.31x10 1 F V = V During ventricular fibrillation the heart muscles contract ranomly, preventing the coorinate pumping of bloo. A efibrillator can often restore normal bloo pumping by ischarging the charge on a capacitor through the heart. Pales are hel against the patient s chest, an a 6µF charge capacitor is ischarge in several millisecons. If the capacitor energy is 50 J, what potential ifference was use to charge the capacitor? How much electric charge is store by the capacitor? U C = C V V = q C A parallel plate capacitor has square metallic plates of ege length 10 cm separate by 1.0 mm (air). V = U C C V = 918 V q = V C q = mc a. Calculate the capacitance of this evice. b. Calculate the electric fiel between plates if this capacitor stores 1.061nC of electric charge. c. What potential ifference was use to charge the capacitor? 33

34 C = κa 4πk E = 4πkq κa C = 88.4 pf E = 1,000 V m E = F q q U q = q V F q = kqq E = kq V = q C V = 1 V E = V x V = kq ΔV = q C POINT LIKE CHARGES CAPACITORS C = κε 0A C = κa 4πk E = kq E = 4πkq κa E = V E = 4πkq κa E = q κε 0 A V = kq V = q C U C = q V U C = U C = q C C V U q = q V U C = q V 34