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T. Norah Ali Almoneef1 Lightning Becomes very “negative” Becomes very “positive”

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2 T. Norah Ali Almoneef1 Lightning Becomes very “negative” Becomes very “positive”

3 Arbitrary numbers of protons (+) and electrons (-) on a comb and in hair (A) before and (B) after combing. Combing transfers electrons from the hair to the comb by friction, resulting in a negative charge on the comb and a positive charge on the hair. 2T. Norah Ali Almoneef If a positively charged rod is brought near a trickle of water, the water moves towards it. What happens if we use a negatively charged rod? 2

4 T. Norah Ali Almoneef3 Electric Charge Types: – Positive Glass rubbed with silk Missing electrons – Negative Rubber/Plastic rubbed with fur Extra electrons Arbitrary choice – convention attributed to ? Units : amount of charge is measured in [Coulombs] Empirical Observations: – Like charges repel – Unlike charges attract

5 T. Norah Ali Almoneef4 In the process of rubbing two solid objects together, electrical charges are NOT created. Instead, both objects contain both positive and negative charges. During the rubbing process, the negative charge is transferred from one object to the other and this leaves one object with an excess of positive charge and the other with an excess of negative charge. The quantity of excess charge on each object is exactly the same.

6 T. Norah Ali Almoneef5 Charge Properties CONSERVATION OF ELECTRIC CHARGE – Charge is not created or destroyed, only transferred., however, it can be transferred from one object to another. The net amount of electric charge produced in any process is zero. Quantization – The smallest unit of charge is that on an electron or proton. (e = 1.6 x C) It is impossible to have less charge than this It is possible to have integer multiples of this charge – A coulomb is the charge resulting from the transfer of 6.24 x of the charge carried by an electron. – The magnitude of an electrical charge (q) is dependent upon how many electrons (n) have been moved to it or away from it. Mathematically,

7 Electric Charge and Electrical Forces: Electrons have a negative electrical charge. Protons have a positive electrical charge. These charges interact to create an electrical force. – Like charges produce repulsive forces – so they repel each other (e.g. electron and electron or proton and proton repel each other). – Unlike charges produce attractive forces – so they attract each other (e.g. electron and proton attract each other). T. Norah Ali Almoneef6 It is impossible to have less charge than this It is possible to have integer multiples of this charge How many electrons constitute 1  C?

8 7 Electric Force - Coulomb’s Law Consider two electric charges: q 1 and q 2 The electric force F between these two charges separated by a distance r is given by Coulomb’s Law The constant k is called Coulomb’s constant and is given by T. Norah Ali Almoneef Coulomb law – The electrical force between two charged bodies is directly proportional to the charge on each body and inversely proportional to the square of the distance between them.

9 8 The coulomb constant is also written as  0 is the “electric permittivity of vacuum” – A fundamental constant of nature According to the superposition principle the resultant force on a point charge q equals the vector sum of the forces exerted by the other point charges Q i that are present: T. Norah Ali Almoneef The force between two charges gets stronger as the charges move closer together. The force also gets stronger if the amount of charge becomes larger.

10 T. Norah Ali Almoneef9 the resultant force on any one of them equals the vector sum of the forces exerted by the various individual charges. For example, if four charges are present, then the resultant force exerted by particles 2, 3, and 4 on particle 1 is or

11 T. Norah Ali Almoneef10 example A positive charge of 6.0 x C is 0.030m from a second positive charge of 3.0 x C. Calculate the force between the charges. = (8.99 x 10 9 N m 2 /C 2 ) (6.0 x C) (3.0 x C) ( 0.030m ) 2 = (8.99 x 10 9 N m 2 /C 2 ) (18.0 x C) (9.0 x m 2 ) = x N

12 T. Norah Ali Almoneef11 Three point charges, q 1 = - 4 nC, q 2 = 5 nC, and q 3 = 3 nC, are placed as in the Fig.

13 Coulomb's Law The force between charges is directly proportional to the magnitude, or amount, of each charge. Doubling one charge doubles the force. Doubling both charges quadruples the force. The force between charges is inversely proportional to the square of the distance between them. Doubling the distance reduces the force by a factor of 2 2 = (4), decreasing the force to one-fourth its original value (1/4).. 12T. Norah Ali Almoneef

14 13 Compare electrical to gravitational force in a hydrogen atom Electron and proton attract each other times stronger electrically than gravitationally

15 T. Norah Ali Almoneef14 An object, A, with x C charge, has two other charges nearby. Object B, -3.5 x C, is m to the right. Object C, * C, is m below. What is the net force and the angle on A?V

16 T. Norah Ali Almoneef Example What is the magnitude of the electric force of attraction between an iron nucleus (q=+26e) and its innermost electron if the distance between them is 1.5 x m.The magnitude of the Coulomb force is F = kq 1 q 2 /r 2 = (9.0 X 10 9 N · m 2 /C 2 )(26)(1.60 X 10 –19 C)(1.60 X10 –19 C)/(1.5 X10 –12 m) 2 = 2.7 X 10 –3 N. T. Norah Ali Almoneef15 Example We have a charge of 5 C at X = 2 m. What will the force on a charge of -4 C be at X = 0 m. F = kq 1 q 2 / r 2, F = 5 C x (4C) x 9x10 9 / 4 m 2 F = 45 x10 9 N 15

17 T. Norah Ali Almoneef16 Given that q = -13 µC and d = 13 cm, find the direction and magnitude of the net electrostatic force exerted on the point charge q 1 in Figure. F 12 (force on q 1 from q 2 )= k (q 1 )(q 2 )/(d) 2 = (9x10 9 N m 2 /C 2 ) (13x10 -6 C)(26x10 -6 C)/(0.13m) 2 = 180 N towards q 2 (attractive) F 13 (force on q 1 from q 3 )= k(q 2 )(q 3 )/(2d) 2 = (9E9 N m 2 /C 2 ) (13x10 -6 C)(-39x10 -6 C)/(0.26m) 2 = N away from q 3 (repulsive) Net force = (180 – 67.4)=112 N Towards q 2 q 1 =+q q   2q Q 3 =+3q example

18 T. Norah Ali Almoneef17 Two point charges lie on the x axis. A charge of -9.5 µC is at the origin, and a charge of µC is at x = 10.0 cm. (a)What is the net electric field at x = -2.0 cm? Electric field from charge –9.5x10 -6 C is E 1 = (9x10 +9 Nm 2 /C 2 .5x10 -6 C)/(0-(-0.02m)) 2 E 1 = 2.14E8 N/C field points towards charge –9.5x10 -6 (positive x-direction) Electric field at x=-0.02 m from charge 2.5x -6 C is E 2 = (9x10 +9 Nm 2 /C 2 .5x10 -6 C)/(0.10m-(-0.02m)) 2 E 2 = 1.56E6 N/C field points away from charge 2.5x10 -6 (negative x-direction) Net electric field E = 2.14E8 i N/C  1.56E6 i N/C = [2.12e+08] i N/C example

19 T. Norah Ali Almoneef18 Three point charges, q1 = -1.2 x 10-8 C, q2 = -2.6 x 10-8 C and q3 = +3.4 x 10-8 C, are held at the positions shown in the figure, where a = 0.16 m example

20 T. Norah Ali Almoneef19 example Two electrostatic point charges of μC and –30.0 μC exert attractive forces on each other of –145 N.What is the distance between the two charges?

21 T. Norah Ali Almoneef20 Three point charges lie along the x axis as shown in Figure. The positive charge q 1 ! 15.0 *C is at x ! 2.00 m,the positive charge q 2 ! 6.00 *C is at the origin, and the resultant force acting on q 3 is zero. What is the x coordinate of q 3 ? Solution Because q3 is negative and q1 and q2 are positive, For the resultant force on q 3 to be zero, F 23 must be equal in magnitude and opposite in direction to F 13. Setting the magnitudes of the two forces equal,

22 T. Norah Ali Almoneef21T. Norah Ali Almoneef21 Consider two charges located on the x axisThe charges are described by q 1 = 0.15  Cx 1 = 0.0 m q 2 = 0.35  Cx 2 = 0.40 m Where do we need to put a third charge for that charge to be at an equilibrium point? At the equilibrium point, the forces from the two charges will cancel. third charge to be at an equilibrium point when = 0.12m or = 0.72m X

23 T. Norah Ali Almoneef22 Example Suppose two charges having equal but opposite charge are separated by 6.4 × m. If the magnitude of the electric force between the charges is 5.62 ×10 –14 N, what is the value of q

24 T. Norah Ali Almoneef23 example Charged spheres A and B are fixed in position, as shown, and have charges of +7.9 x C and -2.3 x C, respectively. Calculate the net force on sphere C, whose charge is +5.8 x C.

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26 25 Example: What is the force between two charges of 1 C separated by 1 meter? F= Example: What is the electric force between a +5 mC charge and a –3 mC charge separated by 3 cm? F = (9 x 10 9 Nm 2 /C 2 )(.005C)(.003C) / (.03 m) 2 = 150,000,000 N T. Norah Ali Almoneef

27 26 How far apart would two objects, each with a charge of 1Coulomb, have to be so that they only exerted a 1 Newton electric force on one another?

28 T. Norah Ali Almoneef27 Object A has a charge of +2 μC, and object B has a charge of +6 μC. Which statement is true about the electric forces on the objects? F AB = –3F BA F AB = –F BA 3F AB = –F BA F AB = 3F BA F AB = F BA 3F AB = F BA From Newton's third law, the electric force exerted by object B on object A is equal in magnitude to the force exerted by object A on object B and in the opposite direction.

29 T. Norah Ali Almoneef28 Where do I have to place the + charge in order for the force to balance, in the figure at right? Force is attractive toward both negative charges, hence could balance. Need a coordinate system, so choose total distance as L, and position of + charge from  q charge as x. Force is sum of the two force vectors, and has to be zero, so A lot of things cancel, including Q, so our answer does not depend on knowing the + charge value. We end up with Solving for x,, so slightly less than half-way between.A qq  q x L

30 T. Norah Ali Almoneef29 A charged particle, with charge Q, produces an electric field in the region of space around it A small test charge, q o, placed in the field, will experience a force Electric Field Mathematically, Use this for the magnitude of the field The electric field is a vector quantity The direction of the field is defined to be the direction of the electric force that would be exerted on a small positive test charge placed at that point For a point charge

31 T. Norah Ali Almoneef30 The force (F e ) acting on a “test” charge (q o ) placed in an electric field (E) is F e = q o E Note the similarity of the electric force law to Newton’s 2 nd Law (F=ma) Formal definition of electric field: – the electric force per unit charge that acts on a test charge at a point in space or E = F e /q o The electric field exists whether or not there is a test charge present The Superposition Principle can be applied to the electric field if a group of charges is present

32 T. Norah Ali Almoneef31 Electric Field The electric force on a positive test charge q 0 at a distance r from a single charge q : The electric field at a distance r from a single charge q:

33 T. Norah Ali Almoneef32 A positive charge is released from rest in a region of electric field. The charge moves: a) towards a region of smaller electric potential b) along a path of constant electric potential c) towards a region of greater electric potential Example 9 A positive charge placed in an electric field will experience a force given by Therefore Since q is positive, the force F points in the direction opposite to increasing potential or in the direction of decreasing potential

34 T. Norah Ali Almoneef The electric field at a given point P is the sum of the electric fields due to every point charge P A charge distribution

35 T. Norah Ali Almoneef34 The electric force on a charge q is which, together with Newton’s 2 nd Law, can be used to calculate the motion of an electric charge, of mass m Newton’s 2 nd Law for an electric charge can be written as If E is constant, both in direction and magnitude, so to is the acceleration of the charge. Note that the acceleration depends on the charge to mass ratio.

36 T. Norah Ali Almoneef35 Electric field & line Point charge The lines radiate equally in all directions For a positive source charge, the lines will radiate outward For a negative source charge, the lines will point inward

37 T. Norah Ali Almoneef36 – A map of the electrical field can be made by bringing a positive test charge into an electrical field. No two field lines can cross. You can draw vector arrows to indicate the direction of the electrical field. This is represented by drawing lines of force or electrical field lines, These lines are closer together when the field is stronger and farther apart when it is weaker. Field lines must begin on positive charges (or from infinity) and end on negative charges (or at infinity). The test charge is positive by convention the magnitude of the electric force will increase proportionally with an increase in charge and/or and increase in the electric field magnitude The line must be perpendicular to the surface of the charge The number of lines drawn leaving a positive charge or approaching a negative charge is proportional to the magnitude of the charge

38 T. Norah Ali Almoneef37 Shark T. Norah Ali Almoneef37 Fish to detect object

39 T. Norah Ali Almoneef38 Electric Field lines

40 T. Norah Ali Almoneef39 example

41 T. Norah Ali Almoneef40 Which of the following statements about electric field lines associated with electric charges is false? 1) Electric field lines can be either straight or curved. 2) Electric field lines can form closed loops. 3 )Electric field lines begin on positive charges and end on negative charges. 4 ) Electric field lines can never intersect with one another. example

42 T. Norah Ali Almoneef41 4 E 0 (1) 4 E 0 2 E 0 (2) 2 E 0 E 0 (3) E 0 1/2 E 0 (4) 1/2 E 0 4 E 0 (5) 1/4 E 0 You are sitting a certain distance from a point charge, and you measure an electric field of E 0. If the charge is doubled and your distance from the charge is also doubled, what is the electric field strength now? Doubling the chargefactor of 2doubling the distancefactor of 4 factor of 1/2 Remember that the electric field is: E = kQ/r 2. Doubling the charge puts a factor of 2 in the numerator, but doubling the distance puts a factor of 4 in the denominator, because it is distance squared!! Overall, that gives us a factor of 1/2.

43 T. Norah Ali Almoneef dd 1) 2) 3) the same for both Between the red and the blue charge, which of them experiences the greater electric field due to the green charge? same electric field same point in space Both charges feel the same electric field due to the green charge because they are at the same point in space!

44 T. Norah Ali Almoneef dd 1) 2) 3) the same for both Between the red and the blue charge, which of them experiences the greater electric force due to the green charge? electric field is the same force magnitude of that specific charge The electric field is the same for both charges, but the force on a given charge also depends on the magnitude of that specific charge.

45 T. Norah Ali Almoneef44 example Find electric field at point P in the figure o

46 T. Norah Ali Almoneef45 Example In figure shown, locate the point at which the electric field is zero? Assume a = 50cm E 1 = E 2 d = 30cm Example Find the electric field at point p in figure.due to the charges shown.

47 T. Norah Ali Almoneef46 Example In figure shown, locate the point at which the electric field is zero? Assume a = 50cm E 1 = E 2 d = 30cm Example Find the electric field at point p in figure.due to the charges shown.

48 T. Norah Ali Almoneef47 Example In figure shown, locate the point at which the electric field is zero? Assume a = 50cm E 1 = E 2 d = 30cm Example Find the electric field at point p in figure.due to the charges shown. E x = E 1 - E 2 = -36  10 4 N/C E y = E 3 = 28.8  10 4 N/C E p = [(36  10 4 ) 2 +(28.8  10 4 ) 2 ] = 46.1N/C  = 141 o

49 T. Norah Ali Almoneef48 Example 30 ° E1E1 E2E2 E3E3 Three identical charges (q = –5.0 mC) lie along a circle of radius 2.0 m at angles of 30°, 150°, and 270°, as shown. What is the resultant electric field at the center of the circle ?

50 T. Norah Ali Almoneef49 Example A +100  C point charge is separated from a -50  C charge by a distance of 0.50 m as shown below. (A) First calculate the electric field at midway between the two charges. (B) Find the force on an electron that is placed at this point and then calculate the acceleration when it is released. + Q1Q1 _ Q2Q2 In part A we found that E = 2.1x10 7 N/C and is directed to the right. ( to left )

51 T. Norah Ali Almoneef50 the Field on Electric Dipole The total field at P is When r is much greater than a we can neglect a in the denominator then Electric Field due to arrangements of charges 16.3 Electric Field due to arrangements of charges

52 T. Norah Ali Almoneef51 Note the dipole electric field reduces as 1/r 3, instead of 1/r of a single charge. although we only calculate the fields along z-axis, it turns out that this also applies to all direction. p is the basic property of an electric dipole, but not q or d. Only the product qd is important. E E p A set of two (equal and opposite) charges separated by a distance

53 T. Norah Ali Almoneef52 Example Find the electric field due to electric dipole shown in figure along x-axis at point p which is a distance r from the origin. then assume r >> a Solution When x>>a then

54 T. Norah Ali Almoneef53 A test charge of +3 μC is at a point P where an external electric field is directed to the right and has a magnitude of 4 × 10 6 N/C. If the test charge is replaced with another test charge of –3 μC, the external electric field at P A )is unaffected B )reverses direction C )changes in a way that cannot be determined There is no effect on the electric field if we assume that the source charge producing the field is not disturbed by our actions. Remember that the electric field is created by source charge(s) (unseen in this case), not the test charge(s).

55 T. Norah Ali Almoneef54 A Styrofoam ball covered with a conducting paint has a mass of 5.0 × 10-3 kg and has a charge of 4.0 μC. What electric field directed upward will produce an electric force on the ball that will balance the weight of the ball? (a) 8.2 × 102 N/C (b) 1.2 × 104 N/C (c) 2.0 × 10-2 N/C (d) 5.1 × 106 N/C The magnitude of the upward electrical force must equal the weight of the ball. That is:

56 T. Norah Ali Almoneef55 An electric dipole consists of two equal and opposite charges The high density of lines between the charges indicates the strong electric field in this region Two equal but like point charges At a great distance from the charges, the field would be approximately that of a single charge of 2q The bulging out of the field lines between the charges indicates the repulsion between the charges The low field lines between the charges indicates a weak field in this region

57 T. Norah Ali Almoneef 56 Since the 4 C charge feels a force, there must be an electric field present, with magnitude: E = F / q = 12 N / 4 C = 3 N/C Once the 4 C charge is replaced with a 6 C charge, this new charge will feel a force of: F = q E = (6 C)(3 N/C) = 18 N Q 1) 12 N 2) 8 N 3) 24 N 4) no force 5) 18 N In a uniform electric field in empty space, a 4 C charge is placed and it feels an electrical force of 12 N. If this charge is removed and a 6 C charge is placed at that point instead, what force will it feel?

58 T. Norah Ali Almoneef57

59 T. Norah Ali Almoneef58 Determine the point (other than infinity) at which the total electric field is zero we will call the position of the negative charge x = 0, which means the positive charge is at x = 1 m. We will call the position where electric field is zero x. The distance from this point to the negative charge is just x, and the distance to the positive charge is 1 + x. Now write down the electric field due to each charge: example We wrote down the distance x the distance to the left of the negative charge. A negative value of x is then in the wrong direction, in between the two charges, which we already ruled out. The positive root, x = 1.82, means a distance 1.82m to the left of the negative charge. This is what we want.

60 T. Norah Ali Almoneef59 A charge q 1 = 7.0 µC is located at the origin, and a second charge q 2 =5.0 µC is located on the x axis, 0.30 m from the origin.Find the electric field at the point P, which has coordinates (0, 0.40) m.

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62 T. Norah Ali Almoneef61T.Norah Ali Almoneef61

63 T. Norah Ali Almoneef62 Electric fields of Concentric spherical shells If the smaller shell has a radius R : (Out side) Surface area of the sphere is :

64 63T. Norah Ali Almoneef

65 64 Planar Symmetry  Two conducting plates with charge density  1  All charges on the two faces of the plates  For two oppositely charged plates placed near each other, E field outer side of the plates is zero while inner side the E-field= 2  1 /  0 For two oppositely charged plates placed near each other, E field outer side of the plates is zero while inner side the E-field = E = 2π KQ / A

66 T. Norah Ali Almoneef65T. Norah Ali Almoneef65 E field of a single uniformly charged plate Charged infinite plane: E = 2  k  (  : Q/A) E = 2π KQ / A E =4π KQ / A For two oppositely charged plates placed near each other, E field outer side of the plates is zero while inner side the E-field

67 T. Norah Ali Almoneef66 The electrostatic force is a conservative (=“path independent”) force It is possible to define an electrical potential energy function with this force Work done by a conservative force is equal to the negative of the change in potential energy There is a uniform field between the two plates As the positive charge moves from A to B, work is done W AB =F d=q E d ΔPE =-W AB =-q E d – only for a uniform field

68 SI Unit of Electric Potential: joule/coulomb=volt (V) T. Norah Ali Almoneef67 Potential Difference (=“Voltage Drop”) The potential difference between points A and B is defined as the change in the potential energy (final value minus initial value) of a charge q moved from A to B divided by the size of the charge ΔV = V B – V A = ΔPE /q Potential difference is not the same as potential energy The electric potential V at a given point is the electric potential energy U of a small test charge q 0 situated at that point divided by the charge itself: If we set at infinity as our reference potential energy, The electric potential difference between any two points i and f in an electric field.

69 T. Norah Ali Almoneef68 Another way to relate the energy and the potential difference: ΔPE = q ΔV Both electric potential energy and potential difference are scalar quantities A special case occurs when there is a uniform electric field V B – V A = -Ed Gives more information about units: N/C = V/m The electric potential energy U and the electric potential V are not the same. The electric potential energy is associated with a test charge, while electric potential is the property of the electric field and does not depend on the test charge. It is equal to the difference in potential energy per unit charge between the two points. the negative work done by the electric field on a unite charge as that particle moves in from point i to point f. A larger charge would involve a larger amount of PEe, but the ratio of that energy to the charge is the same as it would be if a smaller charge was in that same place.

70 T. Norah Ali Almoneef69 Energy and Charge Movements A positive charge gains electrical potential energy when it is moved in a direction opposite the electric field If a charge is released in the electric field, it experiences a force and accelerates, gaining kinetic energy – As it gains kinetic energy, it loses an equal amount of electrical potential energy A negative charge loses electrical potential energy when it moves in the direction opposite the electric field When the electric field is directed downward, point B is at a lower potential than point A A positive test charge that moves from A to B loses electric potential energy It will gain the same amount of kinetic energy as it loses potential energy

71 T. Norah Ali Almoneef70 Electric field always points from higher electric potential to lower electric potential. A positive charge accelerates from a region of higher electric potential energy (or higher potential) toward a region of lower electric potential energy (or lower potential). – it moves in the direction of the field, Its electrical potential energy decreases, Its kinetic energy increases A negative charge accelerates from a region of lower potential toward a region of higher potential. – It moves opposite to the direction of the field – Its electrical potential energy decreases – Its kinetic energy increases

72 T. Norah Ali Almoneef71 If a charged particle moves perpendicular to electric field lines, no work is done. If the work done by the electric field is zero, then the electric potential must be constant Thus equipotential surfaces and lines must always be perpendicular to the electric field lines. if d  E General Considerations

73 T. Norah Ali Almoneef72 Electrical Potential Energy in a Uniform Electric Field If a charge is released in a uniform electric field at a constant velocity there is a change in the electrical potential energy associated with the charge’s new position in the field.  PEe = -qE  dThe unit: Joules The negative sign indicates that the electrical potential energy will increase if the charge is negative and decrease if the charge is positive. The  V in a uniform field varies with the displacement from a reference point.  V = E  d The displacement is moved in the direction of the field. Any displacement perpendicular to the field does not change the electrical potential energy.

74 T. Norah Ali Almoneef73 POTENTIAL ENERGY IN A UNIFORM FIELD The Electric Field points in the direction of a positive test charge. + Charge- Charge Along ELoses PEeGains PEe Opposite EGains PEeLoses PEe

75 T. Norah Ali Almoneef74

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77 T. Norah Ali Almoneef76 Example : An electron accelerates from rest through an electric potential of 2 V. What is the final speed of the electron? Answer: The electron acquires 2 eV kinetic energy. We have and since the mass of the electron is m e = 9.1 × 10 –31 kg, the speed is

78 T. Norah Ali Almoneef77 example An electron (mass m = 9.11× kg ) is accelerated in the uniform field E (E = 1.33×10 4 N/C ) between two parallel charged plates. The separation of the plates is 1.25 cm. The electron is accelerated from rest near the negative plate and passes through a tiny hole in the positive plate, as seen in the figure. With what speed does it leave the hole? F = qE = ma a = qE/m V f 2 = v i 2 + 2a(d) V f 2 = 2ad = 2(qE/m)d = 2 (1.9 x C) ( 1.33×10 4 N/C) (1.25m) / 9.11× kg = 8.3 x 10 6 m/s m = 9.11×10-31kg E = 1.33×104 N/C d = 1.25 cm GIVEN:

79 T. Norah Ali Almoneef78 “Electric field lines always point in the direction of decreasing electric potential” The unit: V m -1 Or NC -1

80 T. Norah Ali Almoneef79 e-e-

81 T. Norah Ali Almoneef80 But a more useful concept is the electric potential energy of each charge e-e- e-e- e-e- e-e- The greater the magnitude of the charge, the greater is the electric potential energy

82 T. Norah Ali Almoneef81 Clicker Question In the figure, a proton moves from point i to point f in a uniform electric field directed as shown. Does the electric field do positive, negative or no work on the proton? A: positive B: negative C: no work is done on the proton

83 T. Norah Ali Almoneef82 Electric Potential Energy with a Pair of Charges A single point charge produces a non-uniform electric field. PE e = kq 1 q 2 /r The reference point for PE e is assumed to be at infinity. The ground is usually the reference point for PE g. The PE e is positive for like charges and negative for unlike charges. T. Norah Ali Almoneef82 Work and Electrical Potential Energy In order to bring two like charges near each other work must be done. (W = Fd) In order to separate two opposite charge, work must be done. Remember: whenever work gets done, energy changes form. The potential energy will change to kinetic energy.

84 T. Norah Ali Almoneef83 Since the electrical potential energy can change depending on the amount of charge you are moving, it is helpful to describe the electrical potential energy per unit charge. Electric potential: the electrical potential energy associated with a charged particle divided by the charge of the particle. A larger charge would involve a larger amount of PEe, but the ratio of that energy to the charge is the same as it would be if a smaller charge was in that same place.

85 T. Norah Ali Almoneef Electrical Potential Energy of Two Charges V 1 is the electric potential due to q 1 at some point P 1 The work required to bring q 2 from infinity to P 1 without acceleration is q 2 E 1 d=q 2 V 1 This work is equal to the potential energy of the two particle system If the charges have the same sign, PE is positive – Positive work must be done to force the two charges near one another – The like charges would repel If the charges have opposite signs, PE is negative – The force would be attractive – Work must be done to hold back the unlike charges from accelerating as they are brought close together

86 T. Norah Ali Almoneef85 Potential Difference Near a Point Charge An electric potential exists at some point in an electric field regardless of whether there is a charge at that point. The electric potential at a point depends on only two quantities: the charge responsible for the electric potential and the distance (r)from this charge to the point in question.  V = k c q/r Voltage is a way of using numbers (quantitative) to describe an electric field. Electric fields are measured in volts over a distance. This means the larger the E the larger the V. When an E is attracting or repelling an object, we instead could say that the object is being driven by the voltage.

87 T. Norah Ali Almoneef86 Calculate the electric potential, V, at 10 cm from a -60  C charge. Equation: Answer: V = -5.4x 10 6 V Calculate the electric potential, V, at the midpoint between a 250  C charge and a -450  C separated by a distance of 60 cm. Equation:Answer:V = -6.0x10 6 V Example:

88 T. Norah Ali Almoneef87 If you want to move in a region of electric field without changing your electric potential energy. You would move a)Parallel to the electric field b)Perpendicular to the electric field Example

89 T. Norah Ali Almoneef88 General Points for either positive or negative charges The Potential increases if you move in the direction opposite to the electric field and The Potential decreases if you move in the same direction as the electric field Electric Potential Electric Potential is a scalar field it is defined everywhere but it does not have any direction it doesn’t depend on a charge being there

90 T. Norah Ali Almoneef89 What is the potential difference between points A and B? Δ V AB = V B - V A a) Δ V AB > 0b) Δ V AB = 0c) Δ V AB < 0 E A B C Example Points A, B, and C lie in a uniform electric field. Since points A and B are in the same relative horizontal location in the electric field there is on potential difference between them The electric field, E, points in the direction of decreasing potential

91 T. Norah Ali Almoneef90 E A B C Point C is at a higher potential than point A. True False Example Points A, B, and C lie in a uniform electric field. As stated previously the electric field points in the direction of decreasing potential Since point C is further to the right in the electric field and the electric field is pointing to the right, point C is at a lower potential The statement is therefore false

92 T. Norah Ali Almoneef91 Compare the potential differences between points A and C and points B and C. a) V AC > V BC b) V AC = V BC c) V AC < V BC E A B C Example Points A, B, and C lie in a uniform electric field. In Example 4 we showed that the the potential at points A and B were the same Therefore the potential difference between A and C and the potential difference between points B and C are the same Also remember that potential and potential energy are scalars and directions do not come into play

93 T. Norah Ali Almoneef92 If a negative charge is moved from point A to point B, its electric potential energy a) Increases.b) decreases.c) doesn’t change. E A B C Example Points A, B, and C lie in a uniform electric field. The potential energy of a charge at a location in an electric field is given by the product of the charge and the potential at the location As shown in Example 4, the potential at points A and B are the same Therefore the electric potential energy also doesn’t change

94 T. Norah Ali Almoneef93 The Volt The commonly encountered unit joules/coulomb is called the volt, abbreviated V, after the Italian physicist Alessandro Volta ( ) With this definition of the volt, we can express the units of the electric field as For the remainder of our studies, we will use the unit V/m for the electric field. The Joule The joule is a measure of work accomplished on an object. It is also a measure of potential energy or how much work an object can do. In the English system the unit of work and energy is the ft x lb. F = m x a For a falling object a = g, so F = m x g Energy is force x distance. E = F x d For a falling object d=h (h=height) E = F x h PE= m x g x h

95 T. Norah Ali Almoneef94 The Electron Volt The electron volt (eV) is defined as the energy that an electron (or proton) gains when accelerated through a potential difference of 1 V 1 V=1 J/C  1 eV = 1.6 x J There is an additional unit that is used for energy in addition to that of joules A particle having the charge of e (1.6 x C) that is moved through a potential difference of 1 Volt has an increase in energy that is given by

96 T. Norah Ali Almoneef95 Potential energy Calculate the work done bringing a 250  C charge and a -450  C from infinity to a distance of 60 cm apart. Equation: = 9x10 9 (250x10 -6 )(-450x10 -6 ) 60x10 -2 PE = -1.7x10 3 J

97 T. Norah Ali Almoneef96 Example A proton is placed between two parallel conducting plates in a vacuum as shown. The potential difference between the two plates is 450 V. The proton is released from rest close to the positive plate. What is the kinetic energy of the proton when it reaches the negative plate? + - = V(+)-V(  ) The potential difference between the two plates is 450 V. The change in potential energy of the proton is  U, and  V =  U / q (by definition of V), so  U = q  V = e[V(  )  V(+)] =  450 eV

98 T. Norah Ali Almoneef97 Example Suppose an electron is released from rest in a uniform electric field whose magnitude is 5.90 x 10 3 V/m. (a) Through what potential difference will it have passed after moving 1.00 cm? (b) How fast will the electron be moving after it has traveled 1.00 cm? (a)  V| = Ed = (5.90 x 10 3 V/m)( m) = 59.0 V (b) q  V| = mv 2 /2  v = 4.55x10 6 m/s Example An ion accelerated through a potential difference of 115 V experiences an increase in kinetic energy of 7.37 x 10 –17 J. Calculate the charge on the ion. qV= 7.37x J, V=115 V q = 6.41x C

99 T. Norah Ali Almoneef98 Example A particle has a mass of 1.8x10 -5 kg and a charge of +3.0x10 -5 C. It is released from point A and accelerates horizontally until it reaches point B. The only force acting on the particle is the electric force, and the electric potential at A is 25V greater than at C. (a) What is the speed of the particle at point B? (b) If the same particle had a negative charge and were released from point B, what would be its speed at A?

100 T. Norah Ali Almoneef99 Example The work done by the electric force as the test charge (+2.0x10 -6 C) moves from A to B is +5.0x10 -5 J. (a)Find the difference in EPE between these points. (b)Determine the potential difference between these points. (a) (b)

101 T. Norah Ali Almoneef100 Example 3: A proton is moved from the negative plate to the positive plate of a parallel- plate arrangement. The plates are 1.5cm apart, and the electric field is uniform with a magnitude of 1500N/C. a)How much work would be required to move a proton from the negative to the positive plate? b)What is the potential difference between the plates? c)If the proton is released from rest at the positive plate, what speed will it have just before it hits the negative plate? 1

102 T. Norah Ali Almoneef101 Example: Two 40 gram masses each with a charge of -6µC are 20cm apart. If the two charges are released, how fast will they be moving when they are a very, very long way apart. (infinity)

103 T. Norah Ali Almoneef 102 Problem : An electron is released from rest in an electric field of 2000N/C. How fast will the electron be moving after traveling 30cm? _ v = 0m/s _ v = ? 30cm

104 T. Norah Ali Almoneef 103 F = qE  q=+0.045·10  C > 0, Force parallel to E  Charge accelerates to left  Work done on charge = qEd = Change in Kinetic Energy  W = (0.045·10  C)(1200 V/m)(0.05m) = 54. ·10  V ·C =0.054J  (K f -K i ) = W  (1/2) mv f 2 – 0 = J  v f 2 = 2 (0.054 J) / ( 3.5 ·10  kg)= 30.8 m 2 /s 2  v f =5.6 m/s F=qE Ignore triangle on sketch.

105 T. Norah Ali Almoneef104 Example : An electron accelerates from rest through an electric potential of 2 V. What is the final speed of the electron? The electron acquires 2 eV kinetic energy. We have and since the mass of the electron is m e = 9.1 × 10 –31 kg, the speed is

106 T. Norah Ali Almoneef105 An  - particle, mass = 6.7x kg, initially at rest travels 25 cm through a uniform electric field of 250 N/C. Calculate the potential difference across the 25 cm path Equation : V =Ed Answer: 62.5 V What is the a-particle’s speed after 25 cm of travel? Equati on: qEd = ½ mv 2 Answer: 7.7x10 4 m/s

107 T. Norah Ali Almoneef106 How much energy (work) is necessary to bring three point charges from infinity to the vertices of the right triangle shown below. 2.0  C 3.0  C -  C 5.0 cm 3.0 cm 4.0 cm Equation: PE = kq 1 q 2 Answer: PE = -2.16J PE = 1.8 J PE = -1.8 J PE = -2.16J r

108 T. Norah Ali Almoneef107 Calculate the work done bringing a 250  C charge and a -450  C from infinity to a distance of 60 cm apart. = 9x10 9 (250x10 -6 )(-450x10 -6 ) 60x10 -2 PE = -1.7x10 3 J example : The potential difference between to charge plates is 500V. Find the velocity of a proton if it is accelerated from rest from one plate to the other. 500 V High Potential Low Potential Positive charges move from high to low potential Negative charges move from low to high potential example

109 T. Norah Ali Almoneef108 Which of the following statements is false? A.The total work required to assemble a collection of discrete charges is the electrostatic potential energy of the system. B.The potential energy of a pair of positively charged bodies is positive. C.The potential energy of a pair of oppositely charged bodies is positive. D.The potential energy of a pair of oppositely charged bodies is negative. E.The potential energy of a pair of negatively charged bodies is negative. The figure depicts a uniform electric field. Along which direction is the increase in the electric potential a maximum? example

110 T. Norah Ali Almoneef109 Potential Difference in a Uniform field +Q AB C d ||

111 T. Norah Ali Almoneef110 The potential at a point due to a unit positive point charge is found to be V. If the distance between the charge and the point is tripled, the potential becomes A. V/3. B. 3V. C. V/9. D. 9V. E. 1/V 2. example The figure shows two plates A and B. Plate A has a potential of 0 V and plate B a potential of 100 V. The dotted lines represent equipotential lines of 25, 50, and 75 V. A positive test charge of 1.6 × 10 –19 C at point x is transferred to point z. The electric potential energy gained or lost by the test charge is A.8 × 10 –18 J, gained. B.8 × 10 –18 J, lost. C.24 × 10 –18 J, gained. D.24 × 10 –8 J, lost. E.40 × 10 –8 J, gained. example

112 T. Norah Ali Almoneef111 Two parallel metal plates 5.0 cm apart have a potential difference between them of 75 V. The electric force on a positive charge of 3.2 × 10 –19 C at a point midway between the plates is approximately A.4.8 × 10 –18 N. B.2.4 × 10 –17 N. C.1.6 × 10 –18 N. D.4.8 × 10 –16 N. E.9.6 × 10 –17 N. example When +2.0 C of charge moves at constant speed from a point with zero potential to a point with potential +6.0 V, the amount of work done is A.2 J. B.3 J. C.6 J. D.12 J. E.24 J. example

113 The concept of “potential difference" or “voltage" in electricity is similar to the concept of "height" in gravity, or “pressure” in fluids 112T. Norah Ali Almoneef

114 113 A positive charge of 6 nC attracts a negative charge of 5 nC from a distance of 4 mm. What is the work done? What does the sign mean? A point charge of +3 μC is located at the origin of a coordinate system and a second point charge of -6 μC is at x = 1.0 m. At what point on the x-axis is the electrical potential zero? m m m m

115 T. Norah Ali Almoneef114 Example : Compute the energy necessary to bring together the charges in the configuration shown below: Calculate the electric potential energy between each pair of charges and add them together.

116 T. Norah Ali Almoneef115 Two point charges of values +3.4 and +6.6 μC respectively, are separated by 0.20 m. What is the potential energy of this 2-charge system? J J J J What will be the electrical potential at a distance of 0.15 m from a point charge of 6.0 μC? x 10 4 V x 10 5 V x 10 6 V x 10 7 V

117 T. Norah Ali Almoneef116T.Norah Ali Almoneef116 Example 1. The potential difference between the two terminals on a battery is 9 volts. How much work (energy) is required to transfer 10 coulombs of charge across the terminals? V = 9.0 V W= ? q= 10 c 9.0 = W/10 W = 90 J 2. The work required to transfer 30 coulombs of charge across two terminals is 50 joules. What is the potential difference? V= ? W= 50 J q= 30 c V=50/30 V =1.7 V Example

118 T. Norah Ali Almoneef117 Problem Show that the amount of work required to assemble four identical point charges of magnitude Q at the corners of a square of side s is 5.41k e Q 2 /s.

119 T. Norah Ali Almoneef118 Example Consider a positive and a negative charge, freely moving in a uniform electric field. True or false? (a) Positive charge moves to points with lower potential. (b) Negative charge moves to points with lower potential. (c)Positive charge moves to a lower potential energy position. (d)Negative charge moves to a lower potential energy position –Q+Q0 +V –V (a) True (b) False (c) True (d) True

120 T. Norah Ali Almoneef119 PROBLEM A proton is released from rest in a uniform electric field with a magnitude of 8 E 4 V/m. The proton is displaced 0.5 m as a result. A) Find the potential difference between the proton’s initial and final positions. B) Find the change in electrical potential energy of the proton as a result of this displacement. E = 8 E 4 V/m q = C d = 0.5 m A)  V = -Ed   V = - (8 E 4 V/m)(0.5 m) =  V = V B)  V =  PEe/q   PEe =  V q  ( V)( C) = E-15 J

121 T. Norah Ali Almoneef120 The Potential due to a Point Charge Although only changes in potential are physically relevant, it is often convenient to choose the location of the zero of the potential. For a car battery, this is typically the car’s chassis; for an electrical outlet it is the ground. For an isolated point charge, it is convenient to choose the potential to be zero at infinity

122 T. Norah Ali Almoneef121 The Potential of a Point Charge The potential difference between two points A and B from a point charge can be re-written as When r A = infinity the last term vanishes. We are free to choose V(A) as we please, e.g., V(A) = 0. With this choice, the potential of a point charge becomes

123 T. Norah Ali Almoneef122T. Norah Ali Almoneef122 Superposition principle applies The total electric potential at some point P due to several point charges is the algebraic sum of the electric potentials due to the individual charges – The algebraic sum is used because potentials are scalar quantities Electric Potential of Many Point Charges Electric potential is a SCALAR not a vector. Just calculate the potential due to each individual point charge, and add together! (Make sure you get the SIGNS correct!) q1q1 q5q5 q4q4 q3q3 q2q2 r1r1 r2r2 r3r3 r4r4 r5r5 P

124 Potential from more than one charge Principle of superposition Total Potential is sum of all individual potentials Each potential is Thus Total potential is Which can be written 123T. Norah Ali Almoneef

125 124 Electric Potential of a single charge + if V = 0 at r A =  E B A It can be shown that Remember that so This looks a bit like the formulae for the potential in a Uniform Field Potential energy Arbitrary shape Potential difference Arbitrary shape

126 T. Norah Ali Almoneef125 Potential Energy in 3 charges Q2Q1Q3 Energy when we bring in Q2 Now bring in Q3 So finally we find

127 T. Norah Ali Almoneef126 Example: Pictures from Serway & Beichner How many electrons should be removed from an initially uncharged spherical conductor of radius m to produce a potential of 7.5 kV at the surface? Substituting given values into V = = 7.50 x 10 3 V = N = 1.56 x electrons

128 T. Norah Ali Almoneef127 Example 1: The electron in the Bohr model of the atom can exist at only certain orbits. The smallest has a radius of.0529nm, and the next level has a radius of.212m. a)What is the potential difference between the two levels? b)Which level has a higher potential? +e r1r1 r2r2 r 1 is at a higher potential.

129 T. Norah Ali Almoneef128 Example: Finding the Electric Potential at Point P (apply V=k e q/r). 5.0  C -2.0  C Superposition: V p =V 1 +V 2 V p =1.12  10 4 V+(-3.60  10 3 V)=7.6  10 3 V

130 T. Norah Ali Almoneef129 PROBLEMS As a particle moves 10 m along an electric field of strength 75 N/C, its electrical potential energy decreases by 4.8 x10 – 16 J. What is the particle’s charge? What is the potential difference between the initial and final locations of the particle in the problem above? An electron moves 4.5 m in the direction of an electric field of strength 325 N/C/ Determine the change in electrical potential energy. 1) E = 75 N/C PEe = x J d = 10 m q = ? Must be a + q since PEe was lost PEe = - qEd  q = (- 4.8 x )/(-(75)(10)) = Q = +6.4 E -19 C 2)  V =  PEe/q  (-4.8 x J)/(6.4 E -19 C)  V = V 3) q = x C d = 4.5 m E = 325 N/C  PEe = ? PEe = - qEd  -(-1.6 x C)(325 N/C)(4.5 m) = 2.3 x J

131 T. Norah Ali Almoneef130

132 T. Norah Ali Almoneef131 Example: Consider three point charges q 1 = q 2 = 2.0  C and q 3 = -3  C which are placed as shown. Calculate the net force on q 1 and q 3

133 T. Norah Ali Almoneef132 The force on q1 is F1 = F12 + F13.

134 T. Norah Ali Almoneef133 Similarly, F3 = F31 + F32

135 T. Norah Ali Almoneef134 Example At locations A and B, find the total electric potential.

136 T. Norah Ali Almoneef135 Example A charge +q is at the origin. A charge –2q is at x = 2.00 m on the x axis. For what finite value(s) of x is (a) the electric field zero ? (b) the electric potential zero ? x x – 4.00 = 0 (x+4.83)(x  0.83)=0  x = m (other root is not physically valid)  x = m and x= m

137 T. Norah Ali Almoneef136 In the drawing on the right, q = 2.0 μC and d = 0.96 m. Find the total potential at the location P, assuming that the potential is zero at infinity.

138 T. Norah Ali Almoneef137 The Potential due to a Point Charge:

139 T. Norah Ali Almoneef138

140 T. Norah Ali Almoneef139

141 T. Norah Ali Almoneef140 The three charges in Fig., with q 1 = 8 nC, q 2 = 2 nC, and q 3 = - 4 nC, are separated by distances r 2 = 3 cm and r 3 = 4 cm. How much work is required to move q 1 to infinity? Example

142 T. Norah Ali Almoneef141

143 T. Norah Ali Almoneef142 Calculate the electric potential, V, at 10 cm from a - 60  C charge. Equation: Answer: V = -5.4x 10 6 V Calculate the electric potential, V, at the midpoint between a 250  C charge and a -450  C separated by a distance of 60 cm. Equation: Answer:V = -6.0x10 6 V Example

144 T. Norah Ali Almoneef143

145 T. Norah Ali Almoneef144 If a negative charge is moved from point A to point B, its electric potential energy a ) Increases.b) decreases.c) doesn’t change. E A B C Example Points A, B, and C lie in a uniform electric field. The potential energy of a charge at a location in an electric field is given by the product of the charge and the potential at the location As shown in Example 4, the potential at points A and B are the same Therefore the electric potential energy also doesn’t change 144

146 T. Norah Ali Almoneef145 E. Potential & Potential Energy vs Electric Field & Coulomb Force If we know the potential field this allows us to calculate changes in potential energy for any charge introduced Coulomb Force is thus Electric Field multiplied by charge Electric Field is Coulomb Force divided by test charge D Potential is D Energy divided by test charge D Energy is D Potential multiplied by test charge

147 T. Norah Ali Almoneef146 Electric Potential Difference The electric potential energy depends on the charge present We can define and electric potential V which does not depend on charge by using a “test” charge Change in potential is change in potential energy for a test charge divided by the unit charge Remember that for uniform field

148 T. Norah Ali Almoneef147 Parallel-Plate Capacitor The capacitor consists of two parallel plates Each have area A They are separated by a distance d The plates carry equal and opposite charges When connected to the battery, charge is pulled off one plate and transferred to the other plate The transfer stops when  V cap =  V battery 147

149 T. Norah Ali Almoneef148 Capacitors Storing a charge between the plates Electrons on the left plate are attracted toward the positive terminal of the voltage source This leaves an excess of positively charged holes The electrons are pushed toward the right plate Excess electrons leave a negative charge _ + _

150 Capacitance of parallel plates +Q-Q The bigger the plates the more surface area over which the capacitor can store charge C  A Moving plates togeth`er Initially E is constant (no charges moving) thus  V=Ed decreases charges flows from battery to increase  V  C  1/d Never Ready + E VV 149T. Norah Ali Almoneef Capacitance – is a measure of the capacitor’s ability to store electric energy where: q – magnitude of the charge V – potential/potential difference C - capacitance

151 T. Norah Ali Almoneef150 In SI units, q – coulomb (C) V – volt (V) C – coulomb/volt = A Farad is very large Often will see µF or pF farad (F) T. Norah Ali Almoneef150 The capacitance of a device depends on the geometric arrangement of the conductors For a parallel-plate capacitor whose plates are separated by air: 150 (C does not depend on Q or V) [V = Ed, E=Q/(A  0 ), V = Qd / (A  0 )] For any geometry, capacitance scales as Area/distance

152 T. Norah Ali Almoneef151 Electric Field A uniform electric field ( The field strength is the same magnitude and direction at all points in the field) can be produced in the space between two parallel metal plates. The plates are connected to a battery. E The field strength at any point in this field is: E = field strength (Vm -1 ) V = potential difference (V) d = plate separation (m)

153 Capacitance The constant of proportionality C is the capacitance which is a property of the conductor Experiments show that the charge in a capacitor is proportional to the electric potential difference (voltage) between the plates. 152T. Norah Ali Almoneef To increase C, one either increases , increases A, or decreases d. 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 that they will spark across the gap and destroy the capacitor. Thus capacitors have a maximum voltage!

154 T. Norah Ali Almoneef153 Example The plates of the capacitor are separated by a distance of m, and the potential difference between them is V B -V A =-64V. Between the two equipotential surfaces shown in color, there is a potential difference of -3.0V. Find the spacing between the two colored surfaces.

155 T. Norah Ali Almoneef154 The two plates of a capacitor hold  C and  C, respectively, when the potential difference is 200V. What is the capacitance? Equation: How much charge flows from a 12 V battery when connected to a 20  F capacitor? Equation: q = CV 240  C q =20  F x12V =

156 T. Norah Ali Almoneef155 Example How strong is the electric field between the plates of a 0.80  F air gap capacitor if they are 2.0 mm apart and each has a charge of 72  C? Find the capacitance of a 4.0 cm diameter if the plates are separated by 0.25 mm. Example

157 T. Norah Ali Almoneef156 Example When a potential difference of 150 V is applied to the plates of a parallel-plate capacitor, the plates carry a surface charge density of 30.0 nC/cm2. What is the spacing between the plates? What is the charge on a 250 microfarad capacitor if it has been charged to 12 V? q = CV = (250E-6 F)(12 V) = 3E-3 Coulombs Example

158 T. Norah Ali Almoneef157 Circular parallel plate capacitor rr s r = 10 cm A =  r 2 =  (.1) 2 A =.03 m 2 S = 1 mm =.001 m p = pico = Show Demo Model, calculate its capacitance, and show how to charge it up with a battery.

159 T. Norah Ali Almoneef 158 Example: Thundercloud Suppose a thundercloud with horizontal dimensions of 2.0 km by 3.0 km hovers over a flat area, at an altitude of 500 m and carries a charge of 160 C. Question 1: – What is the potential difference between the cloud and the ground? Question 2: – Knowing that lightning strikes require electric field strengths of approximately 2.5 MV/m, are these conditions sufficient for a lightning strike?

160 T. Norah Ali Almoneef159 Example: Question 1 We can approximate the cloud-ground system as a parallel plate capacitor whose capacitance is The charge carried by the cloud is 160 C, which means that the “plate surface” facing the earth has a charge of 80 C 720 million volts … …

161 T. Norah Ali Almoneef160 Question 2 We know the potential difference between the cloud and ground so we can calculate the electric field E is lower than 2.5 MV/m, so no lightning cloud to ground – May have lightning to radio tower or tree….

162 T. Norah Ali Almoneef161 A parallel-plate capacitor has an area of 5.00 cm 2 and the plates are separated by a distance of 2.50 mm Calculate the capacitance. What is the capacitance if it has.014 V across it when it has a charge of 2.13x C? C = q/V = (2.13E-15 C)/(.014 V) = 1.52x F Problem

163 T. Norah Ali Almoneef162 If a spark jumps across a 1mm gap when you reach for a doorknob, what was the potential difference between you and the knob and how much charge was on you? Assume your finger and the knob form a parallel plate capacitor with area 1 cm 2 and breakdown electric field = 3MV/m. V = Q/C V = Ed Q = CV=CEd = (A  0 /d)(E d) = A  0 E Q = (0.01m) 2 (8.85  F/m)(3.06 V/m) Q = 2.7  (C/V)(Vm 2 /m 2 ) = 2.7 femto-Coulomb Problem

164 T. Norah Ali Almoneef 163 Problem (a)What plate area is required if an air-filled, parallel plate capacitor with a plate separation of 2.6 mm is to have a capacitance of 12 pF? (a)C = A  0 /d, A = Cd/  0 = (12·10  12 F)(2.6 ·10  3 m) / (8.85  10  12 F/m) A = 3.5 ·10  3 m 2 = (6cm) 2 A 0.25 μ F capacitor is connected to a 400 V battery. What is the charge on the capacitor? x C x C C C Problem

165 T. Norah Ali Almoneef164 The two plates of a capacitor hold  C and  C, respectively, when the potential difference is 200V. What is the capacitance ? Equation: Answer 25 mF How much charge flows from a 12 V battery when connected to a 20 mF capacitor ? Equation: q = CV Answer: 240  C Problem


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