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Electrostatics

Electrostatics. W. Sautter 2007. Electrons, Protons & Charged Objects. Electrostatics is the study of the effects of stationary charges on each other in their surroundings. Charges are created by the transfer of electrons to or from one body to another. (Protons are NEVER transferred.).

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Electrostatics

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  1. Electrostatics W. Sautter 2007

  2. Electrons, Protons & Charged Objects Electrostatics is the study of the effects of stationary charges on each other in their surroundings. Charges are created by the transfer of electrons to or from one body to another. (Protons are NEVER transferred.) Objects with equal numbers of protons and electrons are neutral. They have no net charge. Objects with more electrons than protons are charged negatively Objects with less electrons than protons are charged positively.

  3. Unlikes attract Unlikes attract Likes repel Likes repel - - - - - - + + + + + + Fundamental Laws of Electrostatics

  4. 6.25 x 10 18electrons 1 coulomb Measuring Charge Charge on 1 electron = - 1.6 x 10 –19 coulombs Charge on 1 proton = + 1.6 x 10 –19 coulombs

  5. F = G x m x m 1 2 g 2 r 2 2 -11 G = 6.67 x 10 N m / kg ' w = m x g Law of Universal Gravitation Fg = gravity force between m1 and m2 separated by a distance r G is the Universal Gravitational Constant The weight of an object is its mass times g’, the gravity value at location r

  6. F = k x q x q 1 2 e 2 r 2 2 9 k = 9.0 x 10 N m / C COULOMB'S LAW Fe = electrostatic force Between m1 and m2 separated by a distance r k is Coulomb’s Constant Coulomb’s Law is an inverse square law similar to the Law of Gravitation It is dissimilar in that electrostatic forces can be attraction or repulsion. Gravity is attraction only. Electrostatic force is strong, gravity is very weak.

  7. Static Discharge from Van Degraff Generator

  8. 2 2 -11 G = 6.67 x 10 N m / kg 9 2 2 k = 9 x 10 N x m / C COULOMB'S LAW Gravity is classified as a weak force because huge amounts of mass are required to create a reasonably large force. Note the very small value of the force constant. Electric forces is classified as a strong force because small charge quantities can to create large forces. Note the very large value of the force constant

  9. 2 2 9 k (Coulomb's constant 9 x 10 Newtons x meters / coulombs ) Electrostatic Quantities and Symbols F (force in Newtons) q (charge in Coulombs) r (separation distance in meters) E ( electric field strength in Newtons / coulomb or volts / meter) v ( potential in volts 1 volt = 1 joule / coul)

  10. Identifying Electric Fields Electric charges are detected by the presence of an electric field (E). Recall that a gravity field can be detected by its influence (attractive force) on a mass (often called a test mass). Electric fields are detected by their influence (attractive or repulsive forces) on a charge (often called a test charge). Electrically charged bodies can be created by physically rubbing electrons off one object on to another. For example, rubbing a rubber rod with fur will transfer electrons from the fur to the rod which becomes negatively charged (it has extra electrons). Rubbing a glass rod with silk will transfer electrons from the rod to the silk. The rod is charged positively (it has missing electrons).

  11. - - - - - - - - - - Moveable leaves Charging an Electroscope • - - • - - Charged rod - - - An Electroscope is an instrument used to measure the presence of an Electric field (presence of charged bodies). Leaves of electroscope Diverge (like charges repel)

  12. - - - - - - - - - - - - - - - - ELECTROSTATIC CHARGING CHARGING BY CONDUCTION NEUTRAL OBJECT OBJECT IS NOW NEGATIVELY CHARGED - - - - - WHEN OBJECT IS TOUCHED BY THE CHARGED ROD ELECTRONS MOVE FROM THE ROD TO THE SPHERE UNTIL ELECTROSTATIC EQUILIBRIUM IS REACHED -

  13. CHARGED RUBBER ROD (EXCESS ELECTRONS) ELECTROSTATIC CHARGING INDUCTIVE CHARGING Electrons on sphere move to the opposite side due to repulsion of electrons on the charged rod

  14. Electric Field Direction Electric Field Charged Plates E is always out of plus (+) into minus (-)

  15. Electric Fields vs Gravity Fields Recall that a gravity field (g) is measured by dividing the force acting on it (its weight(w) in Newtons ) by the mass quantity (m) in kilograms g = w / m g in N / Kg An electric field (E) is measured by dividing the force acting on it (in Newtons ) by the charge quantity (q) in coulombs (C ). E in N / C E = F / q

  16. Measuring Electric Field Strength - - - - - - - - + + E E Scale Scale As Electric field strength increases, the Force on a test Charge increases. This is similar to the weight of a Mass increasing in a stronger Gravity field Test charge Test charge

  17. Charged Plates Measuring a Uniform Electric Field Electric Field Force + E = Force (N) / Charge (Coul)

  18. Electric Field of a Point Charge + Lines of Flux spread over a greater area as distance from charge increases and field strength weakens E = k q / r2

  19. Electric Field at a Distance from a Point Charge (1) E = F/q (by definition) (2) F = kq1 q2 / r2 (Coulomb’s Law) (3) E = kq1 q2 / r2 q2 ( by substitution) E = kq / r2 q = point charge in coulombs k = coulomb’s constant r= distance from charge in meters E = electric field strength at that point (N / C)

  20. Electrical Potential Electrical Potential is defined as the work required to move a charge over a distance in an electric field. Electrical potential is measured in volts. One volt equals one joule (work) divided by charge (coulombs) In a uniform electric field (a field between two parallel charged metal plates) , potential (V) equals work (W) divided by charge (q). Therefore, since work equal force times distance and force equals the electric field strength (N/C) times charge in coulombs (C) , potential equals electric field times distance the charge moves. V = W / q = (F x d) / q = (F / q) x d = E x d V (volts) = E ( N/C) x d (m)

  21. work work + + Charged Plates Potential Difference in a Uniform Electric Field Electric Field x Point B x Point A V = Work (Joules) / Charge (coul)

  22. Electrical Potential In an electric field issued from a point charge the field varies inversely with the square of the distance from the charge, therefore the work required to move the charge is continuously changing. This is similar to the varying gravity field surrounding a mass. The work required to move a mass in that field also continuously changing. To find the work required to move the charge in a varying field we must integrate the force times distance equation using calculus. The following frames show the use of calculus to find work form force versus distance relationships. If your course does not require calculus, skip these frames

  23. WORK - Finding Area Under a Non linear Force vs Displacement Curve BOX METHOD F O R C E (N) WORK = AREA UNDER THE CURVE W =  F  X (SUM OF THE BOXES) AS THE NUMBER OF BOXES INCREASES, THE ERROR DECREASES! AREA MISSED - INCREASING THE NUMBER BOXES WILL REDUCE THIS ERROR! WIDTH OF EACH BOX =  X X1 X2 DISPLACEMENT (M)

  24. Finding Area Under Curves Mathematically • Areas under force versus distance graphs (work) can be found mathematically. The process requires that the equation for the graph be known and integral calculus be used. • Recall that integration is also referred to as finding the antiderivative of a function. • The next slide reviews the steps in finding the integral of the basic function, y = kxn.

  25. n Y = k X n n + 1 n   d y = k X dx = k X + C --------------- n + 1 INTEGRATION – THE ANTIDERIVATIVE INTEGRATION IS THE REVERSE PROCESS OF FINDING THE DERIVATIVE. IT CAN ALSO BE USED TO FIND THE AREA UNDER A CURVE. THE GENERAL FORMAT FOR FINDING THE INTEGRAL OF A SIMPLE POWER RELATIONSHIP, Y = KXn ADD ONE TO THE POWER n + 1 DIVIDE THE EQUATION BY THE N + 1 + C ---------------  is the symbol for integration n + 1 ADD A CONSTANT

  26. 3 3 , dy = 5 X dX Y = 5 X 4  d y = = 5 X --------------- 4 n n + 1 n   d y = k X dx = k X + C --------------- n + 1 APPLYING THE INTEGRAL FORMULA GIVEN THE EQUATION FORMAT TO FIND THE INTEGRAL 3 + 1 3 5 X + C --------------- 3 + 1 Integration can be used to find area under a curve between two points. Also, if the original equation is a derivate, then the equation from which the derivate came can be determined.

  27. 4 = 5 X + C --------------- 4 4 4 = 5 (5) 5 (2) + C + C --------------- --------------- 4 4 APPLYING THE INTEGRAL FORMULA Find the area between x = 2 and x = 5 for the equation y = 5X3. First find the integral of the equation as shown on the previous frame. The integral was found to be 5/4 X4 + C. The values 5 and 2 are called the limits. each of the limits is placed in the integrated equation and the results of each calculation are subtracted (lower limit from upper limit) 5 Area 2 Area = 761.25 -

  28. r r   W= Fdr = ( k q1 q2 / r2 )dr Coulomb’s Law 0 0 0 0 0 0 W = k q1 q2 r-1 = - k q1 q2 / r r | Work to Move a Charge in an Electric Field From Infinity ( a great distance) to a point in the field ( 1/ infinity ~ 0 ) therefore W = - k q1 q2 / r

  29. Work = - G m1 m2 Work = - k q1 q2 r r Work to move a Charge in an Electric Field q Recall: Work to move a charge in a gravity field: By Analogy: Work to move a charge in an electric field:

  30. Electrical Potential in a Point Charge Field Dividing the work equation obtained in the previous slides by the charge quantity of the point charge we get a relationship which describes absolute potential at point in the field Absolute potential is measured by using the Work to move a charge from infinity to a point

  31. work work + + Electrical Potential in a Point Charge Field Point A x + Work to move a charge from infinity to a point in an electric field is called Absolute Potential 1 volt = 1 joule/ 1coul V = k q / r

  32. Potential = Work / Charge Volts = Joules / Coulomb VAbsolute = ( - k q1 q2 / r ) / q2 Absolute Potential in an Electric Field Recall from previous slide: Absolute potential is measured from infinity to a point r in the field W = - k q1 q2 / r q1 = point charge r= distance from the point charge VAbsolute = - k q1 / r

  33. V = W = - k q1 q2 / q1 r q V = V - V Potential Difference From point A to B Absolute at point B Absolute at point A Since we now know: = k q / r \ \ V = k q / r Where V is the absolute potential at a point (r) in the electric field caused by charge (q) , we can now find the potential difference between two points in the field.

  34. Capacitors Capacitors are electrical devices used to store electrical energy. They are not to be confused with batteries which create electrical energy via chemical reaction. The structure of a capacitor is shown on the next slide. Essentially, electrons are pumped onto one of the metal plates shown and pushed off the other plate by the electric field developed on the first plate. The electrons are pumped be a battery, power supply or other voltage source. Once the charge is isolated on the metal plates it represents stored energy since when the capacitor is connected to an electrical circuit, electrons will flow from the plate of high electron concentration through the connecting device, thus releasing energy and back to the plate of low electron concentration until charge equilibrium is reached.

  35. Capacitance + + + + + + + + + + - - - - - - - - - - Electrons in Capacitors Store electrical energy Although it may seem illogical, in conventional current the + plate has the excess electrons ! C (farads) = q (coul) / V (volts) A Farad is a unit of capacitance 1 F = 1 coul / 1 volt Large capacitor can hold more charge at lower voltages and can store more energy

  36. Capacitors CAPACITOR TRENDS (1) Large capacitors can hold more charge and larger amounts of energy than smaller ones. (2) The closer the plates are together, the more energy that can be stored since the electric field between the plates becomes stronger. (3) The larger the dielectric constant the less the tendency of electrons to “jump the gap” (short out) between the plates at higher voltages and more energy can be stored.

  37. Capacitors Mounted in a Typical Circuit

  38. Metal plates (A) in meters2 + + + + + + + + + + - - - - - - - - - - Dielectric (non conductive material) Plate separation (d) in meters Capacitors C = e0kA / d e0 = permittivity constant 8.85 x 10-12 Coul2 / N m2 K = dielectric constant (air = 1.0) It depends on the nature of the dielectric As plate area increases, C increases As plate separation increases, C decreases

  39. Energy Stored in a Capacitor Area under the graph gives the work to charge the capacitor Charging a Capacitor The slope of the graph line equals the capacitance (C ) C = q / V C H A R G E (C) C W = ½ qV, Also since q = CV, by Substitution W = ½ CV2, and since V = q / C W = ½ q 2/ V V VOLTAGE (V)

  40. + + + + + + + + + - - - - - - - - - Capacitors on Parallel Effectively, the capacitors in parallel become one larger capacitor C(total) = C (1) + C (2) + C (3) Since each individual capacitor is directly connected to the voltage source (battery) each bears the same voltage The charge on each capicator is dependent on its individual capacitance. (Plate size, separation, dielectric)

  41. + + + + + + + + + - - - - - - - - - Capacitors on Series 1/C(total) = 1/ C (1) + 1/ C (2) + 1/C (3)

  42. Capacitors Capacitors in series have different characteristics than Those connnected in parallel • Since they are not all directly connnected to the • voltage source, the voltage on each varies according • to its individual capacitance. Larger ones have • lower voltage than smaller ones. (2) Each capacitor, despite its size, has the same charge. 1/C(total) = 1/ C (1) + 1/ C (2) + 1/C (3)

  43. The End

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