1. WHY?

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3. Electromagnetism I Lecture 2 Coulomb’s Law and the Concept of Electric Field

4. Learning Objectives To present a mathematical expression to the experimentally observed force To describe Coulomb’s Law Quantifying the Forces between Charges – superposition principle To introduce the concept of the Electric Field - a great way to deal with forces!

5. Charles Augustin de Coulomb Born: 14 June 1736 in Angoulême, France Died: 23 Aug 1806 in Paris, France He worked on applied mechanics, but he is best known for his work on electricity and magnetism, and in particular the Coulomb Law

6. Coulomb studied Friction Friction is commonly studied by rubbing two surfaces. And rubbing two surfaces sometimes can create charges. Create is the wrong word. Charge transfer

7. Coulomb’s Law Like charges repel, unlike charges attract The force acts along the line joining two charged particles The force between two charged particles is proportional to the magnitude of each charge The force between two charged particles is inversely proportional to the square of the distance between the particles

8. Mathematically

9. q 1 and q 2 are measured in Coulombs r 21 is in metres F 21 is measured in Newtons In SI Units  0 is called the permittivity of free space = 8.854  10 -12 Coulomb 2 Newton -1 m -2 (simplifies equations)

10. For two protons this is 10 34 smaller than the repulsive force between two proton. Gravitational force is only significant because it is always attractive! Gravitational Force

11. Worked Example: assume that the positive and negative charge within a 1 pence coin do not cancel, but there is an excess of 0.0001% of electrons. What force would two such pennies exert on each other if they are placed 1 m apart? 1p 1 m

12. 1p 1 m One pence  3 g copper T&M lectures: Molar mass of Cu 63.5 g One pence = (3/63.5)  6 x 10 23 = 2.84  10 22 Cu atoms One Cu atoms contains 29 electrons. Hence one pence contains 8.25  10 23 electrons This corresponds to 131956 C. 0.0001% of that is 0.132 C 0.132 C at 1 m gives F = 157,000,000 N (17,662 Tons) So when we say that macroscopic objects are neutral, we mean it!

13. Compute the ratio of the electric force to the gravitational force exerted by a proton on an electron at a separation of 10 6 m. m p = 1.67x10 -27 kg m e = 9.11x10 -31 kg e = 1.60x10 -19 C G = 6.67x10 -11 N m 2 kg -2 Classwork

15. (a)Did you really need to know how far apart the two particles are? (a)Why is gravity the dominating force between large objects such as planets and stars?

16. Forces between many charges - the principle of superposition What if you have more than two positive charges? q2q2 q3q3 F 12 F 21 q1q1 F 31 F 13 F total Consider the charges one pair at a time Vector sum of forces Procedure

17. In general the force on a charge q due to a collection of charges is the vector sum of all the individual forces due to all the others. This is known as the principle of superposition of forces

18. The Electric Field Concept of the Electric Field Electric Field Strength E Electric Field Lines Calculation of E at a point The Electric Dipole

19. Definition of Electric Field Consider a small test charge, q 0, placed at a point in a distribution of point charges. The force on this test charge is: Define electric field E as

20. Thus, the E at a certain point is equal to the electric force per unit charge at that point

21. F = qE Once we know E at a certain point we can calculate the force on any charge q placed there, because E is the force acting on a unit charge.

23. A better known example of a field; the gravitational force F g that the earth exerts on a mass m 0 : We can interpret g as the gravitational field

24. These are lines drawn in space so that the tangent to them at any point gives the direction of E at that point. Visualisation of E Use field lines of force

25. The E field near to a point charge

26. The density of the field lines is proportional to the strength of E The absolute number of lines is not important. You cannot obtain the field strength by counting the number of lines. Adding lines must follow rules.

27. Field lines only start on positive charge, they can only end on negative charge Example: E field lines around an electric dipole The tangent at any point gives the direction of E at that point.

29. Field lines do not cross one another! Field lines are continuous,only stop at charges!

30. Field can be measured without knowing the detailed configuration of the charges which produce the field E Test charge

31. The E-field due to an Electric Dipole - Calculation To simplify the calculation, we will only compute the field along the axis r E -q+q a a/2

32. This simplifies to: For r >> a The quantity qa is the electric moment of the dipole, p

33. For r >>>>>>>> a -q+q a r For large r, r+ar P To the distant point P, the field looks like that from -q and +q from the same point.

34. electric dipole moment = aq=p; note r -3 dependence

35. Review and Summary Coulomb’s law gives the force on a charged particle placed in a distribution of charges Forces must be superposed as vectors - superpositon principle Field lines provide a graphical representation of electric fields An electric dipole is a pair of electric charges of equal magnitude q but opposite sign separated by a distance a Attention must be paid to direction, it depends on the signs of charges and geometry of configuration

37. Coulomb’s Law Definition of an Electric Field

38. The effects of gravitational attraction are negligible, and hence do not need to be considered when discussing atomic (or molecular) interactions. Gravity is the dominating force between large objects because these objects are electrically neutral – containing equal numbers of positive and negative electrical charges – the net electrical force is zero because there are attractive and repulsive forces. The gravitational force is attractive only.

39. Next Lecture Calculating E-fields of continuous charge distributions