Early scientists identified positive charges as the charge carriers in metals; however, - PowerPoint Presentation

1. Early scientists identified positive charges as the charge carriers in metals; however, the discovery of the electron led to the introduction of “conventional” current direction. Was this a suitable solution to a major shift in thinking? What role do paradigm shifts play in the progression of scientific knowledge?

2. Understandings:ChargeElectric fieldCoulomb’s lawElectric currentDirect current (dc)Potential differenceElectric fields….

3. The law of charges…How can you charge an insulator?Create a demonstration that shows it is charged?What occurs during the demonstration?Equipment:Insulating rodClothHanging plastic sphere painted in conducting paint

4. These are called monopoles.

5.  A Coulomb of charge is transported by a current of 1A in 1s.From IGCSE…..

6. http://www.physicsclassroom.com/Physics-Interactives/Static-Electricity/Coulomb-s-Law

7. Investigate the relationship between electrostatic force and distance:

8. Create an equation for electric force based on what you have discovered….

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10. Electric fields….The definition of a field in Physics is; a region in space in which a force is experienced. In a magnetic field around a magnet another magnet experiences a force. In a gravitational field around a mass another mass experiences a force.An electric field is a region around a charged object in which another charge experiences an force.

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12. Two unlike charges – attraction.Two like charges – repulsion.A charge and an oppositely charged plate.Two oppositely charged plates.

13. Rules to drawing E-field vectors…Field lines are always directed away from +ve and towards –ve.Field lines are always drawn at right angles to the charged objects they leave and meet.Field lines cannot cross.The density of the field lines are an indication of the relative strength of the field they represent.There is no field inside a hollow conductor.The field between parallel plates is uniform in intensity.

14. Inside the sphere the e-field vectors all sum to zero…

15. Electric Field Strength, E..The strength of an electric field, E at a point is defined as:‘The force experienced per unit charge by a small positive test charge placed at that point.’, C E is a vector with same direction as the electric force.

16. Using Coulomb’s law to find Electric field strength….   If +ve charge then E is positive and E acts outwards, if it is –ve it acts towards the charge.

17. Vector addition of electric charges…..If our test charge is in an electric field due to multiple charges each exerts a force.The resultant force per unit charge (FE/q) gives the resultant field strength at the particular position of our test chargeWe can consider 3 scenarios:

18. Our test charge experiences two forces F1 = qE1 & F2 = qE2 The resultant F is simply F = F1 + F2The resultant field strength E = F/q = (qE1 + qE2) /qE = E1 + E2Test charge +q+Q2 point charge-Q1 point chargeF1F2Forces in the same direction :

19. Our test charge experiences two forces F1 = qE1 & F2 = qE2 The resultant F is simply F = F1 - F2The resultant field strength E = F/q = (qE1 - qE2) /qE = E1 - E2Test charge +q+Q2 point charge+Q1 point chargeF1F2Forces in the opposite direction

20. Standard resolving techniques... From Pythagoras FR2 = F12 + F22Electric Field Strength ER2 = E12 + E22Trigonometry can be used to find the resultant directionTest charge +q-Q2 point charge+Q1 point chargeF1F2Forces at right angles:

21. ABCConsider below line AB = BC = r, what is the electric field strength at B given that a charge of 2Q is placed at A and -3Q is placed at C?As the fields are acting in the same direction it must be an addition of the charges regardless of the sign!!! 

22. -3.75 x 108 NC-1 towards Q15.6 x 10-3 NC-1 towards Q1((k x 25 x 10-6)/(20 x 10-3)2) – ((k x 100 x 10-6)/(40 x 10-3)2) = 0

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24. F13 = 1.08*10-8NF23 = -5.625*10-9NFR = 7.15*10-9NΘR = 24.95⁰ From verticalCalculate the electric field strength at q3 due to both q1 and q2?

25. Combining fields questionCalculate the resultant force and electric field strength experienced by test charge + q of magnitude 2pC in the situations shown opposite.Both Q1 & Q2 have a charge of magnitude of 4μCIn situations (a) and (b) q is 3cm from Q1 and 4cm from Q2In situation (c) q is 4cm from Q1 and 3cm from Q2Remember that both force and electric field strength are vectors.

26. Combining fields answers(c) F1 = 4.5 x 10 -5 N UPWARDS TO THE RIGHTF2 = 8.0 x 10 -5 N DOWNWARDS TO THE RIGHTΣF = 9.18 x 10 -5 N θ = 151⁰ from Q2q lineE1 = 2.25 x 107 NC-1 UPWARDS TO THE RIGHTE2 = 4.0 x 107 NC-1 DOWNWARDS TO THE RIGHT ΣE = 4.59 x 10 7 NC-1 θ = 151⁰ from Q2q line(a) F1 = 8.0 x 10 -5 N LEFTF2 = 4.5 x 10 -5 N LEFTΣF = 12.5 x 10 -5 N LEFTE1 = 4.0 x 10 7 NC-1 LEFTE2 = 2.25 x 10 7 NC-1 LEFTΣE = 6.25 x 10 7 NC-1 LEFT(b) F1 = 8.0 x 10 -5 N RIGHTF2 = 4.5 x 10 -5 N LEFTΣF = 3.5 x 10 -5 N RIGHTE1 = 4.0 x 10 7 NC-1 RIGHTE2 = 2.25 x 10 7 NC-1 LEFTΣE = 1.75 x 10 7 NC-1 RIGHT

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32. Electrical Conduction….+ve-veElectric field direction from high to low potential.Electrical conduction can also happen in liquids and gases….

33. Electrical Current…. Electric current is defined as rate of charge flow past a point in electrical circuit.

34. How fast do electrons move?…..AL    

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36. Drift Velocity…..EHT and potassium permanganate.

37. Potential Difference….QqABFor a charge to move a distance near another charge work must be done.Potential difference, V for the above would be work done in moving the point charge from AB. 

38. Finding the potential difference between parallel plates…. Here cosθ = 1 as θ = 0.We also know that:  

39. Power, P, Voltage, V and Current, I…. 

40. The electronvolt…..It is useful in physics for us to use a unit of energy that is more manageable when we are discussing individual particles and the energy they carry as this value can be extremely small.This is particular important in Nuclear, quantum and particle physics.