Genphy2_Wk1-2_Electric Force and Fields.pptx

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1. Electricity General Physics 2

2. Prayer We give thee thanks for all blessings you have given us this day. May we be able to use these blessings to honor your name AMEN.

3. Objectives: describe using a diagram charging by rubbing and charging by induction explain the role of electron transfer in electrostatic charging by rubbing describe experiments to show electrostatic charging by induction calculate the net electric force on a point charge exerted by a system of point charges;

4. Styrocup to Balloon Rub the styrocup to the inflated balloon. Then place the balloon at the top of the small pieces of papers. What did you observe? Plastic cup to Balloon Rub the plastic cup to the inflated balloon. Then place the balloon at the top of the small pieces of papers. What did you observe? Styrocup to dry hair Rub the styro cup to the dry hair. Then place the styro cup at the top of the small pieces of papers. What did you observe?

5. Plastic cup to Dry Hair Balloon to Dry Hair Rub the balloon to the dry hair. Then place the balloon at the top of the small pieces of papers. What did you observe? Comb to Dry Hair Rub the comb to the dry hair. Then place the comb at the top of the small pieces of papers. What did you observe? Rub the plastic cup to the dry hair. Then place the plastic cup at the top of the small pieces of papers. What did you observe?

6. Electricity General Physics 2

7. Electric charges

8. Electricity Type of energy that involves the flow of charges

9. Opposite charges: Same charges: ATTRACT REPEL + – F 2 1 Q 1 Q 2 + + Q 1 Q 2 – – Q 1 Q 2 F 1 2 F 2 1 F 1 2 F 2 1 F 1 2

10. Charging Objects Materials that allow the movement of electrons from one region to another are called conductors of electricity . M aterials that do not allow the flow of electrons are called insulators. How can we charge objects? ELECTROSCOPE – metal knob, foil, Metal rod

11. Charging by conduction

12. Charging by induction

13. Opposite charges: Same charges: ATTRACT REPEL + – F 2 1 Q 1 Q 2 + + Q 1 Q 2 – – Q 1 Q 2 F 1 2 F 2 1 F 1 2 F 2 1 F 1 2

14. Coulomb’s Law + – F e F e Q 1 Q 2 R   where: k = 9  10 9 Nm 2 /C 2 ( Coulomb’s constant)

15. Coulomb’s Law Identify the forces acting on Q 2 . + F 3 2 + – F 1 2 Q 1 Q 2 Q 3

16. Example Compute the electric force between two charges of 5×10 −9 C and −3×10 −8 C which are separated by d = 10 cm

17. Practice Two point charges are located in xy coordinate system. A charge 2.00 𝑥 10 −9 𝐶 is located at (0,4.00 cm) and the other charge −3.00 𝑥 10 −9 𝐶 is located at (3.00 cm, 4.00 cm). If the third charge 5.00 𝑥 10 −9 𝐶 is placed at origin, find the resultant force at the third charge.

18. Assessment

19. General Physics 2 Week 2 Electric Field and Capacitance

20. Prayer We give thee thanks for all blessings you have given us this day. May we be able to use these blessings to honor your name AMEN.

21. Objectives: Describe an electric field as a region in which an electric charge experiences a force Calculate the electric field due to a system of point charges using Coulomb’s law and the superposition principle

22. Recall: What are the three kinds of charges in an object? How to charge an object? What does the Coulomb’s Law States?

23. Electric Field The electric field is a region in space in which an electric charge’s force can be observed. The field’s direction at a point in this field is the direction that the test charge (which is normally positively charged) would move if placed at this point

24. Electric Field The direction of the electric field lines depends on the polarity of the charge. A positive charge will always deflect a test charge away from it (since the test charge q is normally positive), while a negative charge will have its electric field lines directed towards itself . Placing the test charge higher, will result in a curved electric field line as the test charge will first be pushed off tangential to the +Q then be attracted to the negatively charged particle

25. Electric Field Lines Field lines must begin on positive charges and terminate on negative charges, or at infinity in the hypothetical case of isolated charges. The number of field lines leaving a positive charge or entering a negative charge is proportional to the magnitude of the charge.

26. Electric Field Lines 3. The strength of the field is proportional to the closeness of the field lines—more precisely, it is proportional to the number of lines per unit area perpendicular to the lines. 4. The direction of the electric field is tangent to the field line at any point in space. 5. Field lines can never cross.

27. our electric field strength is now only dependent on the point charge Q. Since the E is the force per unit charge, the unit is Newtons per Coulomb (N/C) Electric Field     Plugging in the value of F according to Coulomb’s Law. Cancelling the test charge we’ll have… E = electric field strength (N/C) k = Coulomb’s Constant Q = charge given   Formula for electric field strength

28. Calculate the strength and direction of an electric field generated by a 3.00x10 -9 C at a distance of 1.00x10 -2 m. Sample Problem

29. What force will a 2.70x10 5 N/C electric field exert on a -2.50x10 -6 C point charge? Sample Problem

30. Find the magnitude and direction of the total electric field due to the two point charges, and , at the origin of the coordinate system Sample Problem

31. Calculate the electric field strength from a 1.25x10 -8 C charge if the test charge is placed 2.50x10 -5 m. Let’s Try!

32. Assessment

33. Assessment

34. Capacitors Devices used for communication, photography, and high-energy accelerators rely on one of the most important electric devices in modern times, capacitors. It is a device composed of conductors separated by an insulator or a vacuum.

35. Capacitors An electric field is also formed. (the physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field) Capacitors are devices formed from two conductors (equal magnitude of charges with opposite signs) separated by an insulator They act like rechargeable batteries . The primary purpose of capacitors is to store electrostatic energy in an electric field and where possible, to supply this energy to the circuit.

36. Capacitors Dielectrics in capacitors serve three purposes: to keep the conducting plates from coming in contact, allowing for smaller plate separations and therefore higher capacitances ; to increase the effective capacitance by reducing the electric field strength , which means you get the same charge at a lower voltage; to reduce the possibility of shorting out by sparkin g (more formally known as dielectric breakdown) during operation at high voltage.

37. The presence of the electric field found between the plates is directly proportional to the charge Q present in the conductors. Therefore, the potential difference, Vab , is also directly proportional to charge Q. Capacitance     Formula for electric field strength when potential difference is given. Formula in finding capacitance in a circuit C = capacitance expressed in C/V Q is the charge expressed in Coulombs (C) V is the potential difference between conductors expressed in Voltage (V) Symbol for Capacitance

38. The area between the two plates, A, is also directly proportional to capacitance C. Capacitance     Formula for electric field strength when the area in terms of m 2 . Formula in finding capacitance when area and distance is given. Parallel Plate Capacitor with dielectric C = capacitance expressed in C/V A = area in terms of m 2 d = distance in terms of m = permittivity of space in terms of = dielectric constant   Symbol for Capacitance

39. Capacitors in Series 01.

40. Charges have one path way to flow. So, the magnitude of the charge is CONSTANT.

41. To calculate for the total voltage of this circuit we use this formula,   Plugging in the values for V1, V2 and V3, we will arrive at…   We can cancel the Qs at this point giving us the total capacitance as  

42. Sample Problem Find the total capacitance for three capacitors connected in series, given their individual capacitances are 1.000, 5.000, and 8.000 μF .

43. Capacitors in Parallel 02.

44. To find the Ct, you must understand first that the VOLTAGE across capacitors in parallel remains the same as the source.

45. The plates of the capacitors in a parallel connection are directly connected to the source. Hence, there will be no voltage difference as they are all connected directly through a conductor. The total charge Qt on the other hand is the sum of the individual charges as each capacitor will behave as if each of them have been connected individually to the voltage source. Thus Thru cancellation, we obtain the equation for the total capacitance Ct: +   If we use the relationship Q = CV, then we can deduce that Q1 = C 1 V, Q2 = C 2 V and Q3 = C 3 V respectively. Entering these into the previous expressions will give you +   +  

46. Sample Problem Determine the effective capacitance if two capacitors, C 1 = 10F and C 2 = 5F, are connected in parallel.

47. PRACTICE Find (a) the equivalent capacitance of the capacitors in the circuit shown, (b) the charge on each capacitor, and (c) the potential difference across each capacitor.

48. PRACTICE* Determine the net capacitance C of the capacitor combination shown beside. When a 12.0V potential difference is maintained across the combination, find the charge and the voltage across each capacitor.

49. PRACTICE Determine the net capacitance C of the capacitor combination shown beside. When a 12.0V potential difference is maintained across the combination, find the charge and the voltage across each capacitor. ,