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Electricity and Magnetism Unit III. I Electrostatics 0 The study of electric charges at rest and their electric fields and potentials 0 Charges at rest.

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Presentation on theme: "Electricity and Magnetism Unit III. I Electrostatics 0 The study of electric charges at rest and their electric fields and potentials 0 Charges at rest."— Presentation transcript:

1 Electricity and Magnetism Unit III

2 I Electrostatics 0 The study of electric charges at rest and their electric fields and potentials 0 Charges at rest if there is no net transfer of charge A. Microstructure of Matter 0 The smallest unit of matter is an atom 0 Atoms are made up of protons, neutrons and electrons 0 The electron is the fundamental negatively charged particle of matter 0 The proton is the fundamental positively charged particle of matter

3 0 The elementary charge (e) equals 0 The charge on an electron (-e) 0 The charge on a proton (+e) 0 Because all atoms are electrically neutral, atoms contain equal number of protons and neutrons 0 Neutrons have no charge

4 B. Charged Objects 0 Protons and neutrons cannot be removed from an atom 0 Electrically charged objects are formed when neutral objects gain or lose electrons 0 When an atom becomes a charged particle, it is called an ion 0 Excess electrons creates negative ions 0 Loss of electrons creates positive ions 0 Objects with the same sign of charge are repelled 0 Objects with opposite signs are attracted

5 C. Transfer of Charge 0 If a system only contains neutral objects, the total net charge is zero 0 If the objects are rubbed together, electrons are transferred 0 The system as a whole remains neutral

6 Example 3 spheres have initial charges 0 When R and S touch, excess electrons (-8e) move to neutral (0e) 0 When separated each sphere “splits” the excess 0 (-8e + 0e) ÷ 2 spheres 0 Each sphere becomes -4e 0 When S and T touch, excess electrons (-4e) moves to (+6e) 0 When separated each sphere “splits” the excess 0 (-4e + 6e) ÷ 2 spheres 0 Each sphere becomes +1e 0 TOTALS R = -4e S = +1e T = +1e 0e R -8e S +6e T

7 D. Quantity of Charge 0 SI unit of charge is a coulomb (C) 0 1C = 6.25 x 10 18 elementary charges (electrons) 0 The charge on an electron (-e) = 1.6 x 10 -19 C 0 The net charge on a charged object is always a multiple of the charge on an electron

8 E. Coulomb’s Law 0 The electrostatic force between 2 charges is 0 Directly related to the product of the charges 0 Inversely related to the square of the distance between them 0 F…electrostatic force (newtons) 0 q…charge (coulombs) 0 r…distance between charges (meters) 0 k…electrostatic constant (8.99 x 10 9 N m 2 /C 2 ) F = kq 1 q 2 r2r2

9 II. Electric Fields o Region around a charged particle o Exerts a force on other charged particles o Represented by field lines o Direction a positive charge(+) would move o If lines are curved, charge would move tangent to the point on the field line o Begin (+) and end (-) o Never cross o Energy varies inversely squared with distance

10 o Strength Where E = electric field strength (N/C) F = force the charge experiences (N) q = the charge (C) E = F q

11 What is the magnitude of the electric field at a point in a field where an electron experiences a 1.0 N force? E = 6.3 x 10 18 N/C E = F q E = 1.0 N 1.60 x 10 -19 C

12 A. Potential Difference o Work done as charge is moved against a field o Where W is work in joules o q is charge in coulombs o V is potential difference o Joules/coulomb is the derived unit volt o 1 J/C = 1 volt o If the charge is an elementary charge (ex. electron) the potential difference is an electron volt (eV) o 1 eV = 1.60 x 10 -19 J V = W q

13 Moving a point charge of 3.2 x 10 -19 coulombs between two points in an electric field requires 4.8 x 10 -18 joules. What is the potential difference between these two points? V = 15 V q V = W 3.2 x 10 -19 C V = 4.8 x 10 -18 J

14 III Electric Current A. Electric current o The rate that a charge passes a point in a circuit o Represented using the symbol, I o Unit is an ampere (A) o Formula I = q t o I is current in amperes o q = charge in coulombs o t = time in seconds o Current is measured using an ammeter

15 B. Electric circuit o The closed path that a charged particle moves along o Potential difference is also needed for an electric current o Supplied by o Cell o Battery o Measured using a voltmeter o Electron flow determines the direction of current o A switch is used to make, break or change the connections in a circuit

16 C. Resistance o Electrical resistance (R) is the opposition of electron flow o Ratio of potential difference to current flow o R = V I o Where o V = potential difference in volts o I = current in amperes o R = resistance in volts per ampere Ohm (Ω)represents 1 volt per ampere o Resistance is affected by temperature

17 A current of 0.10 amperes flow through a lamp connected to a 12.0-volt source. What is the resistance of the lamp? I = 0.10 A V = 12.0 V R = V I R = 12.0 V 0.10 A R = 120 Ω

18 Factors Affecting Resistance Length of Wire o Increasing length, increases the collisions of electrons with atoms in the wire o Directly related Thickness of Wire o Increasing thickness creates more spaces for electrons to travel through, decreasing the resistance o Inverse relationship with cross-sectional area Resistivity o Characteristic of the material o Good conductors have low resistivities o As temperature increases, resistivity increases o Values found in reference tables

19 Resistance of a Wire R = ρL A Where R is resistance in ohms (Ω) ρ is the resistivity in ohm-meters (Ω-m) L is length in meters (m) A is cross-sectional area in square meters (m 2 )

20 Determine the resistance of a 4.00-meter piece of copper wire that has a diameter of 2.00 mm at 20°C. ρ copper = 1.72 x 10 -2 L = 4.00 m Diameter = 0.002 m Area = πr 2 Area = π(.001) 2 Area = 3.14 x 10 -6 m 2 R = (1.72 x 10 -2 x 4.00) 3.14 x 10 -6 R =2.19 x 10 -2 Ω

21 Resistor Device to have a definite amount of resistance Used to limit current flow Provide a potential drop Symbol

22 Electric Power Power (P) = Work/time Units Work,,,joules Time sec Power watts Electrical power is the product of VI Since V=IR then P = I2R

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