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Outcome 6 Electricity Topic B Coulomb’s Law

Outcome 6 Electricity Topic B Coulomb’s Law. What does Coulomb’s Law Measure?. It measures the electric attraction or repulsion between two charged particles. Coulomb’s Law – Gives the electric force between two point charges. Inverse Square Law.

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Outcome 6 Electricity Topic B Coulomb’s Law

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  1. Outcome 6 Electricity Topic BCoulomb’s Law

  2. What does Coulomb’s Law Measure? It measures the electric attraction or repulsion between two charged particles.

  3. Coulomb’s Law – Gives the electric force between two point charges. Inverse Square Law k = Coulomb’s Constant = 9.0x109 Nm2/C2 q1 = charge on mass 1 q2 = charge on mass 2 r = the distance between the two charges The electric force is much stronger than the gravitational force.

  4. Fundamental Charge: The charge on one electron. e = 1.6 x 10 -19 C Unit of charge is a Coulomb (C) Therefore: One coulomb is the charge of 6.25 x 1018 electrons.

  5. Two types of charge: Positive Charge: A shortage of electrons. Negative Charge: An excess of electrons. Conservation of charge – The net charge of a closed system remains constant.

  6. n n n n n n + + + + + + - - - - - - - - Nucleus Which statement below bests describes the atom pictured? Positive Atom Number of electrons < Number of protons +2e = +3.2 x 10-19C Negative Atom Number of electrons > Number of protons -2e = -3.2 x 10-19C Neutral Atom Number of electrons = Number of protons

  7. F F + + + F F - Electric Forces Like Charges - Repel Unlike Charges - Attract

  8. F F q1 q2 r Example 1 Two charges are separated by a distance r and have a force F on each other. If r is doubled then F is : ¼ of F 2F If q1 is doubled then F is : 16F If q1 and q2 are doubled and r is halved then F is :

  9. 3μC 40g 3μC 40g 50cm Example 2 Two 40 gram masses each with a charge of 3μC are placed 50cm apart. Compare the gravitational force between the two masses to the electric force between the two masses. (Ignore the force of the earth on the two masses)

  10. The electric force is much greater than the gravitational force

  11. Example 3 - 5μC 45º - 4μC 5μC 20cm F1 45º F2 20cm Three charged objects are placed as shown. Find the net force on the object with the charge of -4μC. F1 and F2 must be added together as vectors.

  12. 45° Law of Cosines R 2 = (2.3 N)2 + (4.5 N) 2 - 2(2.3 N)(4.5 N)cos 45° R 2 = 5.29 N 2 + 20.25 N 2 - 14.6 N 2 R 2 = 10.94 N 2 R = 3.31 N 4.5 N R 2.3 N 45° Law of Sines Sin 45°/3.31 N = Sin θ/2.3 N sin θ = 0.49 θ = 29.4° 4.5 N θ 2.3 N 3.31 N Final Answer: 3.31 N at 209.4°

  13. - 2.9 F1 - 1.6 45º 3.31 F2 + 2.3cos45≈1.6 θ 29º 2.3sin45≈1.6 F1 = < - 4.5 , 0.0 > F2 = < 1.6 , - 1.6 > Fnet = < - 2.9 , - 1.6 > 3.31N at 209º

  14. Example 4 (Balloon Lab) Two 8 gram, equally charged balls are suspended on earth as shown in the diagram below. Find the charge on each ball. 20º 10º 10º L = 30cm L = 30cm FE FE 30sin10º q q r r =2(30sin10º)=10.4cm

  15. T FE q Fg = .08N Draw a force diagram for one charge and treat as an equilibrium problem. Tsin80º 80º Tcos80º

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