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Electric Charge and Coulomb’s Law

Electric Charge and Coulomb’s Law. Fundamental Charge: The charge on one electron. e = 1.6 x 10 -19 C. Unit of charge is a Coulomb (C). Two types of charge:. Positive Charge: A shortage of electrons. Negative Charge: An excess of electrons.

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Electric Charge and Coulomb’s Law

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  1. Electric Charge and Coulomb’s Law

  2. Fundamental Charge: The charge on one electron. e = 1.6 x 10 -19 C Unit of charge is a Coulomb (C)

  3. 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.

  4. n n n n n n + + + + + + - - - - - - - - Nucleus Negative Atom Number of electrons > Number of protons -2e = -3.2 x 10-19C Neutral Atom Number of electrons = Number of protons Positive Atom Number of electrons < Number of protons +2e = +3.2 x 10-19C

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

  6. 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.

  7. 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 :

  8. 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)

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

  10. 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.

  11. - 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

  12. Example 4 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

  13. 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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