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SPH4U Notes Unit 7.2 Electric Forces: Coulomb’s Law Page 1 of 6 Physics Tool box Coulomb’s Law –The magnitude of the electric force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. F k k is the Coulomb proportionality Constant k Coulomb’s Law applies when the charges on the two spheres are very small, and the two spheres are small compared to the distance between them. There are similarities between Coulomb’s Law and Newton’s Law of universal Gravitation: o Both are inverse squares laws that are proportional to the product of quantities that characterize the body involved o The gravitational force only attracts, while the electric force and attract or repel. o The universal gravitational constant is very small while the Coulomb constant is very large. q1q2 . r2 1 N m2 . 8.988 109 4 0 C2 o FE kq1q2 , r2 Fg Gm1m2 r2 The electric force between two point charges also depends on the quantity of charge on each body, which is denoted by q. Coulomb found that the forces that two point charges q1 and q2 exert on each other are proportional to each charge and therefore proportional to the product q1q2 of the two charges. Coulomb’s Law –The magnitude of the electric force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. F k q1q2 r2 Where k is the Coulomb Proportionality Constant whose numerical value depends on the system of units used. The absolute value bars are used because the charges q1 and q2 can be either positive or negative, while the force magnitude F is always positive. k 1 N m2 8.988 109 4 0 C2 0 (epsilon-nought) is permittivity of free space = 8.854 1012 C2 N m2 SPH4U Unit 7.2 Electric Forces: Coulomb’s Law Page 2 of 6 Plastic rods and fur (real or fake) are typically used to demonstrate electrostatics. We now that there are exactly two types of charges negative and positive The Laws of Electric Charges – Two positive charges or two negative charges repel each other. A positive charge and a negative charge attract each other Example The magnitude of the electrostatic force between two point like charged objects is 4.0 106 N . Calculate the force for each of the following situations: a) The distance between objects is doubled b) Distance remains same but charge on one object tripled and charge on other is cut by 5. c) Both a and b occur. Solution: kq1q2 2 r1 F2 r22 a) F1 kq1q2 r2 r12 r F2 F1 1 r2 2 1 4.0 10 N 2 1.0 106 N 6 2 SPH4U Unit 7.2 Electric Forces: Coulomb’s Law Page 3 of 6 The magnitude of the force is now 1.0 106 N 1 k 3q1 q2 5 2 F 3 r2 b) 2 kq1q2 F1 5 2 r1 3 5 F2 F1 3 4.0 106 N 5 7 2.4 10 N The magnitude of the force is now 2.4 107 N 1 k 3q1 q2 5 2 F2 3 1 r22 c) kq1q2 F1 5 2 2 r1 F2 F1 3 20 3 4.0 106 N 20 6.0 107 N The magnitude of the force is now 6.0 107 N Example Three objects with charges A: 6.0C , B: 5.0C , and C: 4.0C are placed in a line. Determine the net electric force on charge A. A + Solution: 0.70m The magnitude of the force exerted on A by C FAC kq AqC 2 rAC m2 9 9.0 10 N 6.0 106 C 4.0 106 C 2 C 2 0.70 0.30 m 0.216 N B C - + 0.30m SPH4U Unit 7.2 Electric Forces: Coulomb’s Law Therefore the force on A by C is 0.216 N [left] The magnitude of the force exerted on A by B FAB Page 4 of 6 Charges are same sign so the forces repels. kq AqB 2 rAB m2 9 9.0 10 N 6.0 106 C 5.0 106 C 2 C 2 0.7m 0.5510 N Charges are different sign so the forces attract. Therefore the force on A by B is 0.551N [right] So the net force on A is : F F AC F AB 0.216 N [left ] 0.551N [right ] 0.33N [right ] Example A neutral metal sphere A (0.10kg), hangs from an 2.0 m insulating wire. An identical sphere B, with charge –q, is brought into contact with sphere A. The spheres repel each other so that the angle of the wire is 12 . What was the initial charge on B 12 B A Solution: First we must determine what the forces are that holds a 0.10 kg sphere at an angle of 12 . So let’s draw a force diagram FT 12 Now T cos 12 mg FE T sin 12 FE So we can solve for FE Fg SPH4U Unit 7.2 Electric Forces: Coulomb’s Law Page 5 of 6 FE sin 12 mg cos 12 FE mg tan 12 N 0.10kg 9.8 tan 12 kg 1 2.08 10 N Before we can apply F k q1q2 , we need the distance (r value) r2 r sin 12 2.0m r 2.0m sin 12 12 0.416m 2.0m r The charge on each sphere after touching are now identical so original charge q on B Now, 1 1 q q 2 2 FE k r2 4 FE r 2 2 q k q 4 FE r 2 k 4 2.0 101 N 0.416m 9.0 109 N 3.9 106 C 6 The initial charge on B is 3.9 10 C m2 C2 2 q1 q2 , therefore half of the SPH4U Unit 7.2 Electric Forces: Coulomb’s Law Extra Notes and Comments Page 6 of 6