The effect of charge and distance on electric force

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Presentation transcript:

The effect of charge and distance on electric force Coulomb’s Law The effect of charge and distance on electric force

Coulomb’s Law Coulomb’s law states that the electrical force between two charged objects is directly proportional to the product of the quantity of charge on the objects and inversely proportional to the square of the separation distance between the two objects

Formula for Coulomb’s Law F = k * Q1 * Q2 d2 Q1 = quantity of charge on object 1 (in Coulombs - C) Q2 = quantity of charge on object 2 (in Coulombs - C) d = distance of separation between the two objects (in meters) k = Coulomb’s law constant 9 x 109 N *m2/C2 F = Force in Newtons (N)

Coulomb’s Law Q1 and Q2 can be expressed as positive (+) or negative (-) This does not mean that the number in the calculator needs to be positive or negative If the number is – it means the charge is negative, there are more electrons than protons If the number is + it means the charge is positive and there are fewer electrons than protons

Inverse Square Law If you increase the distance between two objects then the force between them will decrease If you decrease the distance between two objects then the force between them will increase Distance plays a big role with force between charged objects

Inverse Square Law The force between two objects will change inversely with the square of the distance that the charges are separated So if the distance is doubled then the force is decreased by a factor of four, if the distance is tripled then the force is decreased by a factor of 9 In other words F ≈ 1/d2 so doubling the distance would be 1/22 = 1/4 and tripling the distance would be 1/32 = 1/9

Example Problem #1 Determine the force of attraction between two charged balloons. The charge on the balloons is 6.0 x 10-7C and the distance is .50m. F = k * Q1 * Q2 = d2 F = 9 x 109 N *m2/C2 * 6.0 x 10-7C * 6.0 x 10-7C (.50m)2 F = .013N

Example Problem #2 A balloon has a charge of 1.0x 10-6C and a golf tube has a charge of 4.0x10-6C and they are a distance of .50m apart. What is the electrical force of attraction? F = k * Q1 * Q2 = d2 F = 9 x 109 N *m2/C2 * 1.0x 10-6C * 4.0x10-6C (.50m)2 F = .144N