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A metal ball with diameter of a half a centimeter and hanging from an insulating thread is charged up with 1010 excess electrons. An initially uncharged identical metal ball hanging from an insulating thread is brought in contact with the first ball, then moved away, and they hang so that the distance between their centers is 20 cm. (a) Calculate the electric force one ball exerts on the other, and state whether it is attractive or repulsive. (Enter the magnitude of the force.) Entry field with correct answer 1.44e-7 N This force is Entry field with correct answer attractive repulsive (b) Now the balls are moved so that as they hang, the distance between their centers is only 5 cm. Naively one would expect the force that one ball exerts on the other to increase by a factor of 42

User Pkinsky
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Answer:

a) F = 1.44 10⁻⁷ N, the charges are of the same sign the force is repulsive

b) F₂ / F₁ = 16

Step-by-step explanation:

a) When the balls are touched the load is distributed evenly between the two balls, therefore when separating each ball has a load of

q₁ = q₂ = ½ 10¹⁰ 1.6 10⁻¹⁹ C

q₁ = q₂ = 0.8 10⁻⁹ C

As the charges are of the same sign the force is repulsive

To calculate the force let's use Coulomb's law

F = k q₁ q₂ / r²

F = 8.99 10⁹ 0.8 10⁻⁹ 0.8 10⁻⁹ / 0.20²

F = 1.44 10⁻⁷ N

b) Let us seek strength for the distress of

r₂ = 0.05 m

F₂ = 8.99 10⁹ 0.8 0.8 10⁻¹⁸ / 0.05²

F₂ = 2,301 10⁻⁶ N

b) The relationship between these two forces is

F₂ / F₁ = 23.01 10⁻⁷ / 1.44 10⁻⁷

F₂ / F₁ = 16

User Kachina
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