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1. A boy rubs his dog with a balloon. The dog gets 20,000 extra electrons.a. What is the charge on the dog?b. Then the boy rubs a piece of silk on his cat. The cat LOSES 30,000 electrons to the silk. What is thecharge on the CAT?c. Finally, the boy places his cat (20 kg) on frictionless ice and he places his dog (35 kg) standing 2maway on the edge of the ice.i. Calculate the electrostatic force on the CAT from the DOG. Do they ATTRACT or REPEL?ii. What is the initial acceleration of the cat?iii. What is the electrostatic force on the DOG from the CAT? Why is this answer the same as ordifferent from “i”?iv. What is the initial acceleration of the dog? Why is this answer the same as or different from “ii”?

Question

  1. A boy rubs his dog with a balloon. The dog gets 20,000 extra electrons.a. What is the charge on the dog?b. Then the boy rubs a piece of silk on his cat. The cat LOSES 30,000 electrons to the silk. What is thecharge on the CAT?c. Finally, the boy places his cat (20 kg) on frictionless ice and he places his dog (35 kg) standing 2maway on the edge of the ice.i. Calculate the electrostatic force on the CAT from the DOG. Do they ATTRACT or REPEL?ii. What is the initial acceleration of the cat?iii. What is the electrostatic force on the DOG from the CAT? Why is this answer the same as ordifferent from “i”?iv. What is the initial acceleration of the dog? Why is this answer the same as or different from “ii”?
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Solution

a. The charge on the dog can be calculated using the formula Q = ne, where n is the number of electrons and e is the charge of one electron. The charge of one electron is -1.6 x 10^-19 Coulombs. So, the charge on the dog is Q = 20000 x -1.6 x 10^-19 = -3.2 x 10^-15 Coulombs.

b. The charge on the cat can be calculated in the same way. However, since the cat loses electrons, the charge will be positive. So, the charge on the cat is Q = 30000 x 1.6 x 10^-19 = 4.8 x 10^-15 Coulombs.

c. i. The electrostatic force between the cat and the dog can be calculated using Coulomb's law, which is F = kQ1Q2/r^2, where k is Coulomb's constant (8.99 x 10^9 Nm^2/C^2), Q1 and Q2 are the charges, and r is the distance between them. So, the force is F = 8.99 x 10^9 x -3.2 x 10^-15 x 4.8 x 10^-15 / (2^2) = -2.16 x 10^-26 N. The negative sign indicates that the force is attractive.

ii. The initial acceleration of the cat can be calculated using Newton's second law, which is F = ma, where m is the mass and a is the acceleration. So, the acceleration is a = F/m = -2.16 x 10^-26 / 20 = -1.08 x 10^-28 m/s^2.

iii. According to Newton's third law, the force on the dog from the cat is equal in magnitude and opposite in direction to the force on the cat from the dog. So, the force on the dog is also -2.16 x 10^-26 N.

iv. The initial acceleration of the dog can be calculated in the same way as the cat's. So, the acceleration is a = F/m = -2.16 x 10^-26 / 35 = -6.17 x 10^-29 m/s^2. This is smaller than the cat's acceleration because the dog is heavier.

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