An average cat weighs about 3.20 N. If two such cats each carried 1.0 coulomb ex

Question

An average cat weighs about 3.20 N. If two such cats each carried 1.0 coulomb excess charge, one positive and one negative, how far apart would they have to be for the magnitude of the electric force between them to equal their 3.20-N weight? Would the force be an attraction or a repulsion? Suppose that the cats are sitting only 5.00 m from each other. If this is the case, how much excess charge would each need to have (assume both have the same amount) for the electric force between them to be 3.20 N? Based on the calculations, comment: Does it appear that 1.0-C of charge is a rather large or a rather small amount of charge? Would you say that most everyday objects carry more, or less, than 1.0-C of charge when they pick up some electrostatic charge? Explain your reasoning.

Explanation / Answer

a) since they are oppositely charges, the force will be attractive

F = kq^2 / r^2

3.2 = 9*10^9*1^2 / r^2

r = 53033 m = 53 km

b) 3.2 = 9*10^9*q^2 / 5^2

q = 9.42*10^-5 C

c) 1 coulomb charge is rather large charge .

charge on 1 electron = 1.6*10^-19 C

number of electrons = 1/1.6*10^-19 = 6.25*10^18

this means that the number of electrons are very much higher

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