Object A and object B are initially uncharged and are separated by a distance of 1 meter. Suppose 10,000 electrons are removed from object A and placed on object B, creating an attractive force between A and B. An additional 10,000 electrons are removed from A and placed on B and the objects are moved so that the distance between them increases to 2 meters. By what factor does the electric force between them change?

Respuesta :

Answer:

It is the same.

Explanation:

  • Assuming that we can treat to both objects, once charged, as point charges, the attractive force between them, must obey Coulomb's Law.
  • The charge on each object is just the charge of 10,000 electrons:

        Q = e*10⁴ = 1.6*10⁻19C*10⁴ = 1.6*10⁻¹⁵ C

  • As the same charge that we remove from one object in magnitude is the same as the one built on the other, this force can be expressed as follows (in magnitude):

       [tex]F =\frac{k*(10e4e)^{2} }{1m}[/tex]

  • If we increase the charge removing additional 10,000 electrons from A and placed on B, as this incremental charge is equal to the existing charge, this means that the charge on each object will double.
  • If, at the same time, we double the distance between charges, the force between them can be written as follows:

        [tex]Ff = \frac{k*(2*e*10e4)^{2}}{4m} = k*4*\frac{(e*10e4)^{2} }{4m} = k*\frac{(e*10e4)^{2} }{1m} = F0[/tex]

  • So, the force between the objects will keep the same.

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