Question

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?

Answer 1

It is the same.

Step-by-step 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):

`F =(k*(10e4e)^(2) )/(1m)`

• 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:

`Ff = (k*(2*e*10e4)^(2))/(4m) = k*4*((e*10e4)^(2) )/(4m) = k*((e*10e4)^(2) )/(1m) = F0`

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