Respuesta :
Answer:
two positively charged objects
Explanation:
We have 3 possible combinations of charges for the 2 objects in the problem:
1) 2 positively-charged objects: in this case, the force between them is repulsive. Also, the net charge on each object is given by an excess of positive charge
2) 2 negatively-charged objects: in this case, the force between them is repulsive. Also, the net charge on each object is given by an excess of negative charge
3) 1 positive and 1 negatively charged object: in this case, the force between them is attractive. Also, the net charge on the positive object is given by an excess of positive charge, while the net charge on the negative object is given by an excess of negative charge.
We also remind that the magnitude of the electrostatic force between the two objects is:
[tex]F=k\frac{q_1 q_2}{r^2}[/tex] (0)
where k is the Coulomb's constant, q1 and q2 the charges on the two objects, r the separation between the objects.
Here we want to add the same amount of electrons to each object: it means, we add negative charge on both objects. So this is what happens to the 3 combinations:
1) Here we are reducing (in magnitude) the charge on both objects --> so the force will decrease, according to eq.(0)
2) Here we are increasing (in magnitude) the charge on both objects --> so the force will increase, according to eq. (0)
3) Here we are increasing the charge on one object and decreasing the charge on the other object by the same amount, so the outcome depends on the initial charge on the two objects.
So the correct combination is: two positively charged objects.
