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

Answer:

1

Explanation:

Let's call the magnitude of the charge of 10000 electrons $q$.

Because A loses q, it has a net charge $+q$ (electrons are negative; losing them creates positive charges). Conversely, B has gained q, so it has a charge of $-q$. The distance between them is 1 m.

By Coulomb's law, the force between them is

$F_1 = \dfrac{kq \times q}{1^2}$

$F_1 = kq^2$

When A loses an additional 10000 electrons, the charge on it becomes $+2q$ while that on B is $-2q$. The distance is now 2 m.

The force is now

$F_2 = \dfrac{k\times2q \times 2q}{2^2}$

$F_2 = \dfrac{4kq^2}{4}$

$F_2 = kq^2 = F_1$

$\dfrac{F_2}{F_1}=1$