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marysya
3 months ago
11

Object A and object B are initially uncharged and are separated by a distance of 1 meter. Suppose 10,000 electrons are removed f

rom 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?
Physics
1 answer:
kicyunya [3.2K]3 months ago
6 0

Answer:

1

Explanation:

Let’s denote the charge magnitude for 10,000 electrons as q.

When object A loses charge q, it results in a total charge of +q (since electrons are negatively charged, their loss turns A positively charged). On the other hand, B, which gains q, assumes a charge of -q. They are initially 1 meter apart.

According to Coulomb’s law, the force acting between them is

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

F_1 = kq^2

When an additional 10,000 electrons are stripped from A, its charge becomes +2q, while B’s charge becomes -2q. Now, they are 2 meters apart.

The force at this new distance is

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

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