Two charges are placed at a certain distance apart in air. If brass plate is introduced between them, the force between them will
A. increase
B. decrease
C. remain the same
D. become zero

Hint: Coulomb's law states that the force of attraction or repulsion between two charged bodies is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. In order to find the solution of the Given the question we will use the concept of coulomb’s law.

Formula used:
$F = \dfrac{1}{{4\pi {\varepsilon _ \circ }K}}.\dfrac{{{q_1}{q_2}}}{{{r^2}}}$

Complete step-by-step answer:
Coulomb’s law or Coulomb’s inverse square states the force causing between the two electrically charged bodies to attract each other or repel each other.
Mathematically, it can be written as
$F = \dfrac{1}{{4\pi {\varepsilon _ \circ }K}}.\dfrac{{{q_1}{q_2}}}{{{r^2}}}$
Where $\varepsilon $ is absolute permittivity, $K$ is the dielectric constant of the medium in which the two charges are placed.
When two like charges are placed in air, then they will experience a force of repulsion from each other.
So, according to the question if brass plates are introduced between them, then there would be no force between them. Because, if the two charges are positively charged, then both of them will try to attract electrons of the plate along opposite sides. Thus, there will be no net force acting on the electron plate and it will remain unchanged.
If the two charges are negative then both of them will try to repel electrons from both the sides and nothing will happen.
Therefore, an unaffected plate doesn’t affect the force between the two charges. So, the force between them will remain the same.

So, the correct answer is “Option C”.

Note: There are some limitations of Coulomb’s law such as:
1) This law is applicable only for the point charges which are at rest.
2) This law can only be applied where the inverse square law is obeyed.