One brass plate is inserted between two charges. The force between the charges will
A. Remain the same
B. Increase
C. Decrease
D. Fluctuate

Hint: To answer this question, use the formula for electrostatic force which is also known as Coulomb’s law. Electrostatic force gives the relation between the magnitude of force between two charges and the separation between those two charges. So, use this relation between force and separation between the charges to find what will happen to the charges when a brass plate is inserted between the two charges.
Formula used:
$F = \dfrac {1}{4\pi {\epsilon}_{0}}\dfrac{{q}_{1}{q}_{2}}{{r}^{2}}$

Complete answer:

As the brass plate is a conductor, the potential on both the sides of the plate will be the same.
Electrostatic force is given by,
$F = \dfrac {1}{4\pi {\epsilon}_{0}}\dfrac{{q}_{1}{q}_{2}}{{r}^{2}}$ …(1)
Where, r is the distance between two charges
From the above equation, it can be inferred that the electrostatic force is inversely proportional to the distance between the two charges.
$\Rightarrow F \propto \dfrac {1}{{r}^{2}}$…(2)
When we place a brass plate between two charges, the thickness (t) of the brass plate gets subtracted from the effective distance between the two charges. Thus, the value of distance decreases.
Using the equation. (2) we can say, as the distance between the charges increases, the electrostatic force decreases.
Thus, when a brass plate is inserted between two charges, the force between the charges will increase.

So, the correct answer is option B i.e. increase.

Note:
If both the charges are like charges then there will be an attractive force developed between the two charges. If both the charges are unlike charges, the already attractive force will increase by the additional attractive force developed due to induced charges. Students must understand that the electrostatic field is along the line which is joining the two charges. Stronger the strength of the electric field, greater the force experienced by each charge.