If one of the charges is decreased how does that affect the force between the two charges?

Coulomb's Law is one of the basic ideas of electricity in physics. The law looks at the forces created between two charged objects. As distance increases, the forces and electric fields decrease. This simple idea was converted into a relatively simple formula. The force between the objects can be positive or negative depending on whether the objects are attracted to each other or repelled.

Think about a few concepts before you continue reading. Some charges are attracted to each other. Positive and negative charges like to move towards each other. Similar charges such as two positive or two negative push away from each other. You also need to understand that forces between objects become stronger as they move together and weaker as they move apart. You could yell at someone from far away, and they would barely hear you. If you yelled the same amount when you were together, it would be more powerful and loud.

Coulomb's Work

Charles Augustin de Coulomb was a French scientist working in the late 1700's. A little earlier, a British scientist named Henry Cavendish came up with similar ideas. Coulomb received most of the credit for the work on electric forces because Cavendish did not publish all of his work. The world never knew about Cavendish's work until decades after he died.

Coulomb's Law

But you're here to learn about the law. When you have two charged particles, an electric force is created. If you have larger charges, the forces will be larger. If you use those two ideas, and add the fact that charges can attract and repel each other you will understand Coulomb's Law. It's a formula that measures the electrical forces between two objects.

F=kq1q2/r2

"F" is the resulting force between the two charges. The distance between the two charges is "r." The "r" actually stands for "radius of separation" but you just need to know it is a distance. The "q1" and "q2" are values for the amount of charge in each of the particles. Scientists use Coulombs as units to measure charge. The constant of the equation is "k." As you learn more physics, you will see that this formula is very similar to a formula from Newton's work with gravity.

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Fluxgate Magnetometry (NASA/GSFC Video)

Does force between two charges depend on medium?

Yes

The electric field intensity $\overrightarrow E$ (volts/meter) at point 2 due to a point charge $Q_1$ at point 1 is

$$\overrightarrow E=\frac{Q_1}{4πεr^2}\overrightarrow a_{r12}$$

where

$\overrightarrow a_{r12}$ = a unit vector directed from 1 to 2

$r$ = the distance between points 1 and 2

$ε$ = the permittivity of the medium.

The dielectric constant, or relative permittivity $ε_r$ is

$$ε_{r}=\frac{ε}{ε_0}$$

where $ε_0$ = the permittivity of free air or a vacuum. Thus

$$\overrightarrow E=\frac{Q_1}{4πε_{r}ε_{0}r^2}\overrightarrow a_{r12}$$

A charge $Q_2$ placed at point 2 will experience a force $\overrightarrow F$ of

$$\overrightarrow F=Q_{2}\overrightarrow E=\frac{Q_{1}Q_{2}}{4πε_{r}ε_{0}r^2}\overrightarrow a_{r12}$$

So the force experienced by $Q_2$ due to the electric field at point 2 will be inversely proportional to the dielectric constant $ε_{r}$.

From Newton's third law the force experienced by $Q_2$ at point 2 due to the field created by $Q_1$ at point 1 is equal and opposite to the force experienced by $Q_1$ at point 1 due to the field created by $Q_2$.

The third equation also tells us that, for a given $Q_1$ located at point 1, the strength of the electric field at point 2 is inversely proportional to the dielectric constant of the medium.

If instead of point charges we were dealing with a parallel plate capacitor of given plate separation, plate area, and net positive and negative charge on the plates, if we use a material between the plates having a greater dielectric constant, it effectively reduces the strength of the field due to the partial polarization of the molecules of the dielectric as pointed out in the answer of @probably someone. Since the voltage $V$ across the plates is $V=Ed$, the voltage is reduced.

Finally, since the capacitance of the capacitor is related to the charge and voltage by

$$C=\frac{Q}{V}$$

Increasing the dielectric constant increases the capacitance of the capacitor.

Hope this helps.

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