# Does force between two charges depend on medium?

I know that net force on a charge decreases by a factor of K(dielectric constant of medium ) but I wanna know that whether

The force on one charge due to other charge has changed

Or

This decrease of force is due to force exerted by the medium and force between the two charges still remain the same. ( I mean that forces on one charge remain due to the other remains unchanged but the net decrease is due to the medium applying force on the opposite direction ) .

Thanks.

I wouldn't say that the force law has changed. In every medium, the force as a function of electric field is the same:

$$\vec{F}=q\vec{E}$$

The difference is that, in a dielectric medium, the electric field looks different than in a vacuum. This is because the dipoles in the medium are influenced by the electric field, and orient themselves in the configuration of minimum energy (namely, so that their positive end is pointing in the opposite direction of the electric field). The dipoles themselves generate an electric field, and in this configuration, their contribution partially cancels the external electric field. So the net effect of a dielectric medium is that electric fields are smaller within it than in a vacuum.

• A problem regarding this concept was asked in IITJEE Advance 2020 (an engineering exam) where two charged balls were first suspended through two strings in air and then in water, then it is was asked whether the "electric force" between the balls remains the same in air and water or not. The answer was given that the "electric force" between the balls decreases in water, but through your reasoning it seems that the force should remain the same. Can you please help me out on this? Jul 4 at 18:52

Both pictures are valid due to superposition. Let's say that charge $$Q$$ is a source charge at position $$\mathbf{R}$$ creating an electric field $$\mathbf{E}_{Q}$$, and that the dielectric medium creates a polarization field $$\mathbf{E}_{\mathrm{p}}$$, such that the net electric field is $$\mathbf{E}_{\mathrm{net}} = \mathbf{E}_{Q}+\mathbf{E}_{\mathrm{p}}$$. Now we put a test charge $$q$$ at position $$\mathbf{r}$$ (a test charge does not create its own field, it only reacts to the background field). The net force on the charge is $$\mathbf{F} = q\mathbf{E}_{\mathrm{net}}(\mathbf{r}) = q(\mathbf{E}_{Q}(\mathbf{r})+\mathbf{E}_{\mathrm{p}}(\mathbf{r}))$$ so both statements are equivalent since the two electric fields add by superposition.

At a slightly more philosophical level, you could think that there is actually only a single electric field, it just so happens that mathematically we can write the single electric field as a sum of two contributions. The test charge ultimately only reacts to the net field, and doesn't know the difference between whether that net field came from multiple sources or a single source. This is actually really the case in Quantum Electrodynamics, where the vacuum itself is naturally polarized by the presence of charge due to the presence of virtual pairs of oppositely charged particles, i.e. all physical charges (e.g. electrons) are naturally screened by the dielectric effect of the quantum field itself. All we see is the net screened electric field of the charge, and if you do experiments colliding charges so that they get very very close to each other, they will experience less screening and therefore the effective charge increases, this effect is known as the "running of the coupling".

• "The test charge ultimately only reacts to the net field, and doesn't know the difference between whether that net field came from multiple sources or a single source." - This is extremely important, because it makes it very clear that electromagnetism still follows Newton's Third Law (there's a common misconception that it doesn't because people think two charges should form an action-reaction pair; this picture makes it clear that the action-reaction pair should be the charge and the electromagnetic field itself). Sep 30 '20 at 16:12

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.