# Qualitative explanation for reduced force between two charged particles in some medium other than air/vacuum

Suppose I place two charged particles in the medium other than air or vacuum. The net force on one by other is given as

$$\vec F=\frac{q_1 q_2}{4\pi \epsilon r^2}\vec a_{r12}$$

where $$\vec a_{r12}$$ is a unit vector in the direction between positions 1 and 2, $$\vec F$$ is the magnitude of the force each particle experiences due to the other, and $$\epsilon$$ is permittivity of medium between the particles.

Now I was told that $$\epsilon > \epsilon_o$$ , therefore force between two charged particles is less in any medium than it is in free space at same seperation.

I tried to describe the reason for this qualitatively by considering that particles of the medium are polarised by charges.

If medium is water, for example,and we have both the positive charges, then the negative pole of water molecules i.e. oxygen atoms surrounds both the charges, However I can't derive a reasonable explanation.

Can you complete my reasoning ( if it is correct) or tell me the correct reasoning?

Dielectric [is an] insulating material or a very poor conductor of electric current. When dielectrics are placed in an electric field, practically no current flows in them because, unlike metals, they have no loosely bound, or free, electrons that may drift through the material. Instead, electric polarization occurs. The positive charges within the dielectric are displaced minutely in the direction of the electric field, and the negative charges are displaced minutely in the direction opposite to the electric field. This slight separation of charge, or polarization, reduces the electric field within the dielectric (bolding emphasis mine).

Of course saying ‘lower field’ is identical to saying ‘reduced force on a charged particle’. The field from one particle is reduced and the force pushing or pulling on the other particle in response is reduced, by the same proportion (we wouldn’t reduce the fields for each particle, that would be double counting; we just reduce one particle’s field, for either particle as force is same either way, and say, ‘What’s the force on the other charge now?’).

Field is just a way of describing coulomb attraction/repulsion, as different configurations can give same net forces at various points (think of electric field as force per unit charge, force on a unit charge at that location being equivalent to field at that location). It’s useful and tractable.

And the field is reduced because “the quantity of energy stored in an electric field per unit volume of a dielectric medium is greater”.. than the energy stored in the same field in a vacuum. So the fields wouldn’t be the same - that would increase energy not conserve it, just by moving a dielectric into a field.

Field and energy are related by the concept of electric potential, the energy to move a unit-charge through the field, force times distance (actually integral of Fdt), is electric potential, similar to energy in the field. We could say part of the the field is being ‘used’ to keep the dielectric medium polarized, to hold the dipoles that comprise it in alignment with the field, ie to keep those dipoles polarized.

Another way to say all this is that the dielectric constant of air is 1.

Consider a positive point charge $$q$$, and thick shell of dielectric, $$\epsilon$$, with dimensions $$R_I$$ and $$R_O$$.

In side the shell, the field is:

$$E(r) =\frac 1{4\pi\epsilon_0} \frac{q}{r^2},\,\,\ \ \ \ \ \ (r

At $$R_I$$, the is a negative polarization surface charge density cause by $$\nabla P$$. There is also positive, but lesser density (by $$R^2_I/R_O^2$$), on the outer surface of the dielectric.

Inside the dielectric, the outer positive shell contributes no field. The field is entirely due to the positive point charge and the negative shell of surface charge, so that:

$$E(r) =\frac 1{4\pi\epsilon} \frac{q}{r^2},\,\,\ \ \ \ \ \ (R_I < r R_O)$$

Thus, if you work out the polarization divergence, you should find the negative surface charge totals to:

$$q_- = q(\frac{\epsilon_0}{\epsilon}-1) < 0$$

At $$r>R_O$$, the polarization surface charges cancel, and the field returns to:

$$E(r) =\frac 1{4\pi\epsilon_0} \frac{q}{r^2},\,\,\ \ \ \ \ \ (r>R_O)$$

• You gave a purely mathematic answer, I have just started studying about electrostatics and only wanted a qualitative reasoning Commented Jul 29, 2021 at 17:08
• then ignore the math. You have a + and a - minus charge causing a field that is less than the + charge alone. Note that inside a shell of charge (mass), there is no electric (gravitational) field [Gauss's Theorem].
– JEB
Commented Jul 29, 2021 at 17:20