Why net force on two oppositely charged charges when submerged in medium of dielectric constant $k$ decreases?

Question

I want to talk about two scenarios

1. when a dielectric is placed between two oppositely charged charges.

2. when two oppositely charged particles are submerged in medium of dielectric constant $k$.

In my book, it was given about case 1, that force on each charge increases and. in case 2 when submerged the force on charges decreases as dielectric constant is greater than 1

Explanation in this image

Have a look at this image of my book regarding force dependency on medium

According to the equation the force on charges decreases when they are submerged in medium of dielectric constant k but , I wanted to know theoretical explanation , why force decreases when oppositely charged charges are submerged in medium of dielectric constant k . But I can't understand thereotically about this concept. Please tell if I am wrong in understanding this concepts from book. And explain me

I am curious to know to the thereotical explanation not just according to the formula

Addressing comment:(Raad shaikh)

The material is polarised all around the charges so that there should be no effect on charges but in my book it given that net force on oppositely charged charges decreases. why specifically decreases?

Answer 1

The two scenarios mentioned in the question can be re-framed, without any loss of generality, to What happens when a dielectric is placed in an external electric field, say $E_{ext}$.

We must consider two cases, one where the dielectric is polar, and one where it is not.

Non-Polar dielectric

1. Consider a slab initially without any external $E$. Since it is a non polar slab. Its total dipole moment would be zero.
2. When we now place it in an external electric field, $E_{ext}$, all the atoms inside become polarized as shown. We can then consider them to be tiny dipoles, the vector sum of which would be non-zero, as shown.
3. Note that we can just sort of cancel out the positives and negatives, to get the third picture. (Keep in mind that this doesn't happen, and it just for us to simplify our calculations.)
4. We can then consider the left and right sides of the slab to act as charged sheets, and we know the electric field due to it. (Neglecting fringing fields)

[Note, in the figure, the circle with a dot in it represents a nucleus and its electron cloud, which can be seen to be distorted(ellipses) in the second case]

For any particle in the slab, we can write: $$E_{net} = E_{ext} - E_{in}$$ Since the external and induced fields would be opposite to each other.

To simplify our day-to-day calculations, we define: $$E_{net} = \frac{E_{ext}}{K}$$ Where $K$ is the dielectric constant.

Comparing the two equations, we get:

$$E_{in} = E_{ext} \left( 1 - {1 \over K} \right)$$

Or

$$\sigma_b = E_{ext} \epsilon_0 \left( 1 - {1 \over K} \right)$$

Where $\sigma_b$ denotes the bound charge density as shown in figure.

Polar dielectrics

This is very similar to non-polar dielectrics, and gives the same result. Try it out and see if you get it. Hint: Even in polar dielectrics, initial net dipole moment is zero.

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Coming back to your question. Can you now see why there would be another field present in between the two charges, and how this would affect it? We just use a constant $K$ to simplify our calculations, instead of having to deal with polarizability, etc.

When a dielectric is placed versus when the charges are submerged in the medium is basically the same question.

Answer 2

Imagine the dielectric around the submerged objects is composed of many pairs of equal and opposite charges and spaces between the pairs. When polarized, these pairs move apart, but the distance between the two members of the pair still remains smaller than the spaces between the pairs. Now the submerged objects (let's say they are larger in both charge and size than the ones in the pairs) feel the sum of forces from the polarized dielectric: same force as without the dielectric at the remaining spaces plus a force opposite that, in between the members of each pair.

To make an image of this yourself, remember to displace the charges of the pairs from their central positions according to the field direction from the submerged objects. In order to find the force (qualitatively easy to see, it's smaller) on the objects, average the forces over the volume of the object, which must include several "unit cells" inside it as well. Otherwise you get a random direction of force.

Answer 3

For Case 1 (Dielectric Slab Between Charges): The force increases because the dielectric reduces the effective field in a way that enhances the interaction between opposite charges (attractive force).

For Case 2 (Charges in a Medium with Dielectric Constant k): The force decreases because the dielectric constant reduces the effective electric field strength, weakening the interaction between both types of charges (opposite or like).

Answer 4

Basic Concept:

When charges are submerged in a dielectric medium (which has a dielectric constant ( k ) greater than 1), the polarization of the dielectric material comes into play. This polarization affects the electric field between the charges.

Why Does the Force Decrease?

1. Polarization of the Dielectric:

   • A dielectric material is an insulating substance that, when placed in an electric field, becomes polarized. This means that the molecules or atoms in the dielectric material align in such a way that they create opposing electric fields.
   • The dielectric material’s molecules align so that the positive charges in the material are attracted toward the negative charge and the negative charges are attracted toward the positive charge, thereby shielding the charges from each other.

2. Shielding Effect:

   • The polarization causes the dielectric medium to partially cancel out the electric field produced by the charges. This is the crucial point. Instead of the full electric field being felt by the charges, the dielectric creates its own electric field that opposes the field of the charges.
   • The result is that the electric field between the charges is reduced, because the dielectric material is reducing the strength of the field that the charges would otherwise experience if they were in a vacuum (or air, which has a very small dielectric constant, close to 1).

3. Decreased Electric Field:

   • Since the electric field between the charges is weaker, the force experienced between the charges decreases. This is because the electrostatic force ( F ) between two charges is proportional to the electric field strength ( E ) (which is influenced by the dielectric constant ( k )).

4. Mathematical Aspect (for Understanding):

   • In vacuum, the force between charges is given by Coulomb's Law:
     $$F = \frac{1}{4\pi\epsilon_0} \cdot \frac{q_1 q_2}{r^2}$$ where ( \epsilon_0 ) is the permittivity of free space. When charges are submerged in a dielectric, the dielectric constant ( k ) modifies the permittivity of the medium to ( \epsilon = k \epsilon_0 ). This results in a decrease in the force: $$F = \frac{1}{4\pi\epsilon} \cdot \frac{q_1 q_2}{r^2} = \frac{1}{4\pi k \epsilon_0} \cdot \frac{q_1 q_2}{r^2}$$

     As ( k > 1 ), this makes the denominator larger, reducing the force.

Intuitive Reason:

• Think of the dielectric as a barrier or shield around the charges. In a vacuum or air, the charges can directly influence each other with their electric fields. However, when submerged in a dielectric, the material itself "interferes" with this influence, reducing the overall field between the charges. This makes it harder for the charges to exert their force on each other because the dielectric material is "absorbing" some of the field lines and opposing the force.