Electric Field and Coulombic Force: Effect of Halving Electrical Permittivity

[wangdang]

Hi,

I was just wondering if you had an electric field which had the coulombic force acting on the two charges as:
F₁= (1/4πε)*(q₁q₂/d²)
Then what would happen to the force if you placed the charges in a medium of half the electrical permittivity? I tried to solve this using ratios and got:

F₁:F₂ = (1/4πε)*(q₁q₂/d²) / (1/4π(1/2)ε)*(q₁q₂/d²)
= 1/ε / 2 /ε
= 1 / 2

F₁/F₂ = 1/2
2F₁=F₂

However this doesn't sound right. If you halved the electrical permittivity, would the force acting on the two charges double?

 

[yungman]

$$\vec{F} = q \vec{E} = \frac{qq_1\hat{d}}{4\pi \epsilon d^2}$$

If $\epsilon$ is half, the the force is double from my understanding.

 

[nasu]

The force is maximum in empty space. In a dielectric medium the force is reduced due to the electric fields induced in the material.

 

[PrakashPhy]

permitivity here is not the common sense english term. The word doesn't signify how much does the medium 'permits' the force to act.
In fact higher the permitivity lower the force.
Its true that half permitivity doubles coulumbic force

 

[sardar]

Hi there,

Great question! The effect of halving the electrical permittivity on the coulombic force is actually quite significant. As you correctly calculated, the force would double. This is because the electrical permittivity, represented by the Greek letter epsilon (ε), is a measure of the ability of a material to store electrical energy. In other words, it represents how easily an electric field can be established in a material.

When you halve the electrical permittivity, you are essentially reducing the ability of the medium to store electrical energy. This means that the electric field between the two charges will be stronger, resulting in a stronger coulombic force between them. This is why your calculated ratio of 1:2 for F1 and F2 is correct.

In practical terms, this means that if you were to place two charges in a medium with half the electrical permittivity, the force between them would be twice as strong compared to if they were in a medium with the original electrical permittivity. This can have significant implications in various scientific and technological applications, such as in the design of electrical circuits or in understanding the behavior of charged particles in different environments.

I hope this answers your question and provides a better understanding of the relationship between electric field, coulombic force, and electrical permittivity. Keep exploring and asking questions about these fundamental concepts in electromagnetism!