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