# Homework Help: Permittivity ε

1. Jul 16, 2009

### jeff1evesque

Statement:
Permittivity ε = ε_{r}ε_{0}. I understand the definitions of each permitivity, where the former is the relative permittivity (of the dielectric), and the latter is of the free space between the plates of the capacitor. But can someone explain to me, why they are multiplied and not summed?

My thoughts:
Since a capacitor as a whole has two components of permittivity, namely ε_{r}, ε_{0}, shouldn't the actual permittivity ε = ε_{r} + ε_{0}?

Question:
I guess I am trying to picture this capacitor, and I feel like the permittivity components should be added. Could someone enlighten me?

Thanks,

JL

2. Jul 16, 2009

### rock.freak667

The part in bold. Read that over. εr is the relative permittivity. Relative to what you might ask? Well relative to the permittivity of free space!

What does this result in.

$$Relative \ Permittivity = \frac{Actual \ Permittivity}{Permittivity \ of \ Free \ Space}$$

$$\epsilon_r = \frac{\epsilon}{\epsilon_0} \Rightarrow \epsilon = \epsilon_r \epsilon_0$$

it all lies in what εr was defined to be.

3. Jul 17, 2009

### rl.bhat

When you define Coulomb's law of electrostatic force between two charges you write
F = 1/4πεo* q1*q2/d^2 when charges are placed in free space.
When the space between the charges filled be some other material, force will decrease by some factor
Now the force is written by F' = 1/4πεoεr* q1*q2/d^2. When you say factor it must be division. So F' = F/εr. εr is always greater than one.