How Does Relative Permittivity Relate to the Electric Field in Capacitors?

[jeff1evesque]

Homework Statement

An electric field of a capacitor is defined by the following equation,
$$\vec{E} = \frac{\rho_{s} \hat{a_n}}{\epsilon} = \frac{Q\hat{a_n}}{A\epsilon_r\epsilon_0}$$
where $$\vec{E} = (\frac{\rho_s}{\epsilon})\hat{a_n}$$


Question
I understand the first equality, and reviewing my physics book, I think I understand the derivation. But I was wondering if $$\epsilon = \epsilon_r\epsilon_0$$, and a quick explanation why.


Thanks,


JL

 

[rl.bhat]

Homework Statement

An electric field of a capacitor is defined by the following equation,
$$\vec{E} = \frac{\rho_{s} \hat{a_n}}{\epsilon} = \frac{Q\hat{a_n}}{A\epsilon_r\epsilon_0}$$
where $$\vec{E} = (\frac{\rho_s}{\epsilon})\hat{a_n}$$


Question
I understand the first equality, and reviewing my physics book, I think I understand the derivation. But I was wondering if $$\epsilon = \epsilon_r\epsilon_0$$, and a quick explanation why.


Thanks,


JL

Here εο is the permittivity of the vacuum and εr is the relative permittivity of the medium which is introduces between the plates. It is also called as the dielectric constant.

 

[jeff1evesque]

Here εο is the permittivity of the vacuum and εr is the relative permittivity of the medium which is introduces between the plates. It is also called as the dielectric constant.

Actually, do you mind explaining why it was necessary to break the permittivity into those components? And why the product is equivalent to the permittivity?

Thanks again,


JL

 

[rl.bhat]

Permittivity ε is the property of the space. If you introduce anything between the plates of the capacity, the electric field will decrease due to polarization.. Consequently the potential difference between the plates will decrease and hence the capacitance will increase.
The ratio of capacity with medium and capacity without medium is called relative permittivity εr. εr = Cm / Co = (εA/d)/(εoA/d) = ε/εr