A dielectric is a non-conducting material such as glass, rubber, porcelain etc.which does not allow conduction of charge but undergoes polarization in an electric field. Now what happens if an insulating material or dielectric is present between the conductors of a capacitor?
When an electric field is applied to a dielectric, the redistribution of positive and negative charges occurs in the dielectric material called polarization.
In a dielectric the charges are not free to move like in a conductor, therefore they form electric dipoles when subject to an electric field. Initially when there is no electric field, the molecules are distributed randomly in the dielectric.
In case of polar molecules there is a separation between the average portion of negative charges and the average portion of positive charges forming something like lopsided distribution. The molecules that do not have such arrangement are nonpolar molecules.
The electric field applies torque to the polar molecules which tries to align the molecules with the electric field but the polar molecules are not perfectly aligned with the electric field due to thermal agitation. The alignment depends on the electric field and temperature. Generally the alignment increases with increasing electric field and decreasing temperature.
If the dielectric has nonpolar molecules, induced dipoles are formed under the action of electric field which tend to align with the electric field. So it does not matter whether the molecules are polar or not, the dielectric undergoes polarization with the presence of electric field.
In simple words regardless the type of molecules (polar or nonpolar), the electric field pushes positive charge in the direction of electric field and pulls negative charge in the opposite direction causing the induced charges of opposite sign in the corresponding surface of the dielectric.
The opposite electric field when the dielectric is present as shown in Figure 1 is not strong in comparison to the applied electric field.
Every dielectric material has a limit on how much electric field it can withstand. If very strong electric field is applied to a dielectric, the dielectric can permit conduction through the dielectric material called dielectric breakdown.
If a dielectric is present in a capacitor, the potential difference of the capacitor decreases by a factor called dielectric constant \(K\). If the potential difference decreases by the factor \(K\), the electric field must also decrease by the same factor \(K\).
The decrease in potential difference or the electric field is due to polarization of the dielectric material in the capacitor.
If \(V_0\) is the initial potential difference without dielectric and \(V\) is the potential difference with a dielectric, the potential difference \(V\) when the dielectric is present is lesser than the initial potential difference \(V_0\) without the dielectric.
Since the charge is the same with or without a dielectric, the capacitance \(C_0\) without dielectric is lesser than the capacitance \(C\) with dielectric present in the capacitor.
The ratio of \(C\) to \(C_0\) is equal to the ratio of \(V_0\) to \(V\) and called dielectric constant \(K\), that is
\[K = \frac{C}{C_0} = \frac{V_0}{V}\]
When a dielectric completely fills a capacitor, the capacitance increases by the factor \(K\), that is \(C = KC_0\), and
\[C = K \frac{\epsilon_0 A}{d}\]
The capacitance can be increased by decreasing the distance between the plates and increasing the surface area of the plates. Every dielectric material has a strength of maximum electric field it can experience without electric discharge through the dielectric material called dielectric strength.
If the electric field is greater than the dielectric strength, the dielectric material experiences partial ionization and allows conduction through it called dielectric breakdown. The product \(K\epsilon_0\) is called permittivity of the dielectric denoted by \(\epsilon\).
The dielectric constant \(K\) is a number which is always greater than \(1\). The dielectric constant of a capacitor with just the air as the dielectric is about \(1.00059\) which is approximately equal to \(1\). So, a capacitor with just air between the plates is considered to be a capacitor without any dielectric, that is vacuum. So, the the permittivity in vacuum is \(\epsilon = \epsilon_0\) and we also call \(\epsilon_0\) as the permittivity of free space or vacuum.
If the potential difference decreases by the factor \(K\), the electric field also must decrease by the same factor \(K\), that is the electric field decreases when a dielectric is present. If \(E_0\) is the electric field without dielectric and \(E\) is the electric field with dielectric,
\[K = \frac{E}{E_0} \tag{1} \label{1}\]
The magnitude of charge per unit area or surface charge density in each conductor of capacitor without dielectric is \(\sigma\). When a dielectric is present in a capacitor, the magnitude of induced surface charge density is \(\sigma_i\). We know the electric field between two parallel plates is the total surface charge density divided by \(\epsilon_0\), that is \(E = \sigma_\text{total}/\epsilon_0\).
Here \(\sigma \) is the surface charge density without dielectric and \(E_0\) is the corresponding electric field. So, we have
\[E_0 = \frac{\sigma}{\epsilon_0} \tag{2} \label{2}\]
When there is a dielectric present in the capacitor, the electric field is \(E\) and the net surface charge density is \(\sigma - \sigma_i\), so
\[E = \frac{\sigma - \sigma_i}{\epsilon_0} \tag{3} \label{3}\]
Solving for \(\sigma_i\) using Equations \eqref{1}, \eqref{2} and \eqref{3}, we get
\[\sigma_i = \Big ( \frac{K - 1}{K} \Big ) \sigma\]
We know \(K\) is always greater than \(1\) and the magnitude of \(\sigma_i\) is always less than the magnitude of \(\sigma\). In case of vacuum \(K = 1\) (no dielectric) \(\sigma_i = 0\) which is as expected.
