Decreasing the distance between the two parallel plates of a capacitor increases the amount of charge that can be held on each plate. If this is because the charges are attracted to each other and consequently less "focused" on repelling like charges, why do dielectrics increase capacitance?
Wouldn't the electric fields, which face the opposite direction, caused by the dipoles of the atoms of the dielectric decrease the electric field between plates and as a consequence this attractive force?
And if so, does that mean that the charge-dielectric dipole attraction has a greater effect than the original electric field between plates (without a dielectric)?
Best Answer
Assume that the voltage across an isolated capacitor with air ($\approx$ vacuum) between the plates of separation $d$ is $V_{\text{air}}$ and the charge stored on the capacitor is $Q$.
$Q = C_{\text{air}} V_{\text{air}}$
The electric field between the plates is $E_{\text{air}} = \dfrac {V_{\text{air}}} {d} $.
Now put a dielectric of relative permittivity $\epsilon_r$ between the plates.
Charges are induced in the dielectric which produce an electric field in opposition to the electric field produced by the charges on the plates.
The net electric field between the plates decreases as does the force between the plates and the electric field becomes $\dfrac {E_{\text{air}}}{\epsilon_r}$
This in turn means that the potential difference across the plates must also be reduced to $\dfrac {V_{\text{air}}}{\epsilon_r}$
The charge on the capacitor $Q$ is still the same so the capacitance of the capacitor has incresed and is now $\epsilon_r C_{\text{air}}$.