Here is a problem that I have encountered:

A parallel-plate capacitor has the space between the plates filled with a slab of dielectric with constant $K_1$ and one with constant $K_2$, each of thickness $d/2$, where $d$ is the plate separation. Show that the capacitance is: $$C=\frac{2\varepsilon_0A}{d}\left(\frac{K_1K_2}{K_1+K_2}\right)$$ I know that potential is diminished by factor $\frac{1}{K}$ in the case where a single dielectric exists in a parallel-plate capacitor, but I do not understand the mechanism behind why this occurs, so I cannot generalize the result to more than one dielectric. The way I was taught, the potential is simply "observed" to decrease when a dielectric is introduced. I think I vaguely recall the net electric field diminishing due to the polarization of the dielectric which in turn causes the potential to reduce as well (as it is proportional to the electric field), but I don't understand the exact mechanism behind this.

So my question is how exactly is potential reduced when a dielectric is introduced in a capacitor?

  • 1
    $\begingroup$ do you understand why capacitors in parallel add linearly? If so, do you then understand why capacitors in series add harmonically? Finally: do see that your problem is 2 capacitors in series? $\endgroup$
    – JEB
    Commented Apr 20, 2019 at 20:45
  • $\begingroup$ Yes I understand the additive properties of capacitors (which come from the additive properties of potential while keeping charge conserved). And I suppose that this problem is essentially just two capacitors in series, since each half of the system contains a single dielectric, with a certain potential difference and a certain charge. Thanks! $\endgroup$ Commented Apr 21, 2019 at 17:09

1 Answer 1


It is the potential difference between the plates that is diminished. Another term for this difference is "voltage on plates". The potential itself changes differently in different points of space.

This voltage on plates decreases because the electrostatic field between the plates decreases. This field decreases because the field due to charges on the plates (which are assumed to be the same as they were before the slab was inserted) is partially cancelled by the field due to polarization of the dielectric. In most cases of slab capacitor this additional field is weaker than the external field due to plates, so total field has the same direction as the external field, it is just of lower strength.

THe field due to polarization happens because polarization means the ever-present charges in the dielectric matter shift their positions along the field. Inside the dielectric charge density does not change much, but on its surface, new substantial non-zero surface charge density appears due to this global shift. Most dielectric materials behave in such a way that the field due to their polarized state decreases the external field. Some special materials may behave the opposite way, but they are rare to encounter.


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