# Why does potential decrease when a dielectric is introduced?

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?

• 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? – JEB Apr 20 '19 at 20:45
• 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! – Andrew Paul Apr 21 '19 at 17:09