# Dielectric and capacitor

During my course on dielectrics in capacitors, I learned that the dielectric medium if inserted between the plates of a capacitor, reduces the electric field from $$E_0$$ to $$\frac{E_0}{K}$$, $$K$$ being the dielectric constant. Later while attempting to solve the questions, I happened to come across a question which goes like this:

A parallel plate capacitor has a dielectric slab of dielectric constant $$K$$ between the plates that covers $$\frac13$$ of the area of its plates (continuously). The electric field inside the dielectric slab is $$E_1$$ and that without the dielectric is $$E_2$$. What is $$\frac{E_1}{E_2}$$?

Now based on what I learned, I'd say that $$E_1= \frac{E_2}{K}$$ Or the answer is $$\frac1K$$

But the book uses another way to answer this question which also seems reasonably correct. Since the electric field is constant for dielectric (due to less distance between the plates) and $$|\Delta{V}|= E.\Delta{R}$$ Meaning $$E= \frac{\Delta{V}}{\Delta{R}}$$

And these two can be thought of as independent capacitors in parallel and $$V$$ is same and also $$R$$ so electric fields are equal i.e. $$\frac{E_1}{E_2}=1$$

Can someone explain why the second one is right?

• Is the area covered In parallel or in series Jul 6, 2020 at 4:52
• @BlackSusanoo I will attach a link to the diagram because somehow it doesn't let me attach an image
– Rew
Jul 6, 2020 at 4:53
• @BlackSusanoo Check out now.
– Rew
Jul 6, 2020 at 4:56
• Ok see this is a case of parallel Jul 6, 2020 at 5:00

• See Rew if a dielectric is added it is supposed to make the electric field E/K. But as you can see that the plates are connected to a battery and their potentials cant change so the only way that can happen is if there is a stronger electric field. As electric field due to conducting sheet is $\sigma$/$\epsilon$. As this field has to remain same even though $\epsilon$ becomes k$\epsilon$ the charge density increases by a factor of K and so capicitance also increases Jul 6, 2020 at 5:29