Suppose I have two plate electodes with a dielectric material between them with a permitivity of $\epsilon=10$. I now put a voltage V between them. What is the electric field in the dielectric region? Well since the electrodes are plates we have simply:

$E = V/d$ which is independent of $\epsilon$.

On the other hand, my intuition tells me that this should depend on $\epsilon$. What is wrong with the argumentation above?

  • $\begingroup$ Deleted my answer ,what do you mean by putting voltage V ?you mean putting a battery of emf V? $\endgroup$
    – Paul
    Mar 18, 2017 at 11:47
  • $\begingroup$ For a constant electric field we always have V=Ed. However, with the dielectric, it will require more charge on the plates to achieve this potential difference/electric field, since some fraction of it is being screened by the dielectric. $\endgroup$ Mar 26, 2020 at 5:02

3 Answers 3


Yes. The voltage does depend on ε. Your argument is also correct. But you're missing the point of adding a dielectric. It reduces the total voltage of the system and helps to increase the capacitance. The electric field without a dielectric will be given by E = σ/ε, where σ denotes the surface charge density of the plates. When you introduce a dielectric between the plates (I assume here that you intend to fill the entire space between the two plates with the dielectric), the dielectric gets polarised as well, and thus produces an opposing field. In this case, the net electric field between the plates is given by E = (σ-σp)/ε, where σp stands for the surface charge density of the polarised dielectric. Here's a link that explains about polarization of the dielectric: (http://physics.bu.edu/~duffy/semester2/c08_dielectric.html)

σp can be given by σp = σ(1 - 1/k), where σ is the surface charge density of the plates, and k is the dielectric constant of the dielectric. Hope that helps.

  • $\begingroup$ Hmm okay. But what if the plates are hold at constant potential difference. What then happens when I put in the dielectric. Will the charge on the two plates change? $\endgroup$
    – user13514
    Mar 18, 2017 at 13:04
  • $\begingroup$ @user13514 Yes. It will. Consider a parallel plate capacitor of capacitance 'C' connected across a battery of potential 'V'. C is given by Q/V, where Q is the charge on a plate of the capacitor. C is also given by the formula, C=Aε/d (For a parallel plate capacitor only), where A is the area of the plates and d is the distance between them. Hence, Q can be written as Q=AVε/d. If you insert a dielectric, the capacitance changes. But the potential is constant. Hence, by the derived relation, we can conclude that the charge must change. $\endgroup$
    – e27sam
    Mar 18, 2017 at 13:18
  • $\begingroup$ @e27sam How can the charge change like that? The charge can only change by moving electrons between the plates. By changing the dielectric you change the voltage on the plates, not the charge. The charge is constant, not the potential. $\endgroup$ Jan 7, 2020 at 8:40

Premise is correct. The electric field would be the same. But lets see how this works out through an example.

Let's take two parallel plates, separated by vacuum, and connect them to a battery of potential V. The electric field between the plates will be V/d. Let's now introduce a material that has dielectric constant k.

As soon as you insert a dielectric slab between the plates, the slab will get polarised and oppose the external electric field. As a result, the net electric field inside the slab will become E/k. This will cause the potential difference across the plates to become V/k.

The plates and the battery are acting like two oppositely connected cells. Since the potential drop across the plates (V/k) has become less than that of the battery (V), current will flow until the potential difference between the plates becomes V.

As a result, Q(k-1) charge will be drained out of the battery, so that the final potential difference between the plates becomes V again.

In effect, the net electric field and V across the dielectric is the same as that across the vacuum, but the charge stored in the capacitor has become kQ now. So, the capacitance (Q/V) has increased with the introduction of a dielectric.


In general, a dielectric will decrease the $\vec E_{in}$-field inside the plates as this dielectric will become polarized.

You are correct in pointing out that $V=E_{in}\times d$. Here, $\vec E_{in}$ is the net electric field, which is the sum of the external electric field and the polarization, so that the magnitudes are related by $E_{in}=E_{ext}-E_{pol}$.

With a dielectric this net $\vec E_{in}$ would be smaller than without a dielectric, so $V$ would be smaller for a given separation.

If you insist on keeping $V$ fixed, the external electric field $\vec E_{ext}$ can be made larger than if there was no dielectric. This external electric field $\vec E_{ext}$ is determined by the charges present on the plate, i.e. for the same voltage difference you can accumulate greater charge on the cap.

Since $Q=CV$ or $C=Q/V$, if you can increase the charge but still keep $V$ fixed, you have increased the capacitance.

  • $\begingroup$ How would one find the $E$_ pol $\endgroup$
    – Kashmiri
    Oct 5, 2020 at 14:29

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