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? $\endgroup$ – Lapmid Mar 18 '17 at 11:47

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 '17 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 '17 at 13:18

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.


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