# Electric field at a point inside a capacitor

Let's say I have a parallel plate capacitor.

How would I find the electric field at a certain point INSIDE the capacitor (inside the dielectric let's say). From what I understand, the flux of the electric field will be constant everywhere (even if there is more than 1 different dielectric), but the electric field varies. Is this correct?

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## 3 Answers

(Im assuming effectively infinite capacitors ($\sqrt{A}>>d$), otherwise icky fringe fields come into play)

Calculating the net field inside an infinite dielectric sheet is simple. If the field due to the capacitor plates is E, the net field inside the dielectric will be $\epsilon_rE$. This means that the dielectric itself has an induced field of $E(1-\epsilon_r)$ in the opposite direction (this field only exists inside the dielectric). This also applies to a multiple dielectric system. Inside each dielectric sheet, there will be a similar induced field, and we get the same value of net electric field (where $\epsilon_r$ is the dielectric constant of the sheet in question)

So what's actually happening here? A dielectric contains charges that are pretty much bound in place. When you apply a field E, these charges excercise whatever little freedom they have to move, and they form tiny dipoles (I think the electron cloud of the atom/molecule polarizes, but i'm not sure). These tiny dipoles will be randomly scattered, but will all be aligned with the field. We can vector add their dipole moments to make a column of large dipoles. Since dipoles are after all pairs of charges, we can look at this as two sheets of opposite charge embedded on either face of the dielectric. These sheets create no field outside, but they create a net induced field inside (Just how a capacitor creates field only between its plates, as $E=\frac{q}{2A\epsilon_0}$ for a single plate; both Es cancel outside the capacitor)

Now, when talking about the flux, we have a bit of an issue. If we are considering a perfectly infinite capacitor, the flux is infinite, so we cant discuss it.

If we consider an effectively infinite capacitor, the flux should be constant (no net sources/sinks of of flux.). But, since the field is changing, we have a bit of confusion. This is due to boundary conditions. Actually, the field is diverging a bit in a lemon-like way when it reaches the dipole. Near the center, this divergence is small, but it gets large when go near the top/bottom of the capacitor. So, the electric field has reduced in value, but it has spread out as well, so net flux is constant. Draw a little diagram of the field lines, and you'll understand this clearly. Find some diagram of how field lines bend through a dielectric for reference if you want,

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If you have a single dielectric then you can simply preform Gauss's law at the plate. This will entail imagining a Gaussian pill-box. Say the plates have a charge $Q$ and $-Q$, then if we draw the pill-box around the plate with $Q$ and call the down direction positive $\hat{z}$ we have $$\frac{Q}{\epsilon}=\int \vec{E}\cdot d\vec{a} =EA\implies \vec{E}=\frac{Q}{\epsilon A}\hat{z}$$ where $\epsilon$ is the permittivity of the material. You can use the Electric Displacement field $\vec{D}$ to get the same answer quicker with the boundary conditions that $(\vec{D}_2 -\vec{D}_2)\cdot\vec{n}=\sigma_{\text{free}}$, then from that you can use the relation $\vec{D}=\epsilon \vec{E}$ to get $\vec{E}$ also. If there are multiple dielectrics and you want to find the field in the middle of the capacitor, say at the boundary between two dielectrics, you will have to use the boundary conditions on dielectric media $$(\vec{D}_2 -\vec{D}_2)\cdot\vec{n}=\sigma_{\text{free}}$$ $$(\vec{E}_2 - \vec{E}_1 )\times \vec{n}=0$$ along with the fact that the potential is a solution to Laplace's equation. This should be enough for the majority of problems one comes across.

I hope this helps,

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If the capacitor is of the flat plate variety with area A and separation l, the potential increases linearly from one plate to the other. From the potential you can get the electric field by taking the gradient. Note that the dielectric means that there's a difference between E and D but this should be discussed in your textbook.

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