# How do we conclude that polarized dielectric in electric field reduces overall field?

If we have a dielectric inside an electric field, the dipoles in the dielectric will align in the following direction due to the torque on them by the external field. The conclusion then is that the overall field is reduced since positive charges appear on the bottom plate and negative on the top, opposite to the applied field.

But if we consider the field of the electric dipole itself, we know that it points in the same direction as the dipole moment.

The dipole moments clearly point in the same direction as the applied field. So shouldn't the field due to the polarized dielectric point in the same direction as the applied field, thus strengthening it?

• your dipoles are pointing in the opposite direction to the applied field. – Michael Sep 8 '14 at 13:34
• @Michael: No, the field is from the top to the bottom, and so is the dipole moment. The dipole moment vector points from negative to positive charge. – Gerard Sep 8 '14 at 13:53
• True, but semantics none the less. The Dipole Moment is not an electric field. Look at the resultant charge distribution from the dipole alignment: practically a line of -ve charge just under the +ve charge plate, and a line of +ve charge just above the -ve charge plate. The resultant Electric field points in the opposite direction to the applied field, along the Dipole Moment Vector direction. – Michael Sep 8 '14 at 14:13

You need to at least assume a convention for direction of electric fields. I am using the conventions that is used pretty much everywhere, in this convention the electric field points from positive to negative. I have marked the external field in blue and field of dipole in green, it is clearly evident that both the fields are opposing and hence cancel each other.

Therefore, the net field magnitude reduces.

Edit/Update (To explain the question raised in comment) :

I think you understand that the field reduces inside the dielectric medium. But pay attention, if you look at the dielectric material on an atomic scale will you call the dipoles the dielectric material or the spaces between them?

When we look at the dielectric material on atomic scales we need to start looking at the dipoles and not outside them because these dipoles are what make rhe dielectric material.

Moreover for a solid or liquid dielectic material these dipoles are far more than the spaces between them, your picture in this sense is a little misleading. The following picture is a better depiction of what is more likely to be found in dielectrics. • That is the field inside the dipole. But the field outside the dipole is in the same direction as the dipole moment. For example, the field at an axial point is given by: $$\frac{1}{2\pi\epsilon_0}\frac{\vec{p}}{z^3}$$ where $\vec{p}$ is the dipole moment, $z$ is the distance from the dipole. Why do we consider the field inside the dipole and not the one outside? – Gerard Sep 9 '14 at 3:17
• Good question! I will update my answer in a little while. – Rijul Gupta Sep 9 '14 at 7:41
• So, what you are saying is that the field outside the dipole has no empty space to manifest itself. All the space is dominated by the much stronger field inside the dipole. Am I right? – Gerard Sep 9 '14 at 16:21
• No, what I am saying is that reduction in field happens inside dielectric, dipoles are the dielectric not the spaces outside of them, hence we focus just on the field in dipoles and not the spaces outside. – Rijul Gupta Sep 10 '14 at 6:11

Take a piece of Polarised Material with Dipole Moment per Unit Volume $\mathbf{P}$. The associated electrostatic potential is

$$V(\mathbf{r})=\frac{1}{4\pi\epsilon_0}\int_V\frac{\mathbf{\hat{R}}.\mathbf{P}(\mathbf{r}')}{R^2}dV'$$

where $\mathbf{R}=\mathbf{r}-\mathbf{r}'$. Some clever manipulation, mainly noting that $\nabla'(1/R)=\mathbf{\hat{R}}/R^2$ lets us write this as a volume + surface integral

$$V= \frac{1}{4\pi\epsilon_0}\int_V\mathbf{P}(\mathbf{r}').\nabla'\left(\frac{1}{R}\right)dV'=\frac{1}{4\pi\epsilon_0}\left[ \int_V\nabla'.\left(\frac{\mathbf{P}(\mathbf{r}')}{R}\right)dV'-\int_V\frac{\nabla'.\mathbf{P}(\mathbf{r}')}{R}dV' \right]$$

$$V=\frac{1}{4\pi\epsilon_0}\oint_{\partial V}\frac{\mathbf{P}(\mathbf{r}').d\mathbf{S}'}{R}-\frac{1}{4\pi\epsilon_0} \int_V\frac{\nabla'.\mathbf{P}(\mathbf{r}')}{R}dV'$$ which familiarity with electrostatics will tell you is equivalent to the potential had the object had surface charge density $\mathbf{\hat{n}}.\mathbf{P}$ and volume charge density $-\nabla.\mathbf{P}$.

So, the field of a polarised object, dipole moment per unit volume $\mathbf{P}$, will have an induced Electric Field equal to that generated by considering the object to have surface charge $\mathbf{\hat{n}}.\mathbf{P}$ and volume charge $-\nabla.\mathbf{P}$.

Suppose $\mathbf{P}=-P\mathbf{\hat{z}}$ is constant (to roughly match your problem), then immediately $\nabla.\mathbf{P}=0$ so there is no contribution from a volume charge density. On the $+ve$ plate of the example, $\mathbf{\hat{n}}$ (pointing outwards) points vertically upwards i.e. $\mathbf{\hat{n}}=\mathbf{\hat{z}}$ and on the $-ve$ plate we have the opposite $\mathbf{\hat{n}}=-\mathbf{\hat{z}}$. At the $+ve$ plate surface we therefore induce $-P\mathbf{\hat{z}}.\mathbf{\hat{z}}=-P$ (negative charge density), and on the $-ve$ plate surface we induce $-P\mathbf{\hat{z}}.(-\mathbf{\hat{z}})=P$ i.e. positive charge density. The direction of the induced electric field therefore points from the $-ve$ plate to the $+ve$ plate, acting to reduce the applied field.

Due to the external electric field,the molecules (which may be polar or non-polar) becomes a dipole. As a result there is an existence of consecutive dipoles as shown in your figure.

Now these molecules are placed very closely. So in the material the charges can be neglected due to the coexistence of positive and negative charges very closely. But on the surfaces a net layer of charge is formed (this is how molecules with bond charges are polarised). In your figure the upper surface has a negative charge density and downward one has a positive charge density. So there is a creation of an internal field. (Thus what happens inside the dipole does not matter). So field magnitude is reduced.