# Electric field inside a uniformly polarised sphere

If I use gauss law for finding the electric field inside a uniformly polarised sphere (no free charges) using gauss law,

$$\int D.da=Q_{f,enc}=0$$ $$D=\epsilon_0E_0+P=0$$ $$E=-\frac{P}{\epsilon_0}$$

But in griffiths, its given $$E=-\frac{P}{3\epsilon_0}$$

(eqn 4.14)

Why am I getting a wrong answer ?

• maybe the dielectric has a relative permeability of value 3 ? Mar 13, 2017 at 15:33
• No. Thats is not specified. The second answer can be obtained by using Laplace equation @diegobatt Mar 13, 2017 at 15:35

You assumed that $\int D.da=0$ implies $D=0$. This is only true if the polarization was radial. In griffiths example the polarization is along a particular cartesian axis and you do not have radial symmetry.

• why is that so? Mar 13, 2017 at 17:58
• You solved a different problem, a radially polarized sphere. The reason is that you assumed that the radial component of D is constant, otherwise you cannot take D outside the integral.
– user126422
Mar 13, 2017 at 18:25

The electric field in $D=\epsilon_0E_0+P$ is the net electric field. The field in $E=\frac{-P}{3\epsilon_0}$ is due to bound charges.

• What do you mean net electric field? Only the bound charges cause electric field right? Mar 13, 2017 at 17:58
• Okay, perhaps you've seen the example of a dielectric inserted in a capacitor? Well, there's a field due to the free charge on the capacitor plated, and then an induced one due to the bound charges in the dielectric. The sum of the field due to both free and bound charges is the net field and that is the field that the equation $D=\epsilon_0E_0+P$ is talking about. However this is only a secondary(albeit, important) error. The bigger error is the one Hugh Mungus has detailed. Mar 14, 2017 at 4:49

Gauss's Law is not a useful approach here.

Gauss's Law tells you the integrated value of the field component perpendicular to a surface. So you can only use this to solve for the field itself if you can use symmetry arguments to argue what components of the field are zero, and what the surfaces of constant field will look like. And as we will see in a moment, even this is not always enough.

You did not state the surface you integrated over, but it is important to remember: $$\int D \cdot da = 0 \ \ \not \Rightarrow \ \ D = 0$$

Here the polarization is stated to be constant. So let's just choose our coordinate system so that $\hat{z}$ is in the direction of $P$, which should be parallel to the direction of E or D here. This has cylindrical symmetry. So drawing a cylindrical surface inside of the polarized sphere gives us something like this: The sides will not contribute to the integral, and the top and bottom are opposed, which is why the integral is zero even though $D$ is not.

Unfortunately, this doesn't help us solve for $D$. So Gauss's Law is actually not the way to approach this.

Since you are reading from Griffith's, look at Example 4.2 and Example 4.3 for how to solve this. (err... assuming you have the same edition I guess)

One method is to calculate the bound charges from $P \cdot \hat{n}$, which here is $P \cos \theta$, and then calculate the electric field from that. This is a nice straight-forward approach.

The other example uses a slick superposition argument, along with reusing previous calculation of the overlap between two uniformly charged spheres, to almost directly argue what $E$ is, and from the dipole argue what $P$ is. If you have the book, this method is conceptually nice and I recommend a read through, but it may feel a bit hand-wavy compared to the "just calculate" method.