An insulating solid ball with radius $R$ has a uniform charge density $ρ$. Find the electric field vector inside and outside of the ball if it has dielectric constant $K$.

Normally, I'd treat this as a Gauss' Law problem with the field of a uniformly charged ball but I'm unsure as to whether this being an insulating sphere changes that. Also I don't fully understand the significance of the dielectric constant and how it affects this question. If someone would mind explaining this to me, I would be very grateful.

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    $\begingroup$ Are you talking about a sphere (surface) with uniform surface charge density or about a ball (solid) with uniform volume charge density? Where is located in space the material with dielectric constant K? Is it on the surface of the sphere? or the volume of the ball ? or the whole space? $\endgroup$ – Frobenius Apr 22 '16 at 12:44
  • $\begingroup$ That's a good question. Sorry to be unhelpful but the question I posted from doesn't specify either of those things. I think it is uniform volume charge density as other questions in this section dealt with that. As for the location of the dielectric I think it might be on the surface of the sphere but I have no justification for that. $\endgroup$ – BigBoss Apr 22 '16 at 13:05
  • $\begingroup$ It's a linear dielectric sphere carrying uniform ${\it free}$ charge density $\rho$. See my answer. $\endgroup$ – velut luna Apr 22 '16 at 13:07

If you find my other answer too long, this is the short answer.

Step 1: Just ignore the word "dielectric" and consider it as a uniformly charged sphere (that you may have already solved before). Use Gauss's law to find the E field inside the outside.

Step 2: For the field inside, replace $\epsilon_0$ by $K \epsilon_0$.

  • $\begingroup$ @user929304 According to my long answer below, it is still valid in your case because you still have $\nabla \times {\bf P}= {\bf 0}$. $\endgroup$ – velut luna Jan 24 '18 at 10:05
  • $\begingroup$ @user929304 Because your system still have spherical symmetry, from which one can conclude that ${\bf P}$ must be in the radial direction and can depend on $r$ only. Hence $\nabla \times {\bf P} = {\bf 0}$. $\endgroup$ – velut luna Jan 24 '18 at 10:44

Inside matter, in addition to free charges $\rho_f$ (which I suppose is what your $\rho$ means), there are bound charges

$$\rho_b=-\nabla \cdot {\bf P}$$

as well.

One way to solve the problem is to stick to the two equations

$$\nabla \cdot {\bf E} = \rho/\epsilon_0$$ $$\nabla \times {\bf E} = {\bf 0}$$

But then you have to pay attention here that $$\rho = \rho_f + \rho_b$$

This picture is physically clear.

The second method (not so clear physically but mathematically easier) is to define the displacement field

$${\bf D}=\epsilon_0 {\bf E} + {\bf P}$$


$$\nabla \cdot {\bf D}=\rho_f$$ $$\nabla \times {\bf D} = \nabla \times {\bf P}$$

Compared to the two equations of the E field, the first equation is simplified but the second equation becomes more complicated.

However, if somehow (e.g., by symmetry), you can argue that

$$\nabla \times {\bf P} = 0$$

then the second method is preferred.

This is true in your case, because you have linear dielectrics. So

$${\bf P} = \epsilon_0 (K-1){\bf E}$$

Hence, $$\nabla \times {\bf P} = {\bf 0}$$

So you have

$$\nabla \cdot {\bf D}=\rho_f$$ $$\nabla \times {\bf D}={\bf 0}$$

The above (together with symmetry) implies you can use Gauss's law to obtain

$${\bf D}=\frac{\rho_f {\bf r}}{3}$$ for $r \le R$,

and $${\bf D}=\frac{\rho_f R^3}{3 r^3}{\bf r}$$ for $r > R$.

Then inside the sphere, $${\bf E}=\frac{1}{K\epsilon_0}{\bf D}=\frac{\rho_f {\bf r}}{3K\epsilon_0}$$

Outside the sphere, $${\bf E}=\frac{1}{\epsilon_0}{\bf D}=\frac{\rho_f R^3}{3 \epsilon_0 r^3}{\bf r}$$


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