Concentrical conducting spheres with charge distribution and dielectric

[Telemachus]

Hi. I have this problem, which I must solve. It says: two concentric conducting spherical shells, of radius a and b (a<b), are charged at Q and -Q respectively. The space between the spheres is filled at its half by an hemisphere of dielectric with dielectric constante ε.

a)Find the field for every point between the spheres.
b)Compute the charge distribution for the inner sphere.
c)Compute the surface density charge for the dielectric at r=a.

Well, I don't know how to start with this. It bothers me that the dielectric fills just one half of the space between the spheres, specially bothers me the interface between the dielectric and the "empty" space (the space is actually filled by the electric field), the problem is I don't know how to considere the induced charge on that interface, and I don't how ti find the entire electric field.

I was tempted to write that in the space within the dielectric the field is just:
$$E=\frac{Q}{4\pi \epsilon r^2}$$
And the "empty" space has a field:

$$E=\frac{Q}{4\pi \epsilon_0 r^2}$$
But then I wouldn't be having in mind the polarization on the dielectric.

 

[Telemachus]

I've been thinking that as the electric field is radial, there will be no induced surface charge at the interface between the dielectric and the empty space, so perhaps what I thought at first glance is not such a bad answer, I should have in mind the induced surface charges at the dielectric, but just for the interfaces between the dielectric and the conductors. What you think about this? by the way, I feel tempted to think about capacitance, because of the configuration, opposite charged conductors. Maybe I should use capacitance to solve this.

 

[ehild]

Thinking in terms of capacitors is useful.

You have conducting spheres. What is constant all over a single sphere?

ehild

 

[Telemachus]

The potential. I think I see more clearly now why is useful thinking in terms of capacitors. So I'll have two different capacitance depending on one of the angular coordinates, right? I'm not sure this approach will drive to the electric field, I usally work at the inverse, find the electric field, then the potential difference, then the capacitance, but this case is quiet complicated because of the induced charge in the dielectric.

Thanks for posting ehild.

 

[Telemachus]

I think I just could consider Gauss, defining the dielectric in the region
$$0<\theta<\frac{\pi}{2}$$
$$E=\frac{Q}{4\pi \epsilon r^2}$$
And empty space for
$$\frac{\pi}{2}<\theta<\pi$$
$$E=\frac{Q}{4\pi \epsilon_0 r^2}$$
Then that should work. I mean, the electric field must be radial, so the interface doesn't bother at all. Is that wrong?

 

[vela]

It doesn't work because you don't have spherical symmetry in this case because of the dielectric.

As you noted, the potential is constant on each sphere, so the potential difference between the two spheres is the same, regardless of whether the dielectric is there or not. So what does that tell you about the electric field in the two regions?

 

[Telemachus]

The electric field must be weaker in the dielectric than in the empty space. I know the electric field isn't symmetrical, but its radial, isn't it? I mean, the interface space-dielectric doesn't matter at all, and the only thing that changes is the dielectric constant. That's how I reasoned it. I'm not too sure, because of the induced charges, I don't know if that changes anything on the distribution of Q around the conductors.

How can I set this problem to find the electric field using capacitance? would you help me a bit with the set up?

Thanks for posting vela :)

 

[vela]

Actually, the electric field is the same. Remember that $$\Delta V = -\int \vec{E}\cdot d\vec{r}$$As you said earlier, the potential is constant on each sphere, so the LHS is constant if you integrate from a point on one sphere to a point on the other. If you integrate radially outward from the inner sphere to the outer sphere, the path is essentially identical regardless of whether you're going through the dielectric or not, so the electric field must be the same in the two cases. (Yes, a bit hand-wavey.)

As ehild was hinting at, you'll probably find this problem easiest to analyze if you consider the configuration as two capacitors in parallel.

 

[Telemachus]

What you mean with the LHS? sorry, missed that :P

But now I understand what you mean. I see why my solution is wrong. Thanks.