Electrostatics problem: Find charges on the surface of the dielectric

[Sunil Simha]

Homework Statement

An arbitrarily-shaped object is given a charge q and is then surrounded by a dielectric of permittivity ε as shown in the figure below

Find the charges induced in the inner and outer surface of the dielectric.

Homework Equations

I'm not sure but Gauss's law may be relevant here somehow.

The Attempt at a Solution

I initially thought that electric flux inside the dielectric would be the same as that outside the dielectric. So assuming $-q'$ to be the charge on the inner surface, I got $\frac{q-q'}{ε}=\frac{q}{ε_0}$. But upon reflection, I realized that this is wrong because

$q-q'<q$

$\frac{1}{ε}<\frac{1}{ε_0}$
and therefore $\frac{q-q'}{ε}<\frac{q}{ε_0}$.

Could you please help me by telling me how to approach the problem or how to make a right start.

Thanks in advance.
Sunil

 

[FireStorm000]

Not sure, but it may be a trick question, as charge can't actually flow through a perfect insulator (insulator and dielectric tended to be interchangeable when I took this).

If not, I think this may be relevant:
http://en.wikipedia.org/wiki/Permittivity#Explanation
According to the equation I see there, "how much" the dielectric responds to the E field is proportional to relative permitivity of the dielectric.

 

[DrZoidberg]

It's easier to explain with a parallel plate capacitor. If you place a dielectric in between the two plates the field inside will be weakened. If the dielectric has a relative permittivity $ε_r$ of 2 that means the charge induced on the surface of that dielectric is 1/2 of that of the plates. That charge than produces a field of it's own that partially cancels out the field of the plates. If $ε_r = 3$ the induced charge will be 2/3 of q meaning the field inside is reduced to 1/3. So in general the charge induced on the dielectric is $q*(ε_r - 1)/ε_r$

 

[Sunil Simha]

It's easier to explain with a parallel plate capacitor. If you place a dielectric in between the two plates the field inside will be weakened. If the dielectric has a relative permittivity $ε_r$ of 2 that means the charge induced on the surface of that dielectric is 1/2 of that of the plates. That charge than produces a field of it's own that partially cancels out the field of the plates. If $ε_r = 3$ the induced charge will be 2/3 of q meaning the field inside is reduced to 1/3. So in general the charge induced on the dielectric is $q*(ε_r - 1)/ε_r$

Thanks a lot. I got it now.