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.

 

[Nickie]

Hello Sunil,

Your initial thought about the electric flux being the same inside and outside the dielectric is correct. However, your equation needs some adjustments.

First, let's define some variables:

- q is the charge on the arbitrarily-shaped object
- q' is the induced charge on the inner surface of the dielectric
- q'' is the induced charge on the outer surface of the dielectric
- ε_0 is the permittivity of free space
- ε is the permittivity of the dielectric

Now, we can use Gauss's law to relate the electric flux to the charge enclosed by a surface. In this case, we can draw a Gaussian surface that encloses the entire dielectric and the charged object. Since the electric flux is the same inside and outside the dielectric, we can use the same equation for both regions:

Φ = q_enclosed / ε

Where Φ is the electric flux, q_enclosed is the charge enclosed by the Gaussian surface, and ε is the permittivity.

For the region inside the dielectric, the charge enclosed is q + q', since the induced charge q' adds to the original charge q. Therefore, the electric flux inside the dielectric is:

Φ_inside = (q + q') / ε

For the region outside the dielectric, the charge enclosed is just q, since the induced charges do not contribute to the total charge outside the dielectric. Therefore, the electric flux outside the dielectric is:

Φ_outside = q / ε_0

Since the electric flux is the same inside and outside the dielectric, we can set these two equations equal to each other:

(q + q') / ε = q / ε_0

Solving for q', we get:

q' = q * (ε_0 - ε) / ε_0

This gives us the induced charge on the inner surface of the dielectric. To find the induced charge on the outer surface, we can use the fact that the total charge on the dielectric must be zero (since it is a neutral object). Therefore, we have:

q + q' + q'' = 0

Substituting in the value of q' that we just found, we get:

q + q * (ε_0 - ε) / ε_0 + q'' = 0

Solving for q'', we get:

q'' = -q * (