Capacitor, work done

1. Feb 9, 2010

fluidistic

1. The problem statement, all variables and given/known data
Say I have a parallel plates capacitor. The length of the plates are worth l. Both plates are separated by a distance d. Say there's a dielectric material whose permissivity is $$\varepsilon _1$$ such that it's between the plates from 0 to x, if 0 is the left side of the capacitor, that is if you draw a sketch such that the 2 plates are horizontal.
I want to know if I have to do some work in order to displace the dielectric material on a distance x'. I'd like to know the answer if the dielectric material is semi infinite (i.e. if I pull it, I will gradually fill the capacitor with this material) and if the material is finite (i.e. if I pull it, it will only cover a distance x at all times, although it moves).

2. Relevant equations $$Q=CV$$, $$U=\frac{CV^2}{2}$$.

3. The attempt at a solution
I've done some attempt on my draft. However I don't know why I struggle with the E field. I will detail a bit more. I've said that the problem is equivalent to sum up 2 capacitances. One of a capacitor $$C_1$$ filled with the dielectric material, such that the length of the capacitor is x. The other capacitor $$C_2$$ such that it is not filled with anything (i.e. vacuum) and whose length is l-x. I said both capacitors are considered in parallel, so I have that $$C=C_1+C_2$$.
So my task is to calculate $$C_1$$ and $$C_2$$.
For $$C_1$$, $$V=\frac{Q}{C_1}$$.
$$V=-\int _0 ^d \vec E \cdot d\vec l$$. I know that $$\vec E$$ and $$\vec l$$ are parallel so the integral simplifies a lot, but I'm stuck at what value to plug for E. Of course I have to consider E as constant, but what value do I choose?
I've been stuck on a similar question last time... I'd like to continue alone, but I need someone to help me.

By the way, I'm planning on calculating the energy stored in the capacitor before and after having displaced the dielectric material. The difference of these energy would represent the work I have done by displacing the material. Am I right thinking like this?

Also, as $$U=\frac{CV^2}{2}$$, if the dielectric material is semi infinite then C will increase, but I'm unsure about Q and V, so I can't say that the energy stored in the capacitor increases and thus $$\Delta U >0$$ so to make this increment of energy, I had to do some work.
I also believe that if there's an emf connected to the capacitor, the problem changes completely because I've heard it, but I don't realize it. If someone could enlighten me, I'll be extremely glad.

2. Feb 10, 2010

ehild

Is the charge constant on the capacitor or is it connected to an ideal voltage source?

As for the field inside the capacitor: the potential is constant on the plates as they are from metal. The separation between the plates is the same everywhere. So, what do you say about the electric field?

Yes, you can calculate the energy of the capacitor with or without dielectric. If the change of energy is positive, something or somebody has to provide for it.

The energy of the capacitor is U=CV2/2 =Q2/(2V)

ehild

3. Feb 10, 2010

fluidistic

I do not know. What are the differences? The difference of potential of the plates will change if there is a an emf? But if the charge is constant, as the capacitance change, so must change the difference of potential anyway... I'm confused about this.
Ok, what I can say about the E field is E=V/d. Is this right?
Thanks for the rest and for all!

4. Feb 10, 2010

ehild

If the capacitor is connected to a voltage source, the voltage is given and it does not change by moving the dielectric. Calculate the resultant capacitance, and use U=CV2/2 to get the energy.

If the capacitor is isolated but has a charge Q on it, the voltage will change when the dielectric slab is moved. Determine the energy with the formula U=Q2/(2C)

ehild

5. Feb 10, 2010

fluidistic

Ok. The capacitance will increase so the energy increase. Hence I must do some work to move the dielectric.

Ok I will do it.

6. Feb 10, 2010

fluidistic

I'm having problem calculating the capacitance of such a capacitor (with or without the dielectric).
I'm precisely stuck when it comes to calculate the E field. I assume it is constant, but I'm not sure about its value.

7. Feb 10, 2010

ehild

It is not that simple, as you have the source. Work must be done, but the source does work as it moves charges to maintain the voltage across the capacitor plates.

ehild

8. Feb 10, 2010

ehild

The voltage is constant along the capacitor plates. So E=V/d. You know that the number of field lines emerging from a charge q is q/epsilon. So E=surface charge density/epsilon. If A1 is the area of the capacitor with the dielectric and A2 is the same without the dielectric,

$$Q1=\epsilon*A1*E$$
$$Q2=\epsilon_0*A2*E$$
Q1+Q2 = Q (constant).

You can find E from here. But you do not need it. Q is given, the resultant capacitance is determined, U = 1/2 * Q2/C

ehild

9. Feb 10, 2010

fluidistic

Ok, until here I follow you.

I didn't know that. Thanks for the information. I always thought the field lines were imaginary. The more you have, the more intense is the field... I didn't know they actually existed and could be counted.
I'm a bit confused about this formula. I don't know where it comes from. (Is it the E field of an infinite uniformly charged sheet?)

Do you mean that the E field is the same, regardless if there's the dielectric material?
I thank you very much for all your time sir, and I'm sorry for being so confused.

10. Feb 11, 2010

ehild

The field lines are not real, but they are defined in that way. I explained about it to you in a previous post. The charge of a metal plate is entirely on its surfaces. In case of two parallel metal plates with opposite and equal charge, the charge accumulates on the inner surfaces. As E is uniform, the surface charge density is proportional to the dielectric constant.

The E field of a capacitor is normal to the plates and uniform if the size of the plates is very large compared to the distance between them.

The E field is V/d and uniform. But V depends on the dielectric if the charge is given.

ehild

11. Feb 11, 2010

fluidistic

Yes, thank you very much.