Why Does a Dielectric Slab Experience Force When Fully Inserted in a Capacitor?

In summary, the conversation discusses the analysis of a parallel plate capacitor with a dielectric material inserted between the plates. The force on the dielectric is calculated for two cases - constant charge and constant voltage - and is found to have an x-dependence. However, there are some discrepancies in understanding the physical implications of the results, particularly with the force being non-zero when the dielectric is fully inserted and the variation of force with x in the two cases. Further discussion is had on the validity of the capacitance expression for different values of x, as well as the dependence of the force on whether Q or V is held constant.
  • #1
albega
75
0
...partially inserted between the plates of a parallel plate capacitor.

I have a few questions about this - it is section 4.4.4 of Griffiths, 3rd edition.

Imagine a parallel plate capacitor, of plate area l*d, separation d, filled with a linear dielectric of susceptibility X. Now imagine sliding the dielectric out of the plates in the direction that the plates have length l by a distance x. The idea is to find the force on the dielectric.

You can do this for two cases - constant charge and constant voltage. Both give the force as
F=-ε0XwV2/2d along the x direction where V is the p.d between the plates. I understand both of these derivations.

However understanding the results physically is proving to be an issue.

First of all if we let x=0 so the dielectric slab is perfectly between the plates, why is the force non-zero - apparently there is still a force in the negative x direction which does not make sense at all, from the symmetry of the scenario for starters.

Secondly, at constant charge, F=-ε0XwV2/2d but V is not a constant and has x dependence - this makes me happy as I would expect the force to vary with x. However at constant voltage, F=-ε0XwV2/2d but now V is a constant, and so is this telling me the force is the same on the slab wherever it is in the universe (provided it remains in the correct 'slot' between the plates). This just wouldn't make sense surely because as x gets very large, the fringe field causing the force should be vanishingly small.

If anyone can answer these I would be grateful! I have a few more issues about it but not understanding the above probably means I'm not in a position to worry about them just yet.
 
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  • #2
The analysis is based on Griffith's equation (4.62) for the capacitance when the dielectric is displaced ##x## from being fully inserted: $$C = \frac{\varepsilon_0 a}{d}(\varepsilon_r l - \chi_e x)$$
If you think about how this expression is derived, for what range of ##x## does this expression hold?

In particular, is it valid for ##x > l ## ?

Is it valid for ##x << l \,##?

Consider the assumptions made about a parallel plate capacitor when deriving ##C = \frac{\varepsilon A}{d}##.

EDIT: Watch a qualitative demo video here .
 
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  • #3
TSny said:
The analysis is based on Griffith's equation (4.62) for the capacitance when the dielectric is displaced ##x## from being fully inserted: $$C = \frac{\varepsilon_0 a}{d}(\varepsilon_r l - \chi_e x)$$
If you think about how this expression is derived, for what range of ##x## does this expression hold?

In particular, is it valid for ##x > l ## ?

Is it valid for ##x << l \,##?

Consider the assumptions made about a parallel plate capacitor when deriving ##C = \frac{\varepsilon A}{d}##.

EDIT: Watch a qualitative demo video here .

I can see it's not valid for x>l, as the capacitance should then be C=ε0wl/d. So if the capacitance is constant for x>l, does this mean the force on the slab is zero when the slab is out of the capacitor 'slot'? I think the video would sort of agree with that.

For x<<l, I'm not too sure - I'm guessing it's because we require that d is much less than the two dimensions of the capacitor (why exactly do we need this though?), and if x<<l, this isn't the case for the part of the capacitor in the vacuum, invalidating the capacitance expression...

Onto another problem - Griffiths states 'the force on the dielectric cannot possibly depend on whether you plan to hold Q constant or V constant - it is determined entirely by the distribution of charge, free and bound'.

However although the force expressions are the same in each case, F=-ε0XwV2/2d if you consider them in their separate contexts, they appear to be different. For constant voltage, if you pick some voltage V, F is the same for all x. In the constant charge case, if you pick some V, you get the same F as in the constant voltage case but because V is not being held constant, you can only get this value of V at a certain value of x such that V=Q/C (as C depends on x). So then the variation of force with x appears to be different, despite the expressions being the same - so doesn't that contradict the above statement, or am I misunderstanding it?
 
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  • #4
albega said:
I can see it's not valid for x>l, as the capacitance should then be C=ε0wl/d. So if the capacitance is constant for x>l, does this mean the force on the slab is zero when the slab is out of the capacitor 'slot'? I think the video would sort of agree with that.

There will still be some force on the dielectric when it it outside the capacitor. But the force decreases rapidly with distance once its outside. There is a "bowing out" or "fringing" of the field of the plates at the edges of the plates. This field that exists outside of the region between the plates would create some force on the dielectric when it is outside.

For x<<l, I'm not too sure - I'm guessing it's because we require that d is much less than the two dimensions of the capacitor (why exactly do we need this though?), and if x<<l, this isn't the case for the part of the capacitor in the vacuum, invalidating the capacitance expression...

Yes, that's right.

Onto another problem - Griffiths states 'the force on the dielectric cannot possibly depend on whether you plan to hold Q constant or V constant - it is determined entirely by the distribution of charge, free and bound'.

However although the force expressions are the same in each case, F=-ε0XwV2/2d if you consider them in their separate contexts, they appear to be different. For constant voltage, if you pick some voltage V, F is the same for all x. In the constant charge case, if you pick some V, you get the same F as in the constant voltage case but because V is not being held constant, you can only get this value of V at a certain value of x such that V=Q/C (as C depends on x). So then the variation of force with x appears to be different, despite the expressions being the same - so doesn't that contradict the above statement, or am I misunderstanding it?

I agree with what you are saying. How the force varies with x is different for keeping Q constant and keeping V constant. But, as you noted, in both cases you can express the force at any position with the same equation F=-ε0XwV2/2d. Griffiths is right when he says that the force at some position of the dielectric ultimately depends on just the distribution of free and bound charges at that position. For a particular x, the distribution of the charges will generally be different when keeping V constant as compared to keeping Q constant. So, the force at some x will not be the same for the two cases, in general. But in either case, the force can be expressed by the same equation.

I agree with you that Griffiths statement [ 'the force on the dielectric cannot possibly depend on whether you plan to hold Q constant or V constant - it is determined entirely by the distribution of charge, free and bound' ] could be interpreted as implying that the force should vary with x in the same way for the two cases. As you noted, that would be incorrect.
 
  • #5


I can provide some insights into the force on a dielectric slab partially inserted between the plates of a parallel plate capacitor.

Firstly, it is important to understand that the force on a dielectric slab in a parallel plate capacitor is due to the electric field present between the plates. When the dielectric slab is partially inserted, it affects the electric field and thus experiences a force.

Now, let's address your first question about the force being non-zero even when the dielectric slab is perfectly between the plates (x=0). This is because the electric field is not uniform between the plates. The electric field is stronger near the edges of the plates and weaker near the center. Therefore, even when the dielectric slab is perfectly between the plates, it will still experience a net force in the direction of the weaker electric field.

Next, let's look at the two cases - constant charge and constant voltage. In both cases, the force on the dielectric slab is given by F=-ε0XwV2/2d. This means that the force is proportional to the square of the voltage (V2). In the case of constant charge, as you correctly pointed out, the voltage will vary with x and thus the force will also vary with x. However, in the case of constant voltage, the voltage remains constant and thus the force on the dielectric slab will also remain constant regardless of its position between the plates. This may seem counterintuitive, but it is important to remember that the force is proportional to the square of the voltage, not the voltage itself. As you mentioned, the fringe field may decrease as x increases, but the force will still remain constant.

In conclusion, the force on a dielectric slab partially inserted between the plates of a parallel plate capacitor is caused by the electric field present between the plates. The force will be non-zero even when the slab is perfectly between the plates due to the non-uniform electric field. The force will vary with x in the case of constant charge, but will remain constant in the case of constant voltage. These concepts may seem counterintuitive, but they can be explained by understanding the relationship between force, electric field, and voltage.
 

Related to Why Does a Dielectric Slab Experience Force When Fully Inserted in a Capacitor?

1. What is the force exerted on a dielectric slab in an electric field?

The force on a dielectric slab in an electric field is equal to the product of the electric field strength and the surface charge density of the slab.

2. How is the force on a dielectric slab affected by the thickness of the slab?

The force on a dielectric slab is directly proportional to the thickness of the slab. As the thickness of the slab increases, the force also increases.

3. Can the force on a dielectric slab be negative?

Yes, the force on a dielectric slab can be negative. This means that the electric field and the surface charge density are acting in opposite directions, resulting in a net force in the opposite direction.

4. How does the permittivity of the dielectric material affect the force on a dielectric slab?

The force on a dielectric slab is inversely proportional to the permittivity of the dielectric material. This means that as the permittivity increases, the force decreases.

5. Can the force on a dielectric slab be calculated using Coulomb's law?

No, Coulomb's law cannot be used to calculate the force on a dielectric slab. Instead, the force must be calculated using the equation F = ε0εrEΔA, where ε0 is the permittivity of free space, εr is the relative permittivity of the dielectric material, E is the electric field strength, and ΔA is the surface area of the slab.

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