Forces on Dielectrics: Understanding Griffith's Equations

In summary, the argument that the force due to uniform field doesn't work on the dielectric is unfounded as the energy contained in the uniform field gets reduced.
  • #1
Pushoam
962
51
Forces on dielectrics
I found difficulty in understanding this topic from Griffith.

upload_2017-8-13_8-3-23.png

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Here, while calculating F, in eqn. 4.61 , Griffith doesn't bother about energy stored in fringing field.
So, even if I assume that there is no fringing field, then following the above procedure, I can say that there is an attractive force acting on the dielectric and this force must be due to something other than fringing field.

upload_2017-8-13_8-16-22.png

Are the following conclusions correct?
1) The force due to fringing field pulls the dielectric as the force due to the uniform field is perpendicular to the dielectric surface.

2) The energy stored in the uniform field gets changed in this process. The energy stored in fringing field remains constant. This means that work is being done by the force due to uniform field. But the first conclusion says that the force due to uniform field doesn't work on the dielectric.

I need help here in understanding this point.

In eq. 4.7, force due to battery is constant (as both V and ## \frac { dC } { dx} ## are constant ) . So, the surface charge density changes. Right?
 
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  • #2
Interesting point. I believe the issue here is that energy arguments often hide the details of the forces and how the work is being done. There are many examples of this. Take for example how we calculate the speed of a block at the bottom of a frictionless ramp. With an energy argument we ignore everything the ramp does in redirecting the block into horizontal motion, but we still get the right speed.

I think you have the crux of the point when you say that the fringing field doesn't change (at least not until the outer edge of the dielectric gets significantly into the fringing field). The energy calculated without the fringing field is not the total energy, but since the fringe part doesn't change, using just the uniform part of the energy will give the correct change in energy between states, and so it's derivative shows the force. However just because the energy calculation ignored the fringe field doesn't mean that isn't where the force is being generated.

So I think your first statement is correct and the first two sentences of your second statement are correct. However, the conclusion that the uniform field is therefore applying the force seems unwarranted.

I will add that you clearly already see that there is no way around your first statement. The fringe field has to be doing the work.
 
  • #3
Cutter Ketch said:
Interesting point. I believe the issue here is that energy arguments often hide the details of the forces and how the work is being done. There are many examples of this. Take for example how we calculate the speed of a block at the bottom of a frictionless ramp. With an energy argument we ignore everything the ramp does in redirecting the block into horizontal motion, but we still get the right speed.

Thanks for the above insight as this is something I was looking for.
So, what I conclude from the above two posts is:
If the net work done by the non- conservative forces is 0, then ## \vec F = - ∇ U ## gives the net conservative force acting on the system.
It doesn't tell us about the particular forces acting on the system.
Here, on the dielectrics, the force which keeps the dielectric in the fixed vertical position does no work on it. The only force which works on the dielectric is force due to the electric field and hence it is conservative. Thus, ## \vec F = - ∇ U ## gives here the net conservative force acting on the system.
This doesn't tell us whether the force is due to uniform electric field or due to fringing electric field.
Here, Griffith use this argument that since the force due to uniform field is perpendicular to the dielectric surface, it can't do any work. So, ## \vec F ## must be force due to fringing field.
While calculating U, Griffith doesn't consider the work done due to fringing field as he says that the work done by fringing field remains constant (But, he doesn't prove it.). So, to me, this seems an assumption. So, won't it be better if I simply take work done due to fringing field negligible compared to work done due to uniform field ( and hence, work done due to fringing field is ignored in calculating U.)?

Now, the point which I can't make clear is:
It is the force due to fringing field which pulls the dielectric but it is the energy contained in the uniform field which gets reduced.
I think there is something wrong with the following phrase: "It is the force due to fringing field which pulls the dielectric" because if this happens then it means that there is a component of this force which is along the direction in which the dielectric is being pulled. Hence, there is a component of this force in the direction of displacement and hence using ## W = \int _{\vec s_i }^ { \vec s_f} \vec F. d \, \vec s ## I can say that the work done due to the force due to fringing field is non-zero ( as there is no other force opposing this component of force due to fringing field.).
Is this correct?
 

1. What are the fundamental principles behind Griffith's equations?

The fundamental principles behind Griffith's equations are based on the concept of energy minimization and conservation. These equations describe the forces acting on a dielectric material in an electric field and how the material responds to these forces.

2. How do Griffith's equations differ from other equations describing forces on dielectrics?

Griffith's equations take into account the polarization and deformation of the dielectric material, while other equations may only consider the displacement of charges. This makes Griffith's equations more comprehensive and accurate for describing the behavior of dielectric materials.

3. What is the significance of understanding Griffith's equations in the field of materials science?

Understanding Griffith's equations is crucial in the field of materials science as it allows us to predict and manipulate the behavior of dielectric materials in various electric fields. This knowledge is essential for developing new technologies and improving existing ones.

4. Can Griffith's equations be applied to all types of dielectric materials?

Griffith's equations can be applied to most types of dielectric materials, including solid, liquid, and gaseous materials. However, the specific parameters and properties may vary depending on the type of material.

5. How can Griffith's equations be used in practical applications?

Griffith's equations are used in various practical applications, such as designing and optimizing electronic devices, developing new materials for capacitors and insulators, and understanding the behavior of dielectric materials in electric fields for safety purposes.

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