Direction of the force on a dielectric slab while being inserted into a capacitor

  • #1
Krushnaraj Pandya
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Homework Statement


A constant potential is maintained across a parallel plate capacitor. A dielectric slab is slowly inserted between the plates (width of slab=distance between the plates). What will the direction of force applied by capacitor on dielectric be?

Homework Equations


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The Attempt at a Solution


I went through the derivation and followed it logically for the magnitude of the force = 0.5*(epsilon)(L)(k-1)(V^2)/d where L is length of plate, k is relative permitivitty of dielectric, V is potential and d is distance between plates. I can't figure out what the direction of this force will be though...In case of a isolated capacitor, the force would have pulled the slab inside. I'd appreciate insight, thank you
 

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  • #2
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How did you get the magnitude without the direction?
You can use conservation of energy, for example.
 
  • #3
Delta2
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I haven't met a similar problem before, so I might be completely wrong here, but thinking in a kind of naïve way, the direction of the force should be same as the direction of the electric field. I mean what other direction could be, it would be a force due to the electric field between the capacitor plates.

Turns out my naive thinking must be completely wrong, probably the direction of the force is vertical to the direction of the electric field from the plates.

I think we got to think what happens to the capacitance C of the capacitor due to the insertion of dielectric. Then what happens to the energy stored in the capacitor (given that the potential remains (almost) constant), and finally where the extra energy or loss of energy comes from/goes to...
 
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  • #4
Krushnaraj Pandya
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I'll post the picture of the derivation of the magnitude as soon as I'm free (my midterms start tomorrow, so probably in the next 48 hours), Will look into it in more detail then. Thank you :D
 
  • #5
Krushnaraj Pandya
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Here's the derivation. Its not in the coursework (this is extra work I'm doing for fun, and a little more knowledge and so its above my understanding in many ways) Can someone just specify the direction of the force found?
 

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  • #6
Delta2
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Seems to me that the force is perpendicular to the (supposedly parallel) electric field between the plates. I reach to this conclusion because I see that work equation , the work of the force is taken to be ##Fdx##, so we have to assume that the force is parallel to dx.( there is no ##\cos\theta## or ##\sin\theta## term involved).

But I cant explain the exact inner workings of interaction between the electric field and the dielectric that make such direction of the force possible.
 
  • #7
Krushnaraj Pandya
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Seems to me that the force is perpendicular to the (supposedly parallel) electric field between the plates. I reach to this conclusion because I see that work equation , the work of the force is taken to be ##Fdx##, so we have to assume that the force is parallel to dx.( there is no ##\cos\theta## or ##\sin\theta## term involved).

But I cant explain the exact inner workings of interaction between the electric field and the dielectric that make such direction of the force possible.
Is the slab being pulled inside by the force or being pushed out of the capacitor because of it?
 
  • #8
TSny
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Can someone just specify the direction of the force found?
In the derivation that you posted, note that the expression for the work done on the dielectric by the externally applied force F is -Fdx. This implies that F is taken to be positive when it acts toward the right in the diagram. (This is a little tricky. You also have to note that dx is positive when the dielectric moves to the left.) Since the final answer for F comes out positive, the direction of the external force on the dielectric is to the right. This force balances the electrical force on the dielectric. So, the electrical force is toward the left. That is, the electrical force pulls the dielectric into the region between the plates.

The physical reason for the existence of the electrical force and why it has the direction that it does can be traced to the fringing of the field in the regions near the left and right edges of the capacitor. For example see http://te.fisica.edu.uy/fuerza_en_Capacitor.pdf (especially starting on page 4).
 
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  • #9
berkeman
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Is the slab being pulled inside by the force or being pushed out of the capacitor because of it?
As mfb has pointed out, you can figure that out via energy considerations, or else you could just draw a sketch of the electric dipoles in the dielectric slab...

Consider the case where the parallel plate capacitor has the top plate charged + and the bottom plate is equally charged -. Which direction does the E field point between the plates before the slab is inserted?

So start inserting the slab from the left. Which way will the electric dipoles be oriented in the slab due to the external charges? And what happens to the E field near the leading edge of the slab due to the higher dielectric constant of the slab compared to the air in the capacitor gap? And so you have a bent E-field interacting with those oriented dipoles -- which way does the E field point going into the - end of the dipoles? Does that help you see which way the force points? And can you then use energy considerations to validate this force direction? :smile:

EDIT -- @TSny beats me out again!
 
  • #10
Krushnaraj Pandya
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All the answers are very enlightening...however, I regret that I cannot dig too deep into the intricacies since a doctorate level probing here will lead to abandoning my high school syllabus. Just knowing something of this form exists and going through the derivation gives me a much needed advantage over my peers if I want to get into a good college. Once I get to a good research oriented grad school, I'll definitely revisit all my older threads and dive deep into all of them. Thank you everyone for your help :D
 
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  • #11
TSny
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