Stress tensor for a parallel plate capacitor

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Homework Statement:
Given infinitely large capacitors both with charge density sigma and minus sigma use the stress tensor to find the force on each.
Relevant Equations:
None needed
The question is partially taken from Griffith's book. I am confused about the physical meaning of momentum in fields. I have determined the solution and found that in part d the momentum crossing the x-y plane is some value in the positive z direction. I don't however understand the physical interpretation of that. Does that mean that field transfer momentum upwards towards the positive plate? Does the momentum get absorbed by the capacitor? But then wouldn't it be going upwards? That of course makes no sense because the plates are oppositely charged and therefore they should attract.

Capture.PNG
 
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  • #2
TSny
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I have determined the solution and found that in part d the momentum crossing the x-y plane is some value in the positive z direction. I don't however understand the physical interpretation of that. Does that mean that field transfer momentum upwards towards the positive plate?

It can be a nightmare keeping track of signs and directions. Griffiths explains that the quantity ##\left( - T_{ij}\right)## can be interpreted as “the momentum in the ##i## direction crossing a surface oriented in the ##j## direction, per unit area, per unit time".

(a) Consider ##\left(-T_{zz} \right)##. How would you interpret what this quantity means when evaluated at a point on the x-y plane?

(b) Can you express ##\left(-T_{zz} \right)## in terms of the electric field strength between the plates?

(c) Is ##\left(-T_{zz} \right)## positive or negative for points on the x-y plane?

(d) In which direction is z-component of momentum flowing across the x-y plane?

(e) Is the region above the x-y plane gaining or losing z-component of momentum? What about the region below the x-y plane?

Don't feel like you need to answer all of these questions at one time. We can take them one at a time.
 
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Homework Statement:: Given infinitely large capacitors both with charge density sigma and minus sigma use the stress tensor to find the force on each.
Homework Equations:: None needed

Does that mean that field transfer momentum upwards towards the positive plate? Does the momentum get absorbed by the capacitor? But then wouldn't it be going upwards? That of course makes no sense because the plates are oppositely charged and therefore they should attract.

Just a small addition to good advise of TSny
Say the plates are free they are coming closer by getting momentum from electric field. Usually say the plates are fixed mechanically, e.g. by hard dielectric material between the plates. Electric field pour momentum to the plates but it is cancelled by opposite momentum flow generated by the material's pushing on the plates.
 
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It can be a nightmare keeping track of signs and directions. Griffiths explains that the quantity ##\left( - T_{ij}\right)## can be interpreted as “the momentum in the ##i## direction crossing a surface oriented in the ##j## direction, per unit area, per unit time".

(a) Consider ##\left(-T_{zz} \right)##. How would you interpret what this quantity means when evaluated at a point on the x-y plane?

(b) Can you express ##\left(-T_{zz} \right)## in terms of the electric field strength between the plates?

(c) Is ##\left(-T_{zz} \right)## positive or negative for points on the x-y plane?

(d) In which direction is z-component of momentum flowing across the x-y plane?

(e) Is the region above the x-y plane gaining or losing z-component of momentum? What about the region below the x-y plane?

Don't feel like you need to answer all of these questions at one time. We can take them one at a time.

(A) That's the electromagnetic momentum flowing through that point in the xy plane
(B) You could use the relationship between force and electric field i.e. force per area is the electric field times charge density
(C) It is positive
(D) In the same direction as (C)
(E) The region above xy plane is gaining momentum and the region below xy plane is losing momentum
 
  • #5
TSny
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Several of your answers are incorrect. When dealing with the stress tensor, it's very important to think precisely. Let's take one question at a time. Obtaining a precise answer to the first question will really help with most of the other questions.
(a) Consider ##\left(-T_{zz} \right)##. How would you interpret what this quantity means when evaluated at a point on the x-y plane?
That's the electromagnetic momentum flowing through that point in the xy plane
##\left(-T_{zz}\right)## represents the flux of which component of electromagnetic momentum? Also, in which direction is this component of momentum flowing if ##\left(-T_{zz}\right)## is a positive quantity?
 
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Several of your answers are incorrect. When dealing with the stress tensor, it's very important to think precisely. Let's take one question at a time. Obtaining a precise answer to the first question will really help with most of the other questions.

##\left(-T_{zz}\right)## represents the flux of which component of electromagnetic momentum? Also, in which direction is this component of momentum flowing if ##\left(-T_{zz}\right)## is a positive quantity?

That would be the z component of the electromagnetic momentum to be more precise. I think if the quantity ##\left(-T_{zz}\right)## is positive the direction would be downward.
 
  • #7
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It can be a nightmare keeping track of signs and directions. Griffiths explains that the quantity ##\left( - T_{ij}\right)## can be interpreted as “the momentum in the ##i## direction crossing a surface oriented in the ##j## direction, per unit area, per unit time".

(a) Consider ##\left(-T_{zz} \right)##. How would you interpret what this quantity means when evaluated at a point on the x-y plane?

(b) Can you express ##\left(-T_{zz} \right)## in terms of the electric field strength between the plates?

(c) Is ##\left(-T_{zz} \right)## positive or negative for points on the x-y plane?

(d) In which direction is z-component of momentum flowing across the x-y plane?

(e) Is the region above the x-y plane gaining or losing z-component of momentum? What about the region below the x-y plane?


PS: I apologise for taking my time with the response. Just caught up in work.
Don't feel like you need to answer all of these questions at one time. We can take them one at a time.


So I read the chapter more carefully and I'll try these again
(a) The quantity ##\left(-T_{zz} \right)##: given a surface (in this case x-y plane) oriented in z direction, the z component of momentum flowing through the region

(b) -##\sigma^2/2\epsilon##

(c) The quantity ##\left(-T_{zz} \right)## is negative

(d) Is it in positive z direction?

(e) Since ##\left(-T_{zz} \right)## is negative that means that the amount of momentum flowing out of the region is negative. So it is gaining momentum?
 
  • #8
TSny
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That would be the z component of the electromagnetic momentum to be more precise.
Yes.

I think if the quantity ##\left(-T_{zz}\right)## is positive the direction would be downward.
Griffiths says that ##\left( - T_{ij}\right)## is “the momentum in the ##i## direction crossing a surface oriented in the ##j## direction, per unit area, per unit time".

This means that if ##\left( - T_{ij}\right)## is positive, then there is ##i## - component of momentum flowing in the ##j## direction.

So, if ##\left( - T_{zz}\right)## is positive at a point on the xy-plane, then there is z-component of momentum flowing in the positive z-direction across the plane at that point.

If ##\left( - T_{zz}\right)## is negative, then there is z-component of momentum flowing in the negative z-direction across the plane at that point.

In this problem, what is the sign of ##\left( - T_{zz}\right)## for a point on the xy plane?
 
  • #9
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In this problem, what is the sign of ##\left( - T_{zz}\right)## for a point on the xy plane?

Negative
 
  • #10
TSny
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Negative
Yes. So, z-component of momentum is flowing across the xy plane in the downward direction.

1576015340162.png

Let region I enclose the upper plate and region II enclose the lower plate. Which region is gaining z-component of momentum and which region is losing z-component of momentum?
 
  • #11
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Yes. So, z-component of momentum is flowing across the xy plane in the downward direction.

View attachment 253937
Let region I enclose the upper plate and region II enclose the lower plate. Which region is gaining z-component of momentum and which region is losing z-component of momentum?

Region II is gaining z-component of momentum and region I is losing z-component of momentum
 
  • #12
TSny
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Region II is gaining z-component of momentum and region I is losing z-component of momentum
Yes.

Suppose the two capacitor plates are released from rest with no external forces acting on the system. The only interaction between the plates is the electromagnetic interaction due to the charges they carry. As time progresses, region II increases its z-component of momentum. This means that the lower capacitor plate gains z-component of momentum. Thus, you can deduce which way the lower plate will start to move. Apply the same reasoning to deduce the direction that the upper plate moves.
 
  • #13
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Yes.

Suppose the two capacitor plates are released from rest with no external forces acting on the system. The only interaction between the plates is the electromagnetic interaction due to the charges they carry. As time progresses, region II increases its z-component of momentum. This means that the lower capacitor plate gains z-component of momentum. Thus, you can deduce which way the lower plate will start to move. Apply the same reasoning to deduce the direction that the upper plate moves.

Okay that is starting to make more sense now. Just need to clarify a few small things. There is no magnetic field in the question. Therefore the Poynting vector is zero everywhere. So the volume integral of our Poynting vector will be zero. Therefore the electromagnetic momentum at any point in space should be 0. How then can there be a flow of electromagnetic momentum if there is 0 electromagnetic momentum between the two plates.
 
  • #14
TSny
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Okay that is starting to make more sense now. Just need to clarify a few small things. There is no magnetic field in the question. Therefore the Poynting vector is zero everywhere. So the volume integral of our Poynting vector will be zero.
Yes

Therefore the electromagnetic momentum at any point in space should be 0. How then can there be a flow of electromagnetic momentum if there is 0 electromagnetic momentum between the two plates.
The Poynting vector is associated with the momentum stored in the electromagnetic field. Griffiths denotes this field momentum as ##\mathbf p_{\rm em} ##. If you take any volume ##V## of space, the field momentum in this volume is $$\mathbf p_{\rm em} = \mu_0 \epsilon_0 \int_V\mathbf S d\tau$$ where ##\mathbf S## is the Poynting vector. If there are material particles within ##V##, then there can also be momentum due to the motion of the particles. Griffiths refers to this momentum as "mechanical" momentum ##\mathbf p_{\rm mech}##.

The key theorem relating these two types of momenta and the stress tensor ##\mathbf T## is $$\frac{d}{dt} \left(\mathbf p_{\rm mech} + \mathbf p_{\rm em} \right) = \oint_S \mathbf T\cdot d\mathbf a$$ This says that you can get the rate of change of the total momentum inside ##V## by integrating the stress tensor over the surface S that bounds the volume. So, the stress tensor represents a "flow" or "transport" of total momentum across the surface S.

As you noted, for the capacitor ##\mathbf p_{\rm em} = 0##. So, we just have $$\frac{d \mathbf p_{\rm mech}}{dt} = \oint_S \mathbf T\cdot d\mathbf a$$ If the surface S encloses one of the plates, then ##\mathbf p_{\rm mech} = m \mathbf v## where ##m## is the mass of the plate and ##\mathbf v## is the velocity of the plate.
 
  • #15
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Yes

The Poynting vector is associated with the momentum stored in the electromagnetic field. Griffiths denotes this field momentum as ##\mathbf p_{\rm em} ##. If you take any volume ##V## of space, the field momentum in this volume is $$\mathbf p_{\rm em} = \mu_0 \epsilon_0 \int_V\mathbf S d\tau$$ where ##\mathbf S## is the Poynting vector. If there are material particles within ##V##, then there can also be momentum due to the motion of the particles. Griffiths refers to this momentum as "mechanical" momentum ##\mathbf p_{\rm mech}##.

The key theorem relating these two types of momenta and the stress tensor ##\mathbf T## is $$\frac{d}{dt} \left(\mathbf p_{\rm mech} + \mathbf p_{\rm em} \right) = \oint_S \mathbf T\cdot d\mathbf a$$ This says that you can get the rate of change of the total momentum inside ##V## by integrating the stress tensor over the surface S that bounds the volume. So, the stress tensor represents a "flow" or "transport" of total momentum across the surface S.

As you noted, for the capacitor ##\mathbf p_{\rm em} = 0##. So, we just have $$\frac{d \mathbf p_{\rm mech}}{dt} = \oint_S \mathbf T\cdot d\mathbf a$$ If the surface S encloses one of the plates, then ##\mathbf p_{\rm mech} = m \mathbf v## where ##m## is the mass of the plate and ##\mathbf v## is the velocity of the plate.

Ah okay, so all the momentum will go straight into the plates. I think I understand now. Thanks a million.
 

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