Work to move two plates

In summary: Hope this helps!In summary, two parallel plates of area A separated by a distance d have positive and negative area charge densities, \sigma and -\sigma, respectively. To move the plates a distance 2d, the work required is W=\sigma A \frac{\sigma}{2\epsilon_o}d'. When calculating the force needed to move one plate, only the field due to the other plate is used (\frac{\sigma}{2\epsilon}). The energy of the capacitor is given by (1/2)CV^2 = Q^2/(2C) where Q is the charge on the positive plate. The net charge on the capacitor is always 0, but this does not mean the energy of the
  • #1
indigojoker
246
0
two parallel plates of area A is separated by a distance d. One plate has positive charge distribution with area chrage density [itex]\sigma[/itex] and the other plate has an area charge density [itex]-\sigma[/itex]

how much work does it take to move the plates 2d?

W=Fd=QEd'

i'm not sure what to use for the Q or the E. Should it be for both plates or for just one plate?

Assuming that it's just for one plate, we get: [itex] W = \sigma A \frac{\sigma}{2 \epsilon_o} d' [/itex]

and that would be it right?

then we just let d'=2d
 
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  • #2
Just use the energy stored in a capacitor... What is the initial energy stored in the capacitor? What is the final energy?

Final energy - initial energy = work done.
 
  • #3
right, so this gets at my question of whether to use E for one plate or two

we have U=1/2QV=(0.5)QEd

now i can use what i did on the previous post and use Q for one plate [itex]\sigma A[/itex]

or I could use the Q on both plates [itex]2\sigma A[/itex]

same goes for E
 
  • #4
You need to use Q on one plate... ie: Q = [tex]\sigma A[/tex]

The field E is [tex]\frac{\sigma}{\epsilon}[/tex]
 
  • #5
seems like you are using the charge of one plate and the E field due to both the plates.

you would get what I got: [itex] W = \sigma A \frac{\sigma}{2 \epsilon_o} d'[/itex]

so why did I have to use the E field of just one plate for what I did to work?
 
  • #6
indigojoker said:
seems like you are using the charge of one plate and the E field due to both the plates.

you would get what I got: [itex] W = \sigma A \frac{\sigma}{2 \epsilon_o} d'[/itex]

so why did I have to use the E field of just one plate for what I did to work?

you won't get that answer if you use the E field for just one plate.
 
  • #7
do the plates are a distance d apart.

to move them 2d apart, we move a distance d.

Using Q = [tex]\sigma A[/tex]

E = [tex]\frac{\sigma}{\epsilon}[/tex]

we get the work needed is:
[tex] W = \sigma A \frac{\sigma}{ \epsilon_o} d'[/tex]

so the reason why we are using Q of one plate is because we are saying that we want to move one plate a distance of d (instead of two plates a distance of d/2), and this plate is being moving though the E-field that is contributing from both plates, so we use [itex]\frac{\sigma}{\epsilon}[/itex]

So we actually move it d apart, which is why i used d instead of 2d

do i make sense?
 
  • #8
Ah... I think I'm understanding now... yes, when you use Work = Force * d, to get the force on one plate... you need to use the field due to just the other plate (because the field due to a plate does not create a force on itself...) in other words [tex]\frac{\sigma}{2\epsilon}[/tex] is the field needed to get the force on one plate. And the charge is Q (the force on one plate is the field due to the other plate times the charge of the first plate). so the force on one plate is [tex]Q*\frac{\sigma}{2\epsilon} = \sigma A\frac{\sigma}{2\epsilon}[/tex]

So this way you get that factor of (1/2) in there... just like when you use final energy - initial energy.

So work is force*d = [tex]\sigma A\frac{\sigma}{2\epsilon}d[/tex]
 
  • #9
okay, now I'm confused, why is [tex]\frac{\sigma}{2\epsilon}[/tex] used

I convinced myself that it was [tex]\frac{\sigma}{\epsilon}[/tex] because that is the field in which the plates are being moved
 
  • #10
There are two ways to do the problem...

1) Get the final energy of the capacitor and subtract the initial energy of the capacitor.

2) calculate the force need to move the plate. multiply that force by d.

First try the problem using method 1). show your steps.
 
  • #11
why for total energy of both plates are you using the charge for one plate?

learningphysics said:
You need to use Q on one plate... ie: Q = [tex]\sigma A[/tex]

The field E is [tex]\frac{\sigma}{\epsilon}[/tex]
 
  • #12
That's just what the energy of a parallel plate capacitor is... The energy of the capacitor is (1/2)CV^2 = Q^2/(2C) where Q is the charge on the positive plate...

the net charge on the capacitor (both plates) is Q + -Q = 0.
 
  • #13
then the energy of the negative charge is -Q^2/(2C)

in such case, the total energy is zero?
 
  • #14
indigojoker said:
then the energy of the negative charge is -Q^2/(2C)

in such case, the total energy is zero?

no. that doesn't make sense... then the energy of the capacitor is always 0. I just pointed out that the net charge is always 0 for a parallel plate capacitor... so it doesn't make sense to use that for the enrgy...

All the equations for a parallel plate capacitor... Q = CV, E = (1/2)CV^2, E = Q^2/(2C)

In all these equations Q is the charge on the positive plate...

The energy of the capacitor (the entire capacitor, both plates...) is Q^2/(2C) where Q is the charge on the positive plate...
 

1. What is work to move two plates?

Work to move two plates refers to the energy required to move two tectonic plates, which are large sections of the Earth's crust that move relative to each other.

2. Why do tectonic plates move?

Tectonic plates move due to convection currents in the Earth's mantle, which causes the plates to slowly drift and interact with each other.

3. How does work affect the movement of tectonic plates?

The amount of work done to move two plates is directly related to the force applied and the distance the plates move. The more work done, the greater the movement of the plates.

4. Is work to move two plates the same as earthquake energy?

No, work to move two plates is the energy required to move the plates, while earthquake energy is the energy released when the plates suddenly shift or break along a fault line.

5. Can we control the movement of tectonic plates through work?

No, tectonic plates are constantly moving and shifting due to natural forces and cannot be controlled by human intervention or work.

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