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Electric field between parallel plate capacitor

  1. May 24, 2013 #1
    If you have an infinite non-conducting plate, the electric field just outside is equal to sigma / 2*epsilon.

    The electric field just outside a conductor is equal to sigma / epsilon.

    I understand both these results, but why is it than in the formula for the capacitance of a parallel plate capacitor, they use sigma / 2 * epsilon for the electric field contribution from each plate? The plates are metal, so I would think the formula for the electric field between them would use the result for conductors!
  2. jcsd
  3. May 24, 2013 #2

    Jano L.

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    Gold Member

    Because each charged metallic plate has field similar to that of infinite sheet and hence approximately equal to ##\sigma/(2\epsilon_0)##. The total field above the metallic plate is a sum of the fields due to both plates, hence ##\sigma/\epsilon_0##. This confirms the expectation that above finite metallic surface, the total field is equal to ##\sigma/\epsilon_0##.
  4. May 24, 2013 #3


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    Also, it is important to remember that each plate of a capacitor will hold one of the two kinds of charge: let's say that the botton plate will be negatively charged, and the upper plate positively charged. Physically, this means that is that the field lines of upper of the plate point away from it, and the field lines of the botton plate point towards it! This is the underlying reason why the fields are added in between the plates and cancel each other elsewhere.

    Now, if you want to do the math, here it goes:

    Suppose the plates are parallel to the [itex]xy[/itex] plane and, of course, perpendicular to [itex]\hat{z}[/itex]. The mathematicaly rigorous expression for the field produced by an infinite charged plate resting on [itex]z=0[/itex] is:

    [itex]\vec{E} = \frac{\sigma}{2\epsilon_{o}}\frac{ | \vec{z} |}{z}\hat{z}[/itex]

    Where [itex]\vec{z} = z\hat{z}[/itex] and [itex]\frac{ | \vec{z} |}{z}[/itex] is the function [itex]\mathrm{sign}(z)[/itex]. It basically tells us if we are above ([itex]\mathrm{sign}(z) = +1[/itex]) or below ([itex]\mathrm{sign}(z) = -1[/itex]) the plate under consideration. Notice that this expression will change if the plate is not resting at [itex]z=0[/itex], but we can forget about that and just think about the value of the sign function to perform the calculations in a more brief way.

    Now, since the plates have the same area and opposite charges, we can write

    [itex]\sigma_{botton} = -\sigma_{upper}[/itex]

    So, for the region in between the plates, [itex]\mathrm{sign}(z) = +1[/itex] for the botton plate and [itex]\mathrm{sign}(z) = -1[/itex] for the upper plate. But, since their superficial charge distribution are also opposite, this sign difference disappear and we get that the total field given by

    [itex]\vec{E} = \vec{E}_{botton} + \vec{E}_{upper} = \frac{\sigma_{botton}}{2\epsilon_{o}} (+1) \hat{z} + \frac{\sigma_{upper}}{2\epsilon_{o}} (-1) \hat{z} = \frac{\sigma_{botton}}{2\epsilon_{o}} (+1) \hat{z} - \frac{\sigma_{botton}}{2\epsilon_{o}} (-1) \hat{z} = \frac{\sigma_{botton}}{\epsilon_{o}} \hat{z}[/itex]

    You can also try to calculate the field elsewhere, but because of this sign difference, you will find it is zero.


    If the plate is resting at [itex]z=d[/itex], for example, the expression for the field can be easily guessed as

    [itex]\vec{E} = \frac{\sigma}{2\epsilon_{o}}\frac{ | \vec{z} - d\hat{z} |}{z - d}\hat{z}[/itex]

    If you place one of the plates at [itex]z=d[/itex] and the other one at [itex]z=-d[/itex], you can see that you will get to the same answer for the total field.

    Zag ;)
    Last edited: May 24, 2013
  5. May 24, 2013 #4
    Thank you for your answers!

    Now that I thought about it, if you have Q charge on a conducting plate, wouldn't the charge distribute it self equally on both sides on that one plate? That being the case, the effective charge density on the side of the plate would be sigma / 2 and the electric field contribution on one side would become sigma / 2*epsilon.
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