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Electric Field

  1. Sep 14, 2006 #1
    check whether this result of this problem is consistent with this statement
    [tex] \vec{E_{above}} - \vec{E_{below}} = \frac{\sigma}{\epsilon_{0}} \hat{n} [/tex]

    an infinite plan carries a uniform surface charge sigma. Find its electric field

    Draw a Gaussian Pillbox extending above and below the plane. Then
    since [tex] \oint \vec{E} \bullet d\vec{a} = \frac{Q_{enc}}{\epsilon_{0}} [/tex]

    and since[tex]Q_{enc} = \sigma A [/tex]

    By Symmetry E points up and down

    [tex] \int \vec{E} \bullet d\vec{a} = 2A |\vec{E}| [/tex]

    so [tex] \vec{E} = \frac{\sigma}{2\epsilon_{0}} \hat{n} [/tex]

    Now to tackle the question

    Well for an infinite sheet for each side the Electric field points normal to the sheet, right?
    SO the electric field for the top is [itex] \vec{E} = \frac{\sigma}{\epsilon_{0}} \hat{n} [/itex]
    and te bottom is the negative of that

    So when you add those two together you get [tex] \vec{E} = \frac{\sigma}{\epsilon_{0}} \hat{n}[/tex]

    and this is consistent with statement. Easy enough. i just want to know whether this kind of 'proof' for the statement is satisfactory.
  2. jcsd
  3. Sep 16, 2006 #2
    yeah that's right.
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