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Homework Help: 5 Parallel Large Flat Electrodes (Potential/E-Field)

  1. Oct 1, 2011 #1
    1. The problem statement, all variables and given/known data

    See figure attached.

    2. Relevant equations

    3. The attempt at a solution

    My professor came up with this example in lecture and the way he went it about it was very confusing so hopefully you guys can help me clear up some of steps/thought process he took.

    We are asked to find all the electric field vectors between each plate, so he begins to write equations for these electric fields.

    He first notes that,

    [tex]E_{inside conductor} = 0[/tex]

    He then proceeds to write,

    [tex]V = E_{2}d + E_{3}d[/tex]

    I'll put in my thought process for all the work he skipped,

    [tex]V = \int_{l_{1}} \vec{E_{2}} \vec{dl} + \int_{l_{2}} \vec{E_{3}} \vec{dl} [/tex]


    [tex]\vec{E_{2}} \text{ and } \vec{E_{3}} \text{ are parallel to } \vec{dl}[/tex]

    [tex]\Rightarrow V = E_{2} \int_{l_{1}}dl + E_{3} \int_{l_{2}}dl[/tex]

    Since the distance between the plates in the same,(i.e. a distance d)

    [tex]V = E_{2}d + E_{3}d[/tex]

    He then writes another equation,

    [tex]A\epsilon_{0}E_{3} - A\epsilon_{0}E_{2} = Q[/tex]

    Where is he getting this from? I know that's the difference in flux, but it looks like it's coming from Gauss' Law applied to a Gaussian surface around the middle plate that has a charge Q.

    [tex]\oint_{S} \vec{E} \cdot \hat{n}dS = \frac{Q_{enclosed}}{\epsilon_{0}}[/tex]

    It seems as though

    [tex]E = E_{3} - E_{2}[/tex]

    because then,

    [tex]\Rightarrow \left( E_{3} - E_{2} \right)A = \frac{Q}{\epsilon_{0}} [/tex]

    Rearranging gives me his original equation,

    [tex]A\epsilon_{0}E_{3} - A\epsilon_{0}E_{2} = Q[/tex]

    Why is the electric field for the gaussian surface enclosed the middle plate (E3-E2)?

    After writing those 2 equations, it's obvious that we can solve for E2 & E3.

    He then states that,

    [tex]\rho_{S} = \epsilon_{0} \left( \vec{E} \cdot \hat{n} \right)[/tex]

    and denotes the charge on the plate to the right of the leftmost plate as

    [tex]Q_{2}=A\rho_{S2} - A\rho_{S1} = A\epsilon_{0}E_{2} - A\epsilon_{0}E_{1}[/tex]

    (This comes from Gauss' Law around the plate, where the electric field is (E2-E1) [Just like my question above, why is it (E2-E1)?])

    We now have 1 equation, and 2 unknowns (i.e. Q2 and E1).

    Then he explains how the voltage source is going to pull charge off the 2 rightmost plates and place it onto the plate with charge Q2.

    [tex]Q_{2} = A \rho_{S3} = A\epsilon_{0}E_{3}[/tex]

    Since we know E3 he solves for Q2 in terms of E3 and it is found that,

    [tex]E_{1} = E_{2} - E_{3}[/tex]

    The two main points which I'm confused about are,

    • Why is the electric field for the gaussian surface enclosing the middle plate (E3-E2)?
    • Where does he get the equation,
      [tex]\rho_{S} = \epsilon_{0} \left( \vec{E} \cdot \hat{n} \right)[/tex]

    If I anything I said sounds goofy, or if I am misunderstanding anything else please feel encouraged to correct me.

    Thanks again!

    Attached Files:

    Last edited: Oct 1, 2011
  2. jcsd
  3. Oct 2, 2011 #2
    This is what I think is going on, see the attached,

    Attached Files:

  4. Oct 3, 2011 #3
    This is exactly what I had already mentioned in my original thread.

    I am looking to get the points I have bulleted explained.
  5. Oct 3, 2011 #4
    You wrote,

    "Why is the electric field for the gaussian surface enclosing the middle plate (E3-E2)?"

    The integral is E dot dA where dA either points outward or inward, therefore as drawn the minus sign, (E3-E2).

    You wrote,

    "Where does he get the equation,
    ρS=ϵ0(E⃗ ⋅nˆ)"

    This comes from Gauss's Law, in one form

    integral of E dot dA = Q/ε_o

    in simple case you have this is simply

    E times area = Q/ε_o or

    E = rho/ε_o
  6. Oct 3, 2011 #5
    I'm confused.

    For Gauss' Law,

    [tex]\oint_{s} \vec{E}\cdot\hat{n}dS = \frac{Q_{enclosed}}{\epsilon_{0}}[/tex]

    My text explains that [tex]\hat{n}[/tex] is always pointing from the charge outward, so how would it ever point inward?

    The minus sign still isn't clear to me, from the picture it looks as through E2 and E3 are both pointing to the right, thus why the subtraction?
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