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Question about conducting plates

  1. Oct 23, 2009 #1
    1. The problem statement, all variables and given/known data
    Two infinite sheets with surface charge density σ1 and σ2, respectively, are oriented perpendicular to the x-axis. An infinite, conducting slab of thickness a is placed between the charged sheets as shown in the figure. The conducting plate has a net charge per unit area of σC.



    2. Relevant equations

    Calculate the x-component of the net electric field at x = 0.

    (a) Ex = +7.85 × 105 V/m
    (b) Ex = +6.22 × 105 V/m
    (c) Ex = +5.65 × 105 V/m
    (d) Ex = -2.67 × 105 V/m
    (e) Ex = -3.85 × 105 V/m


    QUESTION 15**
    Calculate the surface charge density σR induced on the right hand face of the conducting plate.

    (a) σR = +3.75 μC/m2
    (b) σR = +2.00 μC/m2
    (c) σR = -1.50. μC/m2
    (d) σR = -3.00 μC/m2
    (e) σR = -4.25 μC/m2


    3. The attempt at a solution

    14. for this number, i suppose left part of the conducting plate has the charge density -1.5. that gives the total E at x = 0 equal to 10. i use the formula E = sigma/2Eo = 10/(2*8.85) to get the answer C

    15. I have no idea. I thought it was c, but the answer is B.

    thank you so much!
     

    Attached Files:

  2. jcsd
  3. Oct 23, 2009 #2

    Andrew Mason

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    Homework Helper

    Apply Gauss' law to the Gaussian cylinder of cross-sectional area da enclosing the left surface of the conducting sheet (ie. ending in the middle of the conducting sheet where the field = 0) and x=0. You know that the net field through the left surface of area da is equal to the field at x=0 (ie. the answer to question 14) and the field through the right surface of area da is 0.

    So applying Gauss' law: [itex]E_x\cdot da = dq/\epsilon_0 = \sigma_L da/\epsilon_0[/itex]

    which means that [itex]\sigma_L = E_x\epsilon_0[/itex] (work out the polarity from the direction of the field, which is the direction in which a positive charge would move).

    Then it is just a matter of working out what the charge density on the right is so that the total charge density is [itex]\sigma_c = -3.0 \mu C[/itex]

    AM
     
    Last edited: Oct 23, 2009
  4. Oct 24, 2009 #3

    ideasrule

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    You don't need to suppose. The electric field induced by an infinite sheet is sigma/e0, regardless of the distance to the sheet. That means the conducting plate induces sigma_c/e0 everywhere except inside the plate, so you can just add up the electric fields and get the anser.
     
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