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Potential Difference in a Capacitor

  1. Sep 25, 2008 #1
    1. The problem statement, all variables and given/known data
    [​IMG]

    An isolated large-plate capacitor (not connected to anything) originally has a potential difference of 1030 volts with an air gap of 2 mm. Then a plastic slab 1 mm thick, with dielectric constant 5.1, is inserted into the middle of the air gap as shown in Figure 16.70. As shown in the diagram, location 1 is at the left plate of the capacitor, location 2 is at the left edge of the plastic slab, location 3 is at the right edge of the slab, and location 4 is at the right plate of the capacitor. All of these locations are near the center of the capacitor. Calculate the following potential differences.
    V1 - V2 =
    V2 - V3 =
    V3 - V4 =
    V1 - V4 =
    2. Relevant equations

    deltaV = E (deltaL)

    3. The attempt at a solution

    I assumed that the equation for deltaV would work for this problem, but apparently not. I am completely lost right now. I thought that V1-V2 would be (1030/5.1)(.0005) Where am I going wrong?
     
  2. jcsd
  3. Sep 25, 2008 #2
    anyone?
     
  4. Oct 1, 2008 #3
    cowmoo32,

    Let's review what we have:

    [tex]\Delta V = 1030 = \Delta V_{1 \rightarrow 4}[/tex]

    [tex]K = 5.1[/tex]

    [tex]s_{tot} = .002m[/tex]

    [tex]s_{1} = .0005m[/tex]

    [tex]s_{2} = .001m[/tex]

    [tex]s_{3} = .0005m[/tex]

    with,

    [tex]\Delta V = E \cdot s[/tex]

    and

    [tex]\Delta V_{tot} = \sum^n_{i=1} \Delta V_i [/tex]

    One accounts for the dielectric constant of the insulator in the following fashion:

    [tex]\Delta V_{insulator} = \frac{\Delta V_{space}}{K}[/tex].

    To get the [tex]\Delta V_{1 \rightarrow 2}[/tex], we recognize that

    [tex]s_{1} = .0005m = \frac{s_{tot}}{4}[/tex]

    Hence,

    [tex]\Delta V_{1 \rightarrow 2}= E \cdot s_{1} = E \cdot \frac{s_{tot}}{4} = \frac{\Delta V}{4}[/tex]

    That should be enough information to give you the correct answer. Make sure to account for the dielectric constant of the insulator.

    - Happy Problem Solving -
     
  5. Oct 1, 2008 #4
    [hit submit too many times]
     
    Last edited: Oct 1, 2008
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