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Algebra homework help

  1. Feb 4, 2007 #1
    1. The problem statement, all variables and given/known data
    If I know [tex]Fs^2 = 6*10^{-8}[/tex], and [tex]\frac{F}{V^2} = 10^{-9}[/tex], how can I figure out [tex]\frac{Fs^2}{V^2}[/tex]? F, s, V are all real numbers.

    2. Relevant equations
    [tex]Fs^2 = 6*10^{-8}[/tex] (equation 1)
    [tex]\frac{F}{V^2} = 10^{-9}[/tex], (equation 2)
    3. The attempt at a solution

    I tried dividing equation 1 by equation 2 and vice versa...and mutiplied them...but the best thing I can get is [tex]\frac{F^2s^2}{V^2}[/tex]...:cry: :confused:
    Last edited: Feb 4, 2007
  2. jcsd
  3. Feb 4, 2007 #2


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    What do you want to find out? In your first line you say you want "Fs2/V2," whereas in the last line you say the best you can get is Fs2/V2. Is this not what you want?
  4. Feb 4, 2007 #3
    oh right, sorry. I had a typo...the best I got was [tex]\frac{F^2s^2}{V^2}[/tex]. Lemme correct that.
  5. Feb 4, 2007 #4
    You have two equations. The first is

    [tex]Fs^2 = 6\times 10^8[/tex]

    while the second is

    [tex]\frac{F}{V^2} = 10^{-9}[/tex]

    Rearranging the second equation gives you

    [tex]\frac{1}{V^2} = \frac{1}{F\times10^{-9}}[/tex]

    If you now multiply your first equation by this you obtain

    [tex]\frac{Fs^2}{V^2} = \frac{6\times 10^8}{F\times10^{-9}}
    = \frac{6\times10^{17}}{F}[/tex]

    There's a certain redundancy here. Multiplying both sides by [itex]F[/itex] gives you

    [tex]\left(\frac{Fs}{V}\right)^2 = 6\times 10^{17}[/tex]

    Is this not what you want?
    Last edited: Feb 4, 2007
  6. Feb 4, 2007 #5
    Hmmm not quite :\ [tex]\left(\frac{Fs}{V}\right)^2 = 6\times 10^{17}[/tex] is the same as [tex]\frac{F^2s^2}{V^2}[/tex]...which was what I got before...I wanna get [tex]\frac{Fs^2}{V^2}[/tex]...but I don't know if that's actually possible. Anyway, thanks a bunch for the help! :)
  7. Feb 4, 2007 #6
    Going to the root of the problem, I am actually working on my lab report. I need to determine the value of the permittivity of free space from my experiment...and from my results I plotted F with [tex]V^2[/tex] and I got a linear relationship between them with a slope of 10^-9...which is the number I had up there [tex]\frac{F}{V^2} = 10^{-9}[/tex]. Then I plotted F with [tex]\frac{1}{s^2}[/tex] and I got another relationship...with a slope of [tex]Fs^2 = 6*10^{-8}[/tex].

    My experiment is to deduce the permittivity of free space so I can eventually deduce the speed of light.

    F is the attractive force between two parallel plates (capacitor). A is the area of the plate. s is the distance between the plates. V is the voltage between them. My formula I'm dealing with is [tex] F=\frac{e_{0}AV^2}{2s^2}[/tex]

    The lab requires me to find the "best" value of [tex]\frac{Fs^2}{V^2}[/tex] ...which is where I was stuck at from the beginning. :\

    From then on, once I get [tex]\frac{Fs^2}{V^2}[/tex], find the permittivity of free space and eventually the speed of light.
    Thanks again for the time and effort! :)
    Last edited: Feb 4, 2007
  8. Feb 5, 2007 #7


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    You simply don't have enough information. If you could find [tex]\frac{Fs^2}{V^2}[/tex], since you already know [itex]Fs^2[/itex], you could divide the second by the first to find V then use [itex]\frac{F}{V^2}[/itex] to find F and finally solve for s. You can't expect to be able to solve 2 equations for 3 values.
  9. Feb 5, 2007 #8
    Hmmm...ok...I'll have to check my values again. Thanks!
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