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Capacitor physics help

  1. Aug 1, 2010 #1
    1. The problem statement, all variables and given/known data

    [PLAIN]http://img841.imageshack.us/img841/5497/phys.jpg [Broken]

    Capacitor has two cylinders (conducting), as shown.

    Charged 0V on LHS cylinder and 12V on the other one. d=140mm.

    Show the potential at P is Vp = 6-2.34ln((140-x)/x)

    2. Relevant equations

    E=integralE.dA

    3. The attempt at a solution

    I got as far as using Vp = VPQ + VPR but I cannot seem to get the constant term 6 or the 2.34... could someone please show me the correct method for doing this?

    Thanks
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Aug 1, 2010 #2
    Re: Capacitor

    Firstly, you find the total electric field at P. From this result, you calculate the potential at point P (from a to d-a). When you arrive this point, consider conditions given in problem: VL = 0 (x = a) and VR = 12 (x = d-a). Plug these into the potential you just found to determine some constants.
     
  4. Aug 2, 2010 #3
    Re: Capacitor

    Thanks for your reply. But I can't seem to be able to get this, especially the constant value (ie. 6)... Where does that come from?
     
  5. Aug 2, 2010 #4
    Re: Capacitor

    Sry for this inconvenience. I will try again.
    1/ You should write down the form of electric field caused by a cylinder conductor at point x outside the conductor.
    2/ Find total field at point P (beware the sign of RHS field).
    3/ Find potential (integrate from a to d-a)
    4/ After getting step 3, in the form of V, there is a constant q/2πɛz. This constant contribute with another constant (ln(?)) to be 2 constants in your solution. You will determine it by plugging value V = 12 V at point x = d-a.
     
  6. Aug 2, 2010 #5
    Re: Capacitor

    Thanks for your reply...

    This is where I'm getting stuck :( ... So I should get q/2πɛz = two terms?

    Thanks alot
     
  7. Aug 2, 2010 #6
    Re: Capacitor

    After step 3, the potential at point P is:

    [tex]V = \frac{q}{2 \pi \epsilon_0 z} [\ln \frac{d-x}{x} - \ln \frac{d-a}{a}] [/tex]

    Now plug V = 12 and x = d-a into this formula, and you will have:

    [tex]12 = -2 \frac{q}{2 \pi \epsilon_0 z} \ln \frac{d-a}{a}[/tex]

    This will yield constant "6" in your solution.
     
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