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Last-minute concept problems (before exams!)

  1. May 21, 2003 #1
    Hi. This thred will sort of be "my thread" i suspect for various concepts I still do not get after two semesters of college physics (my final is in two days). Without further ado, I'll ask my first question:

    Here's how the problem reads:
    A long nonconducting cylinder (radius = 12 cm) has a charge of uniform density (5.0 nC/m^3) distributed throughout its column. Determine the magnitude of the electric field 5.0 cm from the axis of the cylinder.

    If they said to determinte it at a radius larger than the radius of the cylinder itself, I would most likely have no problem solvingit; but how do I solve it when they ask for the electric field 5 cm from the axis (i.e., r<R)?
  2. jcsd
  3. May 21, 2003 #2
    The same way as it is outside. Outside, would you not determin it from the axis, rather than the radius? Thus, this is simply some distance from the axis, regardless of where the radius of said cylinder is?
  4. May 21, 2003 #3
    So it makes no difference, then...

    Using Gauss's Law (after being simplified) as follows:


    I still don't come out with the correct answer (these questions are off of one of our previous tests for which we have the answers, but not how to solve them)

    The answer, by the way, is supposed to come out to 14 N/C.

    Am I using the wrong approach?
  5. May 21, 2003 #4


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    What have you tried?
    Hint: Gauss' law, and a cylindrical gaussian surface inside the cilinder.

    Think of what should happen with the E field:

    1. What would you expect the E field to be at the center of the cylinder? (remember how it is defined... how would you expect a test charge to move if you had it at the center?)
    2. You said you can solve it outside the cylinder. What do you get there?
    3. How would you connect both results? (1 and 2).

    This will probably give you a feeling for the kind of result you are looking for.
  6. May 21, 2003 #5


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    Of course it makes a difference!

    It does makes a difference if you are inside or outside the cylinder. Say R is the radius of the cylinder and r the point at which you measure the field:

    Outside, applying Gauss' law yields something like:

    (gaussian integral up to r) = (charge up to R)

    Since R is fixed and r is larger than R, when you solve for E you get

    E = (coefficient) x (charge density) / r

    OTOH, inside the cylinder, you get

    (gaussian integral up to r) = (charge up to r)

    Notice how the right hand side has now an r on it, so that it combines with the r in the left hand side, and you get a different dependence.
  7. May 21, 2003 #6
    got it!! :)

    Thanks. I realized what I was doing wrong. I was using this certain problem as a template for what I was doing, only they were finding the charge of a certain distance from a LINE of charge. Hence, they were using (Lambda)*(L), rather than (ro)*(V), as I was supposed to.

    The final equation, for anyone interested is:


    where r is the distance from the axis, andk k=9x10^9

    Thanks again. I'm sure I'll have more questions shortly.
  8. May 21, 2003 #7
    Ok. I have a new one.

    These circuit problems really get me sometimes. Maybe you can give me some general statements that I haven't caught onto yet to help me think through problems like this. here is goes:

    If Va - Vb = 50V, how much energy is stored in the 54-uF capacitor shown?

    The image that is shown is roughly this:
    Code (Text):

                        ------| 36uF |------
    (a)------| 72uF |----                  -----(b)
                        ------| 54uF |------
    My idea was to first combine the capacitors to find a single equivalent capacitor (which comes out to 80uF). I was trying to use the relationship:

    Together with this, I found the charge on the supposed 80mF capacitor by using the relationship Q=C(delta-V)

    I'm stuck. Any ideas?
    Last edited by a moderator: May 21, 2003
  9. May 21, 2003 #8
    The diagram

    Sorry the diagram didn't come out. it's supposed to be a single line with a 72uF capacitor on it. It then branches into two; one branch has a 36uF capacitor, and the other has a 54uF capacitor. The two then join back together. Hope it's easy enough to understand by just the words.
  10. May 21, 2003 #9


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    I haven't done any of these for a while, but I think you're on the right track: with the equivalent capacitance and deltaV, you can get the total charge. Then, you can get the charge and deltaV on each capacitor, and get the energy on the one you want...

    you'll need to double check,... I'm a bit rusty on this, and about to fall sleep.

    Good luck!
  11. May 21, 2003 #10
    Re: Ok. I have a new one.

    I don't see any 54uF capacitor on your picture.

    Problem just doesn't look posted good enough for an answer.

    Try it again, perhaps I can help

    Make a .BMP and post it!
  12. May 21, 2003 #11


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    Re: Re: Ok. I have a new one.

    It is the one marked "64".
  13. May 21, 2003 #12


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    Re: Ok. I have a new one.


    Code (Text):

                                ------| 36uF |------
    (a)------| 72uF |----                         -----(b)
                                ------| 64uF |------
    Flame: you just need to put your ascii drawing between {CODE}{/CODE}
    (in square instead of curly brackets)
  14. May 21, 2003 #13
    Re: Re: Ok. I have a new one.

    AHA! And now it makes sense!

    I would have to consult my book to get the answer, as this is one area of physics that I didn't remember to remember.

    If you seriously need help, like if you don't get it in 2 days, I can dit it up. But there's probably some people tommorrow that will help ya.

    But I do know the basics. From the change in voltage use forumula to get some value. Consolidate the capacitors. Use said value, plus total capicatance and work backwords. Consolidate first the last two only to get the first one. the opposite.

  15. May 21, 2003 #14
    Don't know why...

    I attach the file, tried both jpg and bmp. Neither one is showing up...
  16. May 21, 2003 #15


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    I think you need to:

    1. get the total equivalent capacitance, C
    2. Obtain the total charge in the circuit (from C and V),
    3. "Unfold" the big C in two capacitances: one is the original C1 (72 uF), and the second one is the equivalent to the remaining two (call it C23).
    4. From there, you can get the voltage difference on each of C1 and C23. IIRC, it goes like V23 = V(C1/(C1+C23)), but you'll need to double check this.
    5. Once with V23 (which is the v. diff. in the system formed by C2 and C3), you know that the voltage across C3 is also V23 (so you can get the charge Q3).
    6. Finally, you can get the energy from Q3 and C3.

    Makes sense?
  17. May 21, 2003 #16
    capacitance = C = q/V
    U = (1/2)CV2
    capacitors sum in series (I always think of "caparallel, plus plus plus")
    Ceq=1/&sum; (1/C)}ser
    you should solve it with Kirchoff's rules:
    input to every junction = output from every junction
    sum of voltages around a closed loop = 0
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