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Test tommorow please help: electic charge

  1. Apr 6, 2005 #1
    Studying help

    An isolated spherical conductor of radius 20cm is charged to 4k V (i.e. 4000 V).
    a. How much charge is on the conductor?
    b. What is the capacitance of the sphere?
    c. how does the capacitance change if the spere is charged to 8 k V (8000)?
     
  2. jcsd
  3. Apr 7, 2005 #2
    Im not sure about a.

    b) Use the relationship Q = CV to find C.
    c) Use the solved equation above to find the effect on C if V = 2V
     
  4. Apr 7, 2005 #3
    Since you have test tomorrow I do not know if this will be of any help, but here it is anyway.
    a=R=20cm

    a.)
    [tex]
    4000 V=- \int_\infty^a \vec{E} dr =-\int_\infty^a \frac{1}{4 \pi \epsilon_0} \frac{Q}{r^2} dr
    [/tex]

    b.)
    [tex]
    Q= C \Delta V
    [/tex]
    [tex]
    \frac{Q}{\Delta V}= C
    [/tex]
    [tex]
    \Delta V =-\int_\infty^a \frac{1}{4 \pi \epsilon_0}\frac{Q}{r^2} dr
    [/tex]
    steps not shown
    [tex]
    C=4 \pi \epsilon_0 R
    [/tex]

    3.) What Whozum did not mention here is that Q and V are related so it is not as easy as he says it is. I provided about an equation that should work just fine for you.
     
  5. Apr 7, 2005 #4
    Davorak:

    With [tex] V = \frac{Q}{4\pi\epsilon_0 R} \mbox{ and } C = \frac{Q}{V} = 4\pi\epsilon_0R [/tex]

    Does it not follow that

    [tex] 2V = \frac{Q}{2\pi\epsilon_0 R} \mbox{ so then } C = \frac{Q}{2V} = 2\pi\epsilon_0R[/tex] ?

    The charges cancel in either case, no?
     
  6. Apr 7, 2005 #5
    You are considering Charge and Voltage independant of each other when they are not.
    Take your second example:

    [tex]
    2V = \frac{2 Q}{4\pi\epsilon_0 R} \mbox{ so then } C = \frac{2 Q}{2V} = 4\pi\epsilon_0 R
    [/tex]

    When you increase Q you increase V and vs versa for this problem.
     
  7. Apr 7, 2005 #6
    So doubling the voltage has no effect on capacitance because doubling the voltage doubles the charge?
     
  8. Apr 7, 2005 #7
    By George I think you’ve got it. Really vs versa though since charge is usually considered fundamental while voltage is not.
     
  9. Apr 7, 2005 #8

    ZapperZ

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    You appear to have posted a SERIES of homework questions on here - many of them actually are related, meaning that if you understand how to solve one, the others should follow along the same line. However, in all of the questions you have posted, not ONCE have you even described your attempt or what you have tried in solving the problem, nor are we told what you do know. I mean, you MUST have some clue on where to start, or else we would just be doing the home work for you and doing you a complete disservice.

    Please READ the STICKY for this section of PF before you post more questions.

    Zz.
     
  10. Apr 7, 2005 #9

    GCT

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    Not to disagree with ZapperZ, nevertheless this is all the OP should need

    [tex]V=k_eq/r[/tex]
     
  11. Apr 7, 2005 #10

    ZapperZ

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    Staff Emeritus
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    And not to disagree with you either, but if the OP does not even know this, then "Houston, we have a problem!"

    Zz.
     
  12. Apr 7, 2005 #11
    Knowing that would certainly help him for spherical cases, but he does not seem to know how to do it general. On another thread he was unaware how to get a electric field in a parallel plate capacitor. This is why I gave the more general form. He seems to be memorizing equations rather knowing how to derive them.
     
    Last edited: Apr 7, 2005
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