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Determining charge on capacitor

  1. Mar 3, 2005 #1
    Please take a look at the attached picture. How would you go about finding the charge on capacitor 2?

    My method which didnt work:

    -combine capacitors 2,3,4 and 5 and get the equivalent capacitance Ceq
    -Since V across C1 is equal to V across Ceq we can find charge on Ceq by V/Ceq

    -Then voltage on Ceq is equal to voltage on C5 and C234(combination of capacitors 2,3 and 4.

    -Then use that voltage to find charge of 2,3,4 combined. Then voltage on 2 and 3 is equal to voltage on 4 since they are 'in parallel' and find this voltage by doing charge on 2,3,4 divided by C234

    -Then charge on 2 and 3 can be found by V23*C23=q23=q2(series)

    I am thankful for any help.
     

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  2. jcsd
  3. Mar 3, 2005 #2

    Curious3141

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    Homework Helper

    You seem to be making this too complicated (plus you have some wrong concepts).

    Just remember these :

    Charges on capacitors in series are EQUAL.

    Voltages on capacitors in parallel are EQUAL.

    Capacitances in series are added like resistors in PARALLEL.

    Capacitances in parallel are added like resistors in SERIES.

    '+' = series, '||' = parallel

    Hints :

    a) You can completely ignore C1 here.

    b) Find combined capacitance of i) (C2+C3) and hence ii) (C2 + C3)||C4 using the rules stated. Let this be Ceq.

    c) Ceq will have the same charge across its "plates" as C5. The voltages across Ceq and C5 will be in inverse proportion to their respective capacitances.

    d) So, treating Ceq and C5 as a voltage divider with voltage 10V applied across it, find the voltage across Ceq alone.

    e) That voltage (found in d) will be equal to the voltage across (C2 + C3). Knowing the capacitance (C2 + C3) found in b), figure out the charge across (C2 + C3).

    f) Observe that the charge across C2 = charge across C3 = charge across the series combination (C2 + C3), found in e) and you're done.

    Looks complicated but it's easy when you work it out since all the capacitances are equal.
     
  4. Mar 3, 2005 #3
    I dont understand how Ceq can have same charge is C5 since capacitors can only have same charge if they in series and are connected by two wires, one directly from top plate of Ceq to bottom of C5 and the other from bottom of Ceq to top of C5. Is that right or am I wrong somewhere?
     
  5. Mar 3, 2005 #4

    Curious3141

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    You're reducing the combination of C2, C3 and C4 to a single equivalent capacitance Ceq. That (virtual) capacitance is indeed in series with C5. Two capacitors placed in series with a voltage source (like a cell) will have the same charge across them. See ?
     
    Last edited: Mar 3, 2005
  6. Mar 5, 2005 #5
    Understood, thanks for your help Curious3141. I didn't quite see that before.
     
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