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Homework Help: Capacitance and electric potential

  1. Jun 7, 2008 #1
    1. The problem statement, all variables and given/known data

    a 100 pico Farad capacitor C_1 is charged to a potential difference of 50 volts. the charging battery is disconnected and the capacitor is then connected to a second uncharged capacitor C_2. if the potential accross C_1 drops to 35 volts, what was the capacitance of C_2?


    2. Relevant equations

    capacitance C = Q/V where Q is charge, and V is electric potential in volts
    units of capacitance is Farad

    3. The attempt at a solution
    let C_1 = Q/V_1 where C_1 = 100 pico farads = 100 *10^-12 farads and V_1 = 50 volts
    solve for Q to get Q = 5*10^-9 coulombs

    let C_2 = Q/V_2 where Q = 5*10^-9 coulombs and V_2 = 35 volts, solve for C_2 to get

    capacitance of C_2 = 1.43*10^-10 Farads

    did i do this correctly?

    thanks
     
  2. jcsd
  3. Jun 7, 2008 #2

    G01

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    Be careful! Does all the charge initially on the first capacitor move onto the second?
     
  4. Jun 7, 2008 #3
    no you are right, all the charge of C_1 does not go to C_2. did soling the C_1 equation for q come into play for this type of problem?

    how do i set up an equation that involves C_1 and C_2, better yet the change in capacitance of C_1?

    or would the equation become C_2 = Q/ deltaV where Q = 5*10^-9 and V = 50 -35 = 25 so that C_2 = 2*10^-10 farads?

    is this the correct approach?
     
  5. Jun 7, 2008 #4

    G01

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    No, your right that C_2 has 35 V across it. Since the two capacitors are in parallel they must have the same voltage across them.

    You know that after the battery is removed: Q_1+Q_2=5nC

    Is there any way you can find Q_1 and then find Q_2?

    HINT: The capacitances of capacitors are "built in quantities." Thus, C_1 is the same before and after the battery is removed.
     
  6. Jun 7, 2008 #5
    didn't i already find Q_1 when i did this calculation:
    " let C_1 = Q/V_1 where C_1 = 100 pico farads = 100 *10^-12 farads and V_1 = 50 volts
    solve for Q to get Q = 5*10^-9 coulombs " ??

    also, how do you know that the capacitors are in parallel? if it is parallel, capacitance C = C_1 + C_2 and C_1 = 100 pico farads, how is the change in voltage factored in?
     
  7. Jun 7, 2008 #6

    nrqed

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    Well, this is a situation where, in some sense, the capacitors can be said to be in parallel and they can be said to be in series so it's a bit confusing.

    Bu the key point is that they are the only two elements in a loop, right? By Kirchoff's voltage law, the potential difference across each capacitor must be equal (in absolute value) so there is a 35 volts potential difference on each of them

    The Q_1 you found (using the 50 volts) gives the total charge on th etwo capacitors after they have been connected. After they are connected Q on capacitor 1 decreases and Q on capacitor two increases so you now have Q_1 + Q_2 = 5*10^-9 coulombs

    So you know the sum of the two charges. And you know that for each capacitor you have

    Q_1 = C_1 V and Q_2 = C_2 V

    (using the fact that they have the same potential difference V).
    So you have three equations fo rthree unknowns (Q_1,Q_2 and C_2). So you can solve
     
  8. Jun 7, 2008 #7
    i am confused about the value of V, is it 35 or 50 volts?
    once of found V is this how i would go about finding capacitance C_2:

    since Q_1 + Q_2 = 5*10^-9 and Q_1 = C_1V and Q_2 = C_2V :

    if i let Q_1 = 5*10^-9 - Q_2
    ----> sub that in for Q_1 so that 5*10^-9 - Q_2 = C_1V
    ----> solve that eq for Q_2 so that Q_2 = -C_1V + 5*10^-9
    ----> sub that into the 'Q_2 = ...' eq and then solve that so that finally

    C_2 = (-C_1V + 5*10^-9)/V

    did i calculate for C_2 correctly?
     
    Last edited: Jun 7, 2008
  9. Jun 7, 2008 #8

    G01

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    In the part where the battery is connected, it is V is 50 volts. In the second part, if one capacitor has 35 volts on it, the other must have 35 volts on it, as nrqed said.

    You equation for C_2 looks good to me.
     
  10. Jun 7, 2008 #9
    so for solving for C_2 using the equation above, i should use V=35volts, C_1 = 100 picocoulombs? and get:

    C_2 = (-C_1V + 5*10^-9)/V = [(100*10^-12)(35) + 5*10^-9]/35 = 2.43*10^-10 farads?

    correct now?
     
  11. Jun 7, 2008 #10

    alphysicist

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    A minor point: I think you dropped the minus sign in front of C_1V.
     
  12. Jun 8, 2008 #11
    okay, other than the sign, my calculation is correct?

    thanks for your help
     
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