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Energy stored in capacitance

  1. Apr 12, 2009 #1
    1. The problem statement, all variables and given/known data

    The giant axon of a squid is 0.5 mm in diameter, 10 cmlong, and not myelinated. Unmyelinated cell membranes behave as capacitors with 1 microF capacitance per square centimeter of membrane area.
    When the axon is charged to the -70 mV resting potential, what is the energy stored in this capacitance?


    2. Relevant equations

    R = (density* length)/Area

    3. The attempt at a solution

    I know how to get R and C is given. and V is also given.
    but how do i solve for E?
     
  2. jcsd
  3. Apr 12, 2009 #2

    Redbelly98

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    I have a few comments:

    1. R is irrelevant here.

    2. The capacitance is not 1 microF, in case that is what you are thinking. Read what it says carefully.

    3. Your text book should have a discussion of energy stored in a capacitor, along with an equation.
     
  4. Apr 12, 2009 #3
    Energy stored in a capacitor, U = 1/2(Q^2/C) = 1/2(Q/V) = 1/2 CV^2
    If we know any two of Q,C, or V we can figure out the energy.
    C is 1 microF per square centimeters. we can get the area using A=(3.14)r^2 and get the C from there.
    and the V is given -70 mV.
    so we can use V and C to get U.
    right?
     
  5. Apr 13, 2009 #4

    Redbelly98

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    Almost. Everything except for A is correct.

    The area is a cylinder, not a circle. Use the surface area of a cylinder to get A, and you'll have it.
     
  6. Apr 13, 2009 #5
    so area for cylinder is, A = 2 pi r ( r + h ).
    = 2(3.14)(2.5*10^-4){(2.5*10^-4)+0.1} m^2
    = 1.57*10^-4 m^2
    = 1.57 cm^2
    so C = (1*10^-6 F)(1.57 cm^2)
    = 1.57*10^-6 F
    and V = -70 mV = -70*10^-3 V

    Energy, E = 1/2 CV^2 = 1/2 (1.57*10^-6 F)(-70*10^-3 V)^2
    = - 4 *10^-9 J

    Does that look right?
     
  7. Apr 13, 2009 #6

    Redbelly98

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    Looks good except that energy is positive. :smile:

    (-70 mV)^2 = +4900 mV^2 > 0​
     
  8. Apr 13, 2009 #7
    oh ya. thanks
     
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