Energy stored in capacitance of a squid

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The discussion centers on calculating the energy stored in the capacitance of a squid's giant axon, which has specific dimensions and a resting potential of -70 mV. The capacitance is derived from the surface area of the axon, calculated as a cylinder rather than a circle, leading to a capacitance of approximately 1.57 microfarads. The energy stored in the capacitor is computed using the formula E = 1/2 CV^2, resulting in a value of 4 nanojoules. A correction is noted regarding the sign of the energy, confirming it should be positive. The final calculations and methodology are affirmed as correct.
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Homework Statement



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?


Homework Equations



R = (density* length)/Area

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?
 
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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 textbook should have a discussion of energy stored in a capacitor, along with an equation.
 
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?
 
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.
 
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?
 
Looks good except that energy is positive. :smile:

(-70 mV)^2 = +4900 mV^2 > 0​
 
oh ya. thanks
 
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