Energy stored in capacitance of a squid

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Homework Help Overview

The problem involves calculating the energy stored in the capacitance of a squid's giant axon, which is characterized by its dimensions and the properties of unmyelinated cell membranes acting as capacitors. The capacitance is given per unit area, and the resting potential is specified.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relevance of resistance in the context of capacitance and energy storage. There are attempts to clarify the capacitance value and its calculation based on the area of the axon. Questions arise regarding the correct formula for energy stored in a capacitor and the proper area calculation for a cylindrical shape.

Discussion Status

Participants are actively engaging with the problem, correcting each other's misunderstandings about capacitance and area calculations. Some have provided guidance on using the correct formula for the surface area of a cylinder and the implications for energy calculations. There is no explicit consensus on the final answer, but the discussion is moving towards a resolution.

Contextual Notes

There are indications of confusion regarding the capacitance value and the assumptions about the geometry of the axon. The discussion reflects a focus on ensuring accurate calculations based on the physical properties described in the problem statement.

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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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