   Chapter 18, Problem 45P

Chapter
Section
Textbook Problem

Using Figure 18.29b and the results of Problems 18.43d and 18.44a, find the power supplied by the axon per action potential.

To determine
The power supplied by the axon.

Explanation

Given Info: The thickness of axon wall membrane is d=1.0×108m , dielectric constant of cell wall is k=3.0 and axon radius is r=1.0×101μm , From the problem 18.44, the capacitance of the parallel plate capacitor is 1.67×108F and the potential difference of the excited state is 3.0×102V , From the problem 18.43 and 18.44 the capacitance of the parallel plate capacitor is 1.67×108F and starting potential is 7.0×102V , The input energy required to return to the resting state is 4.1×1011J and input energy required to charge the capacitor to the excited state is 7.5×1012J .

Explanation:

The following figure shows time duration of an action potential.

Formula to calculate the area of the surface of axon membrane is,

A=2πrL

• A is the area of the axon membrane surface,
• R is the axon radius,
• L is the length of the axon membrane,

Substitute 1.0×101μm for r and 0.10m for L .

A=2(3.14)(1.0×101μm)(106m1μm)(0.10m)=0.628×105m2

Formula to calculate the capacitance of the parallel plate capacitor for model axon,

• C is the capacitance of the parallel plate capacitor,
• k is the dielectric constant,
• ε0 is the permittivity of free space,
• A is the area of the axon membrane surface,
• d is the thickness of axon wall membrane,

Substitute 0.628×105m2 for A, 8.85×1012C2/Nm2 for ε0 , 3.0 for k and 1.0×108m for d to find C .

C=(3.0)(8.85×1012C2/Nm2)(0.628×105m2)(1.0×108m)=1.67×108F

Thus, the capacitance of the parallel plate capacitor is 1.67×108F .

Formula to calculate energy input required to charge the capacitor to the potential difference of the excited state is,

W1=12CΔVf2

• W1 is the energy input required to charge the capacitor,
• ΔVf is the potential difference of the excited state,

Substitute 1.67×108F for C and 3.0×102V for ΔVf

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