Chapter 18, Problem 45P

### College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300

Chapter
Section

### College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300
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=1.0Ã—10âˆ’8â€‰m , dielectric constant of cell wall is k=3.0 and axon radius is r=1.0Ã—101â€‰Î¼m , From the problem 18.44, the capacitance of the parallel plate capacitor is 1.67Ã—10âˆ’8â€‰F and the potential difference of the excited state is 3.0Ã—10âˆ’2â€‰V , From the problem 18.43 and 18.44 the capacitance of the parallel plate capacitor is 1.67Ã—10âˆ’8â€‰F and starting potential is 7.0Ã—10âˆ’2â€‰V , The input energy required to return to the resting state is 4.1Ã—10âˆ’11â€‰J and input energy required to charge the capacitor to the excited state is 7.5Ã—10âˆ’12â€‰J .

Explanation:

The following figure shows time duration of an action potential.

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

A=2Ï€rL

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

Substitute 1.0Ã—101â€‰Î¼m for r and 0.10â€‰m for L .

A=2(3.14)(1.0Ã—101â€‰Î¼m)(10âˆ’6â€‰m1â€‰Î¼m)(0.10â€‰m)=0.628Ã—10âˆ’5â€‰m2

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,
• Îµ0 is the permittivity of free space,
• A is the area of the axon membrane surface,
• d is the thickness of axon wall membrane,

Substitute 0.628Ã—10âˆ’5â€‰m2 for A, 8.85Ã—10âˆ’12â€‰C2/Nâ‹…m2 for Îµ0 , 3.0 for k and 1.0Ã—10âˆ’8â€‰m for d to find C .

C=(3.0)(8.85Ã—10âˆ’12â€‰C2/Nâ‹…m2)(0.628Ã—10âˆ’5â€‰m2)(1.0Ã—10âˆ’8â€‰m)=1.67Ã—10âˆ’8â€‰F

Thus, the capacitance of the parallel plate capacitor is 1.67Ã—10âˆ’8â€‰F .

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

W1=12CÎ”Vf2

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

Substitute 1.67Ã—10âˆ’8â€‰F for C and 3.0Ã—10âˆ’2â€‰V for Î”Vf

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