Explanation Given Info : The thickness of axon wall membrane is d = 1.0 × 10 − 8 m , dielectric constant of cell wall is k = 3.0 and axon radius is r = 1.0 × 10 1 μm , From the problem 18.44, the capacitance of the parallel plate capacitor is 1.67 × 10 − 8 F and the potential difference of the excited state is 3.0 × 10 − 2 V , From the problem 18.43 and 18.44 the capacitance of the parallel plate capacitor is 1.67 × 10 − 8 F and starting potential is 7.0 × 10 − 2 V , The input energy required to return to the resting state is 4 .1 × 10 − 11 J and input energy required to charge the capacitor to the excited state is 7.5 × 10 − 12 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 = 2 π r L 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 × 10 1 μm for r and 0.10 m for L . A = 2 ( 3.14 ) ( 1.0 × 10 1 μm ) ( 10 − 6 m 1 μm ) ( 0.10 m ) = 0.628 × 10 − 5 m 2 Formula to calculate the capacitance of the parallel plate capacitor for model axon, C = k ε 0 A d C is the capacitance of the parallel plate capacitor, k is the dielectric constant, ε 0 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 × 10 − 5 m 2 for A , 8.85 × 10 − 12 C 2 / N ⋅ m 2 for ε 0 , 3.0 for k and 1.0 × 10 − 8 m for d to find C . C = ( 3.0 ) ( 8.85 × 10 − 12 C 2 / N ⋅ m 2 ) ( 0.628 × 10 − 5 m 2 ) ( 1.0 × 10 − 8 m ) = 1.67 × 10 − 8 F Thus, the capacitance of the parallel plate capacitor is 1.67 × 10 − 8 F . Formula to calculate energy input required to charge the capacitor to the potential difference of the excited state is, W 1 = 1 2 C Δ V f 2 W 1 is the energy input required to charge the capacitor, Δ V f is the potential difference of the excited state, Substitute 1.67 × 10 − 8 F for C and 3.0 × 10 − 2 V for Δ V f

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ISBN: 9781305952300

Author: Raymond A. Serway, Chris Vuille

Publisher: Cengage Learning

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Chapter 18, Problem 45P

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The power supplied by the axon.

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Chapter 18, Problem 44P

Chapter 18, Problem 46AP

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Chapter 18 Solutions

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The power supplied by the axon.

Solution Summary: The author explains that the power supplied by the axon is 11nW.

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