   Chapter 18, Problem 44P

Chapter
Section
Textbook Problem

Consider the model of the axon as a capacitor from Problem 43 and Figure P18.43. (a) How much energy does it take to restore the inner wall of the axon to −7.0 × 10−2 V, starting from +3.0 × 10−2 V? (b) Find the average current in the axon wall during this process.

(a)

To determine
The energy restore in the inner wall of 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 , The capacitance of the parallel plate capacitor is 1.67×108F and starting potential is 7.0×102V .

Explanation:

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

(b)

To determine
The average current in the axon wall.

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