Question

A 8.0 nm thick cell membrane undergoes an action potential that follows the curve in the figure.

Part A:What is the strength of the electric field inside the membrane just before the action potential?

Part B:What is the strength of the electric field inside the membrane at the peak of the depolarization?

Answer 1

\((a)\)

Thickness of the cell membrane is,

$$ \begin{aligned} t &=8.0 \mathrm{~nm}\left(\frac{1.0 \mathrm{~m}}{10^{9} \mathrm{~nm}}\right) \\ &=8.0 \times 10^{-9} \mathrm{~m} \end{aligned} $$

The electric fiel d inside the membraner is,

$$ \begin{aligned} E &=\frac{V}{t} \\ &=\frac{\left(70 \times 10^{-3} \mathrm{~V}\right)}{\left(8.0 \times 10^{-9} \mathrm{~m}\right)} \\ &=8.75 \times 10^{6} \mathrm{~V} / \mathrm{m} \end{aligned} $$

(b) The magnitude of elelctric field is the voltage differnce divided

by separation,

$$ \begin{aligned} E=& \frac{V}{t} \\ &=\frac{\left(40 \times 10^{-3} \mathrm{~V}\right)}{\left(8.0 \times 10^{-9} \mathrm{~m}\right)} \\ &=5.0 \times 10^{6} \mathrm{~V} / \mathrm{m} . \end{aligned} $$

The direction of eelctric field from high potential to low potential.