Cell membranes across a wide variety of organisms have a capacitance per unit area of$\mathbf{1}\mathit{\mu }\mathit{F}\mathbf{/}\mathit{c}{\mathit{m}}^{\mathbf{2}}$. For the electrical signal in a nerve to propagate down the axon, the charge on the membrane “capacitor” must change. What time constant is required when the ion channels are open? $\left(a\right)\mathbf{}\mathbf{1}\mathit{\mu }\mathit{s}\mathbf{;}\left(b\right)\mathbf{}\mathbf{10}\mathit{\mu }\mathit{s}\mathbf{;}\mathbf{\left(}\mathit{c}\mathbf{\right)}\mathbf{}\mathbf{100}\mathit{\mu }\mathit{s}\mathbf{;}\mathbf{}\mathbf{\left(}\mathit{d}\mathbf{\right)}\mathbf{}\mathbf{1}\mathit{\mu }\mathit{s}$

Now, it is given that the channels are connected in parallel. Now, for n identical resistor R in parallel, the equivalent resistance is:

$\frac{1}{R_{eq}}=\frac{1}{R}+\frac{1}{R}+......=\frac{n}{R}$

Or,$R_{eq}=\frac{R}{n}$

Now, the current flow in the cell membrane is given by:

$l=jA$

Where the cross-sectional area is given by A , and now from the ohms law we get:

$l=\frac{V}{R_{eq}}$

Substituting the value we get:

$jA=\frac{nV}{R}$

Now, writing for the density of open ion channel in the membrane is$n/A$ :

$\frac{n}{A}=\frac{jR}{V}$

Substituting the values$j=5mA/c{m}^2,V=50V$ and the resistance$R=1.0\times {10}^{1}1\Omega$ and get:

$$\begin{gathered} \frac{n}{A}=\frac{5\times 1.0\times {10}^{1}1}{50} \\ ={10}^{1}1/c{m}^2 \end{gathered}$$

The cell membrane across a wide variety or organism has a capacitance per unit area of . The capacitance of the membrane is the capacitance per area divided by the number density of channels that is:

$$\begin{gathered} C=\frac{1\mu F/c{m}^2}{{10}^{1}1/c{m}^2}=1.0\times {10}^{-10}\mu F \\ =1.0\times {10}^{-10}\mu F \end{gathered}$$

Therefore, the capacitance is$1.0\times {10}^{-10}\mu F$