Question

Cell membranes (the walled enclosure around a cell) are typically about 7.5 nm thick. They are partially permeable to allow charged material to pass in and out, as needed. Equal but opposite charge densities build up on the inside and outside faces of such a membrane, and these charges prevent additional charges from passing through the cell wall. We can model a cell membrane as a parallel-plate capacitor, with the membrane itself containing proteins embedded in organic material to give the membrane a dielectric constant of about 10. (See Fig..)

(a) What is the capacitance per square centimeter of such a cell wall?

(b) In its normal resting state, a cell has a potential difference of 85 mV across its membrane. What is the electric field inside this membrane?

Short answer

a) $1.18\times {10}^{-6}F$

b)$1.133x{10}^{6}N/m$

Step-by-step solution

Concept

$C=\frac{Q}{V_{ab}}={\epsilon }_0\frac{A}{d}$

C =The Capacitance, $V_{ab}$=The potential difference between the plates, Q= Magnitude of charge on each plate, d=Distance between plates, localid="1668318045119" ${\epsilon }_o=8.85\times {10}^{-12}F/m$electricconstant, A= Area of each plate.

Definition of capacitance: The capacitance C of a capacitor is the ratio of the magnitude of the charge Q on either conductor to the magnitude of the potential difference between the conductors. The capacitance depends only on the geometry of the capacitor. The greater the capacitance C of a capacitor, the greater the magnitude Q of charge on either conductor for a given potential difference V and hence the greater the amount of stored energy.

Apply: we have a bunch of variables in the above equations. So, in most problems, we are asked to get one of them by the above equation either directly by substitution or by solving the equation indirectly for this variable.

Dielectric constant:

$K=\frac{C}{C_0}$

K= Dielectric constant, C = Final capacitance, $C_0$= original capacitance

Calculate the capacitance per square centimeter.

$\begin{array}{c}{C}_0=\frac{Q}{V}=\frac{{\epsilon }_{0}A}{d}\\ K=\frac{C}{C_0},\\ C=K{C}_0=K\frac{{\epsilon }_{0}A}{d}\end{array}$

$A=1c{m}^2={10}^4{m}^2\to$as we are told to get the to get the capacitance per square centimeter

$\mathrm{c}=\left(0.01180\mathrm{x}{10}^{-4}\right)\mathrm{F}=1.18\mathrm{x}{10}^{-6}\mathrm{F}$

Calculate the electric field inside the membrane.

$\begin{array}{r}V=KEd\\ E=\frac{V}{Kd}=\frac{85\times {10}^{-3}\mathrm{V}}{10\times 7.5\times {10}^{-9}\mathrm{m}}=1.133\times {10}^6\mathrm{N}/\mathrm{m}\end{array}$