In: Physics

A cell membrane has a resistance and a capacitance and thus a characteristic time constant.

What is the time constant of a 9.0-nm-thick membrane surrounding a 0.040-mm-diameter spherical cell? Assume the resistivity of the cell membrane as 3.6×107 ??m and the dielectric constant is approximately 9.0.

Solutions

Expert Solution

Concepts and reasons The concepts required to solve the problem are resistance, capacitance, and the time constant of an RC circuit. First, using the area and thickness, find the resistance and capacitance of the cell membrane. Then, use the relation between resistance and capacitance, and find the time constant of the cell membrane.

Fundamentals

“The capacitance of a capacitor is directly proportional to the dielectric constant and the area, and is inversely proportional to the thickness". The expression for the capacitance of parallel plate capacitor is, $$C=\frac{k \varepsilon_{0} A}{d}$$

Here, $$k$$ is the dielectric constant, $$\varepsilon_{0}$$ is the permittivity of free space, $$A$$ is the area, and $$d$$ is the separation distance of the plate. The resistance of a material is directly proportional to the length and inversely proportional to the area. The expression for the resistance of the material is, $$R=\frac{\rho L}{A}$$

Here, $$\rho$$ is the resistivity, $$L$$ is the length, $$\mathrm{R}$$ is the resistance, and $$A$$ is the cross-section. The time constant for an RC circuit is, $$\tau=R C$$

Here, $$R$$ is the resistance, and $$C$$ is the capacitance.

The area of the cell membrane is, $$A=4 \pi r^{2}$$

Here, $$r$$ is the radius of the cell.

Replace $$d / 2$$ for $$r$$ in the above equation. $$A=4 \pi\left(\frac{d}{2}\right)^{2}$$

Here, $$\mathrm{d}$$ is the diameter of the cell. Substitute $$0.040 \mathrm{~mm}$$ for $$d$$. $$A=4 \pi\left[\left(\frac{0.040 \mathrm{~mm}}{2}\right)\left(\frac{10^{-3} \mathrm{~m}}{1 \mathrm{~mm}}\right)\right]^{2}$$

$$=5.03 \times 10^{-9} \mathrm{~m}^{2}$$

The equation for capacitance is, $$C=\frac{k \varepsilon_{0} A}{d}$$

Substitute $$9.0$$ for $$k, 8.854 \times 10^{-12} \mathrm{~F} / \mathrm{m}$$ for $$\varepsilon_{0}, 5.03 \times 10^{-9} \mathrm{~m}^{2}$$ for $$A$$, and $$9.0 \mathrm{~nm}$$ for $$d$$. $$C=\frac{(9.0)\left(8.854 \times 10^{-12} \mathrm{~F} / \mathrm{m}\right)\left(5.03 \times 10^{-9} \mathrm{~m}^{2}\right)}{(9.0 \mathrm{~nm})\left(\frac{10^{-9} \mathrm{~m}}{1 \mathrm{~nm}}\right)}$$

$$=4.45 \times 10^{-11} \mathrm{~F}$$

The resistance of the cell membrane is, $$R=\frac{\rho L}{A}$$

Substitute $$3.6 \times 10^{7} \Omega \mathrm{m}$$ for $$\rho, 5.03 \times 10^{-9} \mathrm{~m}^{2}$$ for $$A$$ and $$9.0 \mathrm{~nm}$$ for $$L$$. $$R=\frac{\left(3.6 \times 10^{7} \Omega \mathrm{m}\right)(9.0 \mathrm{~nm})\left(\frac{10^{-9} \mathrm{~m}}{1 \mathrm{~nm}}\right)}{5.03 \times 10^{-9} \mathrm{~m}^{2}}$$

$$=6.44 \times 10^{7} \Omega$$

The RC circuit has both a capacitor and resistor. The capacitance of the cell membrane depends on the dielectric constant, the absolute permittivity, and the area of the cross-section of the material. The resistance of the cell membrane is found from the length and the area of the cross-section.

The equation for the time constant is, $$\tau=R C$$

Substitute $$6.44 \times 10^{7} \Omega$$ for $$R$$ and $$4.45 \times 10^{-11} \mathrm{~F}$$ for $$C$$. $$\tau=\left(6.44 \times 10^{7} \Omega\right)\left(4.45 \times 10^{-11} \mathrm{~F}\right)$$

$$=2.87 \times 10^{-3} \mathrm{~s}$$

$$=2.87 \mathrm{~ms}$$

For RC circuits, the time constant is found from the resistance and the capacitance. The value of the time constant is obtained in milliseconds. The time constant is "the time required to charge the capacitor through the resistor from zero to $$63.2 \%$$ of the applied DC voltage".

Thus, the time constant of the membrane is $$2.87 \mathrm{~ms}$$.