Question

Cell Membranes and Dielectrics Many cells in the body have a cell membrane whose inner and outer surfaces carry opposite charges, just like the plates of a parallel-plate capacitor. Suppose a typical cell membrane has a thickness of 8.5×10⁻⁹ m , and its inner and outer surfaces carry charge densities of -6.3×10⁻⁴ C/m2 and +6.3×10⁻⁴ C/m2 , respectively. In addition, assume that the material in the cell membrane has a dielectric constant of 5.5.

Calculate the potential difference between the inner and outer walls of the membrane.Indicate which wall of the membrane has the higher potential.

Answer 1

Let the thickness of the cell membrane is t = 8.5 × 10^-9 m

Let the surface charge density = 6.3 × 10^-4 C/m²

Dielectric constant k = 5.5

Permittivity of free space ₀ = 8.85 × 10^-12 C²/N m²

From the Gauss law ,

E . ds = q /   

Where is the permittivity of the medium

E . S = q /

E = q / S

We know that q / S = , the surface charge density.

So we get the formula for electric field

E = /

E = / k ₀

A ) For a parallel plate capacitor potential difference between its plates is ∆V = E × d

Similarly the potential difference between the inner and outer walls of the membrane is

∆V = E × t

= ( / k ₀ ) × t

= ( 6.3 × 10^-4 ) × ( 8.5 × 10^-9 ) / ( 5.5 × 8.85 × 10^-12 )

= ( 53.55 × 10^-13 ) / ( 48.67 × 10^-12 )

= 1.100267 × 10^-1

= 110.0267 × 10^-3 V

= 110.0267 mV

So the potential difference between the inner and outer walls of the membrane is ∆V = 110.0267 mV

B) Region of higher potential is the one with more number of positive charges. Since the outer surface of the membrane has positive surface charge density , it is at higher potential and inner wall is at lower potential.