Question

The inner and outer surfaces of a cell membrane carry a negative and positive charge respectively. Because of these charges, a potential difference of about 70 mV exists across the membrane. The thickness of the membrane is 8 nm. If the membrane were empty (filled with air), what would the magnitude of the electric field inside the membrane

Answer 1

Answer:

The magnitude of the electric field inside the membrane is 8.8×10⁶V/m

Explanation:

The electric field due to electric potential at a distance Δs is given by

E=ΔV/Δs

We have to find the magnitude electric field in the membrane

Ecell= -ΔV/Δs

$E_{cell}=-\frac{V_{in}-V_{out}}{s} \\E_{cell}=\frac{V_{out}-V_{in}}{s}\\E_{cell}=\frac{0.070V}{8*10^{-9}m } \\E_{cell}=8.8*10^{6}V/m$

The magnitude of the electric field inside the membrane is 8.8×10⁶V/m