Question

A potential difference of 0.90 V exists from one side to the other side of a cell membrane that is 4.7 nm thick. What is the electric field across the membrane?

Answer 1

We have

$d=4.7nm=4.7 \cdot 10^{-9} m$

$\Delta\phi=0.9 V$

hen the electric field through the membrane is

$E=\frac{\Delta\phi}{d}=\frac{0.9}{4.7 \cdot 10^{-9}}=1.915 \cdot 10^{8} \frac{V}{m}=1.915 \frac{MV}{sm}$