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 0.076 V exists across the membrane. The thickness of the membrane is 8.9 x 10-9 m. What is the magnitude of the electric field in the membrane?

Answer 1

Answer:

E = 8.5 * 10^6 V/m

Explanation:

In general we have the following relation between the Electric Field and the Elecric Potential:

$\int\limits^2_1 {\vec{E} \cdot\vec{dl}} = V_{2} -V_{1}$

Due to the vector nature of the electric filed, we can only know the mean Electric field E across the membrane, and take it out from the integral, that is:

E = (ΔV)/L

Where L is the thickness of the membrane and ΔV is the potential difference.

Therefore:

E = 8.53933*10^6 V/m

rounding to the first tenth:

E = 8.5 * 10^6 V/m