Question

The inner and outer surfaces of a cell membrane carry a negative and a positive charge, respectively. Because of these charges, a potential difference of about 0.076 V exists across the membrane. The thickness of the cell membrane is 7.30 10-9 m.
What is the magnitude of the electric field in the membrane?

Answer 1

Answer:$E=1.041\times 10^7 V/m$

Explanation:

Given

Potential Difference $\Delta V=0.076 V$

thickness of cell membrane $d=7.30\times 10^{-9} m$

Electric Field for this Potential is given by

$E=\frac{\Delta V}{d}$

$E=\frac{0.076}{7.30\times 10^{-9}}$

$E=0.01041\times 10^9 V/m$

$E=1.041\times 10^7 V/m$