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* 10^7 V/m`

Step-by-step explanation:

Given

Potential Difference
`\Delta V=0.076 V`

thickness of cell membrane
`d=7.30* 10^(-9) m`

Electric Field for this Potential is given by

`E=(\Delta V)/(d)`

`E=(0.076)/(7.30* 10^(-9))`

`E=0.01041* 10^9 V/m`

`E=1.041* 10^7 V/m`