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 0.063 V exists across the membrane. The thickness of the membrane is 9.9 x 10-9 m. What is the magnitude of the electric field in the membrane

Answer 1

Answer:

Electric field, E = $6.36\times 10^{6}\ V/m$

Explanation:

Given that,

The potential difference across the membrane, $\Delta V=0.063\ V$

The thickness of the membrane is, $\Delta x=9.9\times 10^{-9}\ m$

We need to find the magnitude of the electric field in the membrane. We know that the relation between electric field and electric potential is given by :

$E=\dfrac{-\Delta V}{\Delta x}\\\\E=\dfrac{0.063\ V}{9.9\times 10^{-9}\ m}\\\\E=6.36\times 10^{6}\ V/m$

So, the magnitude of the electric field in the membrane is $6.36\times 10^{6}\ V/m$.