Answer:
Electric field, E = [tex]6.36\times 10^{6}\ V/m[/tex]
Explanation:
Given that,
The potential difference across the membrane, [tex]\Delta V=0.063\ V[/tex]
The thickness of the membrane is, [tex]\Delta x=9.9\times 10^{-9}\ m[/tex]
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 :
[tex]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[/tex]
So, the magnitude of the electric field in the membrane is [tex]6.36\times 10^{6}\ V/m[/tex].
