Question

Because of these charges, a potential difference of about 0.077 V exists across the membrane. The thickness of the membrane is 7.3 x 10^-9 m. What is the magnitude of the electric field in the membrane?

Answer 1

Electric field is given by $E =\frac{V}{d} = \frac{0.077}{7.3*10^{-9}}= 1.05*10^{7}N/C$