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A cell membrane consists of an inner and outer wall separated by a distance of approximately 10nm. Assume that the walls act like a parallel plate capacitor, each with a charge density of 10−5C/m2, and the outer wall is positively charged. Although unrealistic, assume that the space between cell walls is filled with air.

Respuesta :

Answer:

Electric field, [tex]E=1.12\times 10^6\ N/C[/tex]

Explanation:

It is given that,

Distance between inner and outer wall of a cell membrane, d = 10 nm

Charged density of a parallel plate capacitor, [tex]\sigma=10^{-5}\ C/m^2[/tex]

It is assumed to find the electric field between the membranes. We know that the electric field between the parallel plates is given by :

[tex]E=\dfrac{\sigma}{\epsilon_o}[/tex]

[tex]E=\dfrac{10^{-5}}{8.85\times 10^{-12}}[/tex]

E = 1129943.50 N/C

or

[tex]E=1.12\times 10^6\ N/C[/tex]

So, the electric field between the cell membranes is [tex]1.12\times 10^6\ N/C[/tex]. Hence, this is the required solution.

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