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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?

Respuesta :

Answer:[tex]E=1.041\times 10^7 V/m[/tex]

Explanation:

Given

Potential Difference [tex]\Delta V=0.076 V[/tex]

thickness of cell membrane [tex]d=7.30\times 10^{-9} m[/tex]

Electric Field for this Potential is given by

[tex]E=\frac{\Delta V}{d}[/tex]

[tex]E=\frac{0.076}{7.30\times 10^{-9}}[/tex]

[tex]E=0.01041\times 10^9 V/m[/tex]

[tex]E=1.041\times 10^7 V/m[/tex]

             

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