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Readme [11.4K]
3 years ago
5

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
Physics
1 answer:
sertanlavr [38]3 years ago
7 0

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.

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comparing the two situations, we can see that the current increased when connected in parallel to a ratio of

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The diagram shows the top view of a 65-kg student at point A on an amusement park ride. The ride spins the student in a horizont
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Answer:

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From the question given above, the following data were obtained:

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