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Colt1911 [192]
3 years ago
9

The electric potential difference across the membrane of a body cell is 0.070 V (higher on the outside than on the inside). The

cell membrane is 8.0 x 10-9 m thick. Determine the magnitude and direction of the E field through the cell membrane. Describe any assumptions you made
Physics
1 answer:
12345 [234]3 years ago
5 0

Answer:

The electric field is 8.75 \times 10^{6}~v~m^{-1} and the ditection is from outer to inner side of the membrane.

Explanation:

We know the electric field (\vec{E}) is given by \vec{E} = - \nabla V, 'V' being the potential.

In 1-D, it can be written as

E=\dfrac{V}{d}

where 'd' is the separation of space in between the potential difference is created.

Given, V = 0.070~V~ and the thickness of the cell membrane is d = 8.0 \times 10^{-9}~m.

Therefore the created electric field through the cell membrane is

E = \dfrac{0.07~V}{8 \times 10^{-9}~m} = 8.75 \times 10^{6}~m~s^{-1}

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