A typical cell has an electric potential difference across its cell membrane. The electric potential interior to the cell is 70mV less than that on the exterior. Under certain circumstances, the cell can redistribute charge so that the electric potential inside is 40 mV greater than that outside. Assuming the membrane is 12 nm thick and that the net electric field inside it is uniform, how does that field change in the transition from having an interior that is 70 mV less than that on the exterior to having an interior that is 40 mV more than that on the exterior

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adnansoomro2019 adnansoomro2019 · Answer 1 · 2021-02-17T06:29:46+01:00

Answer:

Answer is explained in the explanation section below

Explanation:

Direction of electric field is on decreasing potential:

Now, let vi = potential at interior and Vo = potential at exterior.

So, in first case:

Vo - Vi = 70mV

Vo >Vi

Let Field be E1,

Then,

E1d = 70mV

or

E1 x (12 x [tex]10^{-9}[/tex] ) = 70m V

Solving for E1,

E1 = 5.833 x [tex]10^{6}[/tex] V/m

and the direction is towards interior.

When Vi>Vo, then

Vi - Vo = 40mV

So,

E2d = 40 mV

E2 x (12 x [tex]10^{-9}[/tex] ) = 40 mV

Solving for E2,

E2 = 3.33 x [tex]10^{6}[/tex] V/m

And direction of E2 is from interior to exterior.

In conclusion, the magnitude of field changes from 5.833 x [tex]10^{6}[/tex] V/m to E2 = 3.33 x [tex]10^{6}[/tex] V/m and

direction changes from being towards cell interior to being towards cell exterior.