Question

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

Answer 1

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

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