Asked by Rony
a potential difference of 0.90 V exists from one side to the other side of a cell membrane that is 5.0 nm thick. What is the electric field across the membrane?
Answers
Answered by
bobpursley
E=Voltage difference/distance in volts/meter
The potential difference you cited is somwhat larger than I have measured, normally, cell potentials in the millivolts exist, and on the order of 200 millivolts, the cell dielectric breaks down, and arcing takes place. So for a construct in physics, the problem is ok, but in reality, .09 volts would have been a better choice.
Now within the cell membrane, the E is much lower, because the dielectric effect of the lipids, the membrane actually acts more as a capacitor than a conductance pathway. Capacitances on the order of 3microFarad/meter^2 are common.
But by any measure, E is high, and certainly facilitates ion transport across cell membranes.
The potential difference you cited is somwhat larger than I have measured, normally, cell potentials in the millivolts exist, and on the order of 200 millivolts, the cell dielectric breaks down, and arcing takes place. So for a construct in physics, the problem is ok, but in reality, .09 volts would have been a better choice.
Now within the cell membrane, the E is much lower, because the dielectric effect of the lipids, the membrane actually acts more as a capacitor than a conductance pathway. Capacitances on the order of 3microFarad/meter^2 are common.
But by any measure, E is high, and certainly facilitates ion transport across cell membranes.
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