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?
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.
To find the electric field across the membrane, we need to divide the potential difference by the thickness of the membrane.
The formula for electric field (E) is:
E = ΔV / d
Where:
E is the electric field,
ΔV is the potential difference, and
d is the thickness.
Given:
ΔV = 0.90 V
d = 5.0 nm
We need to convert the thickness from nanometers (nm) to meters (m) because the SI unit for the electric field is volts per meter (V/m).
1 nm = 1 × 10^-9 m
So, d = 5.0 × 10^-9 m
Now we can substitute the values into the formula:
E = 0.90 V / 5.0 × 10^-9 m
Calculating this expression gives us the electric field across the membrane.