The inner and outer surfaces of a cell membrane carry a negative and positive charge, respectively. Because of these charges, a potential difference of about 0.070 V exists across the membrane. The thickness of the membrane is 0.000000008 m. What is the magnitude of the electric field in the membrane?
To find the magnitude of the electric field in the membrane, we can use the formula:
Electric field (E) = Potential difference (V) / Distance (d)
Given:
Potential difference (V) = 0.070 V
Distance (d) = 0.000000008 m
Substituting these values into the formula, we get:
E = 0.070 V / 0.000000008 m
To simplify the calculation, it's easier to convert the distance from meters to nanometers, as the result will be in volts per nanometer.
1 meter = 1,000,000,000 nanometers
0.000000008 m = 0.000000008 * 1,000,000,000 nm = 8 nm
Now, we can rewrite the formula as:
E = 0.070 V / 8 nm
Calculating further:
E = 0.00875 V/nm
Therefore, the magnitude of the electric field in the membrane is 0.00875 volts per nanometer.