The input end of a human nerve cell is connected to an outpu

The input end of a human nerve cell is connected to an output end by a long, thin, cylindrical axon. A signal at the input end is caused by a stretch sensor, a temperature sensor, contact with another cell or nerve, or some other stimulus. At the output end, the nerve signal can stimulate a muscle cell to perform a function (to contract, provide information to the brain, etc.).
The axon of a so-called unmyelinated human nerve cell has a radius of 5×106m and a membrane that is 6×109m thick. The membrane has a resistivity of about 1.6×107m. The fluid inside the axon has resistivity of about 0.5 m. The membrane wall has proteins that pump three sodium ions (Na+) out of the axon for each two potassium ions (K+) pumped into the axon. In the resting axon, the concentration of these ions results in a net positive charge on the outside of the membrane compared to negative charge on the inside. Because of the unequal charge distribution, there is a 70 mV potential inside compared to outside the axon.
When an external source stimulates the input end of the nerve cell so the potential inside reaches about 50 mV, gates or channels in the membrane walls near that input open and sodium ions rush into the axon. This stimulates neighboring gates to swing open and sodium ions rush into the axon farther along. This disturbance quickly travels along the axon-a nerve impulse. The potential across the inside of the membrane changes in 0.5 ms from 70 mV to +30 mV relative to the outside. Immediately after this depolarization, potassium ion gates open and positively charged potassium ions rush out of the axon, repolarizing the axon. Sodium and potassium ion pumps then return the axon and its membrane to their original configuration.

Find the magnitude of the E  field in the resting axon membrane.

Express your answer to one significant figure and include the appropriate units.

Solution

for a resting axon, the potential difference = -70 mV

The thickness of the membrane = 6x10^-9 m. thus, the electric field is,

E = -dV/dx = -(-70x10^-3V)/(6x10^-9m) = 1.17 x 10⁷ V/m = 1x 10⁷ V/m