If the width of the plasma membrane is approximately 5 nm, and the electric field is constant throughout that width, what is the electric force acting on a sodium cation as it passes from the extracellular space to the intracellular space through a leaky sodium channel at resting potential? (The charge of an electron is approximately 1.602 * 10-19 C.)

The answer to the question is

The electric force acting on a sodium cation as it passes from the extracellular space to the intracellular space through a leaky sodium channel at resting potential is ≈ 2.2*10⁻¹² C

Explanation:

To solve the question, we note the given variables

Width of plasma membrane = 5 nm

Electric field = Constant throughout the width

Resting potential of sodium channel

Electron charge = 1.602*10⁻¹⁹ C

Assumptions

We take the sodium cation as bieng equivalent to one electron

Solution

Change in voltage = Electric field srrength * Distance moved in the field

dV = E * d

Where dV = Change in voltage = Resting potential of sodium channel = - 70 mV

d = distance moved in the electric field = 5 nm

E Electric Field Strength

Therefore E = dV/d = - 70 mV / 5 nm = (7*10⁻² V) / (5*10⁻⁹ m) = 14*10⁵ N/C

The force F acting on a charge Q in a electric field E is given by

F = E * Q

Where Q = Charge of an electron = 1.602 * 10⁻¹⁹ C

Therefore

F = 14*10⁵ N/C * 1.602 * 10⁻¹⁹ C = 2.2428 * 10⁻¹² C ≈ 2.2*10⁻¹² C

F ≈ 2.2*10⁻¹² C