Answer:
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
Step-by-step 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