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

i think the answer would be b

another example of kinetic energy would be,

kinetic energy   >     moving energy

answer   >   letter choice (c),   a hammer falling off a table.

hope that !                                         : )