The Cell Membrane as an Electrical Capacitor

This section explains how we were able to calculate the number of charges necessary to produce the equilibrium membrane potential of -58 mV in the preceding section. The calculation was made by treating the barrier between the two compartments as an electrical capacitor, which is a charge-storing device consisting of two conducting plates separated by an insulating barrier. In Figure 4-2, the two conducting plates are the salt solutions in the two compartments, and the barrier is the insulator. In a real cell, the ICF and ECF are the conductors, and the lipid bilayer of the plasma membrane is the insulating barrier. When a capacitor is hooked up to a battery as shown in Figure 4-3, the voltage of the battery causes electrons to be removed from one conducting plate and to accumulate on the other plate. This will continue until the resulting Figure 4-3 When a battery is connected to a capacitor, charge accumulates on the capacitor until the voltage across the capacitor is equal to the voltage of the battery.

voltage gradient across the capacitor is equal to the voltage of the battery. Basic physics tells us that the amount of charge, q, stored on the capacitor at that time will be given by q = CV, where V is the voltage of the battery and C is the capacitance of the capacitor. A capacitor's capacitance is directly proportional to the area of the plates (bigger plates can store more charge) and inversely proportional to the distance separating the two plates. Capacitance also depends on the characteristics of the insulating material between the plates; in the case of cells, that insulating material is the lipid plasma membrane. The unit of capacitance is the farad (F): a 1 F capacitor can store 1 coulomb of charge when hooked up to a 1 V battery. Biological membranes, like the plasma membrane, have a capacitance of 10-6 F (that is, 1 microfarad, or |F) per cm2 of membrane area.

If the barrier in Figure 4-2 were 1 cm2 of cell membrane, it would therefore have a capacitance of 10-6 F. From q = CV, it follows that an equilibrium potential of -58 mV would store 5.8 x 10-8 coulomb of charge on the barrier. Note that the charge on the membrane barrier in Figure 4-2 is carried by ions, not by electrons as in Figure 4-3. Thus, to know the total number of excess anions on side 2 of the barrier at equilibrium, we must convert from coulombs of charge to moles of ion. This can be done by dividing the number of coulombs on the barrier by Faraday's constant (approximately 105 coulombs per mole of monovalent ion), yielding 5.8 x 10-13 mole or about 3.5 x 1011 chloride ions moving from side 1 to side 2 in Figure 4-2. If the volume of each compartment were 1 ml, then side 2 would contain about 6 x 1020 chloride and sodium ions. These leads to the conclusion stated in the previous section that less than one-billionth of the chloride ions in side 1 cross to side 2 to produce the equilibrium voltage across the barrier.

0 0

Responses

• marina
What is capacitor and conductor in cell membrane?
7 years ago