CHARGING THE MEMBRANE CAPACITANCE

 

(to be read in association with Membrane Potentials for further explanation of how the membrane currents alter the charge across the membrane)

The following are familiar equations from the electricity part in your physics courses.

V is voltage difference in volts, Q is electrical charge in coulombs, C is capacitance in farads, and J is current density or electrical flux in coulombs per seconds per square meter.

V = Q / C,       dV/dt = 1/C dQ/dt,     dQ/dt = J       

In which the units of J are coul @ s-1 @ m-2 and those of C are F @ m-2

The membrane current density is the sum of all the ionic current densities

                JM = E Ji = E - Gi ( VM - Vi )                  Eq 1

                     Eq 2

For the case of a single ion, as potassium,     

                       

                                                 Eq   3

which is of the form y' + Py = Q,   so its solution is 

       

with I = IPdx.

In order to apply this solution to our equation 3 we substitute

 P = GK / C and Q = (GK / C) @VK

Then the solution is

               

Make G /C = "

               

               

The initial condition is that VM = 0 when t = 0. Then, 0 = VK + c  ˆ c = - VK.

                       

which is the equation for an increasing exponential curve with a time constant J = 1/ " = C / GK.  At t = 0  VM = 0, and at t = 4  VM = VK. The figure below shows the time course of VM. The curve should be such that at t = 1J  VM = 0.632 VK. This is so because at t = 1/ "

VM = VK (1 - e -1) = VK (1 - 1/e) = VK (1 - 1/ 2.71828..). Similarly, when t = 3J VM = 0.95 VK and when t = 5J VM = 0.99 VK. 

 

 

 

 

 

 

 

 

 

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