Editing Reversal potential (section)

===Mathematical models and the driving force===
We can consider as an example a positively charged ion, such as [[potassium|K<sup>+</sup>]], and a negatively charged membrane, as it is commonly the case in most organisms.<ref name="molBiol" /><ref name="theoNeuro" /> The membrane voltage opposes the flow of the potassium ions out of the cell and the ions can leave the interior of the cell only if they have sufficient thermal energy to overcome the energy barrier produced by the negative membrane voltage.<ref name="theoNeuro" /> However, this biasing effect can be overcome by an opposing concentration gradient if the interior concentration is high enough which favours the potassium ions leaving the cell.<ref name="theoNeuro" /> 

An important concept related to the equilibrium potential is the '''driving force'''''.'' Driving force is simply defined as the difference between the actual membrane potential and an ion's equilibrium potential <math>V_\mathrm{m}-E_\mathrm{i}\ </math>where <math>E_\mathrm{i}\ </math>refers to the equilibrium potential for a specific ion.<ref name="theoNeuro" /> Relatedly, the membrane current per unit area due to the type <math>i </math> ion channel is given by the following equation: 

:<math>i_\mathrm{i} = g_\mathrm{i} \left(V_\mathrm{m}-E_\mathrm{i}\right) </math>

where <math>V_\mathrm{m}-E_\mathrm{i}\ </math> is the driving force and <math>g_\mathrm{i} </math> is the [[Electrical resistivity and conductivity|specific conductance]], or conductance per unit area.<ref name="theoNeuro" /> Note that the ionic current will be zero if the membrane is impermeable to that ion in question or if the membrane voltage is exactly equal to the equilibrium potential of that ion.<ref name="theoNeuro" />