The inner and outer surfaces of a cell membrane have net negative and positive charges respectivelythese are like sheets of charge. Because of those charges, there would be a net electrical force on charged ions within the membrane, and there's an electric potential difference from one side of the membrane to the other. A typical value for the potential difference is 70 mV, and a typical thickness of the membrane is 8 nm (see Figure 2). Cells can carry ions across a membrane against this potential difference using a va- riety of active transport mechanisms that I don't know much about. But they have to do work on the sodium atom to move it that way. A. Calculate a value for about how much work it takes to move one sodium ion (missing one electron, so its charge is :7 = +1.6 x 1019 Coulombs) from the inside of the membrane to the outside. Moving a single sodium ion from the inside of the membrane to the outside would take that much charge away from the inner surface and add it to the outer surface. That would have a tiny effect on the potential difference, but of course if many sodium ions moved that way the effect could become significant. Because we're moving one bit of positive charge from the inside to the outside, we're making the inside a very tiny bit more negatively charged and the outside a very tiny bit more positively charged. So the potential difference goes up a very tiny bit. B. These ideas lead to our modeling the cell membrane as a capac- itor. Suppose, now, the capacitance of the cell membrane is 104 y F (as you've seen, the Farad is a unit of capacitance, Coulombs per Volt). And the potential difference from one side to the other is 70 mV. How many sodium ions would I have to move across the mem- brane to change the potential difference from 70 mV in one direction to 30 mV in the other direction (i.e. from +70 mV to 30 mV)? C. On what size area of the membrane is charge collecting? Does this make sense to you? Why or why not? Figure 2: Simple model of cell membrane as a capacitor