If the investigator applies a \(1.0 \mu \mathrm{A}\) peak current at \(40 \mathrm{kHz}\) to a cell with twice the membrane area of the cell noted in the passage, what will be the peak out-of-phase voltage?

A. \(0.32 \mathrm{~V}\)

B. \(0.16 \mathrm{~V}\)

C. \(0.080 \mathrm{~V}\)

D. \(0.040 \mathrm{~V}\)

The capacitance of biological membranes is about \(1.0 \mu \mathrm{F}\) per \(\mathrm{cm}^{2}\) of membrane area, so investigators can determine the surface area of a cell membrane by using intracellular electrodes to measure the membrane's capacitive reactance. An investigator applies a \(1.0 \mu \mathrm{A}\) peak current at \(40 \mathrm{kHz}\) to a cell and measures the peak out-of-phase voltage - that is, the component of the voltage due to the capacitive reactance of the cell membrane-to be \(0.16 \mathrm{~V}\).