If the capacitance of a cell membrane is measured to be \(6.0 \times 10^{-11} \mathrm{~F}\), what is the area?

A. \(6.0 \times 10^{-13} \mathrm{~m}^{2}\)

B. \(6.0 \times 10^{-11} \mathrm{~m}^{2}\)

C. \(6.0 \times 10^{-9} \mathrm{~m}^{2}\)

D. \(6.0 \times 10^{-7} \mathrm{~m}^{2}\)

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