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Chapter 23: Problem 38

The potential difference across a cell membrane is \(65 \mathrm{mV}\). On the outside are \(1.5 \times 10^{6}\) singly ionized potassium atoms. Assuming an equal negative charge on the inside, find the membrane's capacitance.

Short Answer

Expert verified
The capacitance of the membrane is obtained by dividing the total charge by the potential difference. After inserting the known values, perform the necessary calculations to get the final result for the capacitance.

Step by step solution

01

Calculate the total charge

First, calculate the total charge on the outside. Since each potassium atom is singly ionized, it implies they each carry a charge of \(1.6 \times 10^{-19} C\). Thus, the total charge on the outside is given by multiplying the single charge by the total number of ions, i.e., Q = \(1.5 \times 10^{6} \times 1.6 \times 10^{-19} C.\)
02

Convert Potential from mV to V

The potential difference is given as \(65 mV\). We need to convert this to volts by dividing by 1,000. So, V=\(65 mV / 1000 = 0.065 V.\)
03

Calculate the Capacitance

Finally, use the formula for calculating the capacitance which is defined as the ratio of the charge \(Q\) to the potential difference \(V\). The formula is accordingly: \(C=Q/V\). Plug in the values of \(Q\) and \(V\) we got from the previous steps.

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