Question

A cell membrane has a surface area of \(1.0 \times 10^{-7} \mathrm{m}^{2},\) a dielectric constant of \(5.2,\) and a thickness of \(7.5 \mathrm{nm}\) The membrane acts like the dielectric in a parallel plate capacitor; a layer of positive ions on the outer surface and a layer of negative ions on the inner surface act as the capacitor plates. The potential difference between the "plates" is \(90.0 \mathrm{mV}\). (a) How much energy is stored in this capacitor? (b) How many positive ions are there on the outside of the membrane? Assume that all the ions are singly charged (charge +e).

Short answer

Answer: The energy stored in this capacitor is approximately \(2.478 \times 10^{-14}\) Joules, and there are approximately \(3.45 \times 10^8\) positive ions on the outer surface of the membrane.

Step-by-step solution

Part (a)

For a parallel plate capacitor with a dielectric material, the capacitance C can be calculated using the formula: \[C = \frac{\epsilon A}{d}\] where \(\epsilon\) is the permittivity of the dielectric material, A is the surface area, and d is the thickness of the dielectric material. The permittivity of the dielectric material can be found using the formula: \[\epsilon = \epsilon_0 \cdot \epsilon_r\] where \(\epsilon_0\) is the vacuum permittivity (\(8.854 \times 10^{-12}\) F/m) and \(\epsilon_r\) is the dielectric constant. In our case, \(\epsilon_r = 5.2\), A = \(1.0 \times 10^{-7} \mathrm{m}^{2}\), and the thickness d = \(7.5 \times 10^{-9}\) meters. First, we will find the permittivity of the dielectric material. \[\epsilon = (8.854 \times 10^{-12}) (5.2) = 4.603 \times 10^{-11} \mathrm{F/m}\] Now, we will use this value to find the capacitance C. \[C = \frac{(4.603 \times 10^{-11}) (1.0 \times 10^{-7})}{7.5 \times 10^{-9}} = 6.138 \times 10^{-10} \mathrm{F}\] Next, we will find the energy stored in the capacitor. The energy U can be found using the formula: \[U = \frac{1}{2} C V^2\] where V is the potential difference between the plates. In our case, V = 90.0 mV = \(90.0 \times 10^{-3}\) V. Now, we will use this value to find the energy stored in the capacitor. \[U = \frac{1}{2}(6.138 \times 10^{-10})(90.0 \times 10^{-3})^2 = 2.478 \times 10^{-14} \mathrm{J}\] So the energy stored in this capacitor is approximately \(2.478 \times 10^{-14}\) Joules.

Part (b)

In order to find the number of positive ions on the outer surface of the membrane, we will first need to find the charge stored on the capacitor. The charge Q can be found using the formula: \[Q = C \cdot V\] Using the values of C = \(6.138 \times 10^{-10}\) F and V = \(90.0 \times 10^{-3}\) V, we will find the charge Q. \[Q = (6.138 \times 10^{-10})(90.0 \times 10^{-3}) = 5.524 \times 10^{-11} \mathrm{C}\] Now, we will find the number of positive ions on the outer surface by dividing the total charge by the charge of a single positive ion (charge +e). The charge of a single positive ion (charge +e) is approximately \[e = 1.6 \times 10^{-19} C\] So, the number of positive ions on the outer surface can be found using the formula: \[N = \frac{Q}{e}\] Using the values of Q = \(5.524 \times 10^{-11}\) C and e = \(1.6 \times 10^{-19}\) C, we will find the number of positive ions, N. \[N = \frac{5.524 \times 10^{-11}}{1.6 \times 10^{-19}} = 3.4525 \times 10^8\] Therefore, there are approximately \(3.45 \times 10^8\) positive ions on the outside of the membrane.