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).

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.

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.