Subfield: Modern Physics

$$

\begin{array}{ll}

g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\

k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\

c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\

N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\

\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\

1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\

m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\

\sin \theta \approx \theta-\theta^{3} / 6 \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\theta^{2} / 2 \text { for }|\theta| \ll 1

\end{array}

$$

Extra Supplementary Reading Materials:

## Strain in the Membrane ${ }^{2}$

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The wall of a neuron is made from an elastic membrane, which resists compression in the same way as a spring. It has an effective spring constant $k$ and an equilibrium thickness $d_{0}$. Assume that the membrane has a very large area $A$ and negligible curvature.



The neuron has "ion pumps" that can move ions across the membrane. In the resulting charged state, positive and negative ionic charge is arranged uniformly along the outer and inner surfaces of the membrane, respectively. The permittivity of the membrane is $\epsilon$.
Context question:
a. Suppose that, after some amount of work is done by the ion pumps, the charges on the outer and inner surfaces are $Q$ and $-Q$, respectively. What is the thickness $d$ of the membrane?
Context answer:
\boxed{$d=d_{0}-\frac{Q^{2}}{2 \epsilon A k}$}


Context question:
b. Derive an expression for the voltage difference $V$ between the outer and inner surfaces of the membrane in terms of $Q$ and the other parameters given.
Context answer:
\boxed{$V=\frac{Q}{\epsilon A}\left(d_{0}-\frac{Q^{2}}{2 \epsilon A k}\right)$}


Context question:
c. Suppose that the ion pumps are first turned on in the uncharged state, and the membrane is charged very slowly (quasistatically). The pumps will only turn off when the voltage difference across the membrane becomes larger than a particular value $V_{\text {th }}$. How large must the spring constant $k$ be so that the ion pumps turn off before the membrane collapses?
Context answer:
\boxed{$\left(\frac{3}{2}\right)^{3} \frac{V_{\mathrm{th}}^{2} \epsilon A}{d_{0}^{3}}$}


Extra Supplementary Reading Materials:

d. How much work is done by the ion pumps in each of the following situations? Express your answers in terms of $k$ and $d_{0}$.

Question: i. $k$ is infinitesimally larger than the value derived in part (c).

Solution:
If $k$ is larger than the value in part (c), then the threshold voltage $V_{\mathrm{th}}>V_{\max }$, and the ion pumps turn off before the membrane thickness $d$ reaches zero. The work $W$ done by the ion pumps is equal to the potential energy of the system relative to the uncharged state (with $Q=0$ and $d=0$ ). That is, $W=\frac{1}{2} k x^{2}+Q^{2} /(2 C)$, where $C=\epsilon A / d$ is the capacitance of the membrane. Writing this equation in terms of $Q$ gives $W=Q^{2} d_{0} /(2 \epsilon A)-Q^{4} /\left(8 \epsilon^{2} A^{2} k\right)$.



$k$ being infinitesimally larger than the critical value means that the ion pumps turn off just as the voltage maximum $V_{\mathrm{th}}$ is reached. At this point the charge $Q$ approaches







$Q_{\mathrm{Vmax}}$, derived in the previous answer. Inserting the value into the expression for $W$ gives



$$

W=\frac{5}{18} k d_{0}^{2}

$$

Final answer: $W=\frac{5}{18} k d_{0}^{2}$