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 ${ }^{2}$

<img_4543>


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{$V=\frac{Q}{\epsilon A}\left(d_{0}-\frac{Q^{2}}{2 \epsilon A k}\right)$}


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?

Solution:
The voltage $V$ first increases and then decreases as a function of $Q$, which implies that there is a maximum voltage $V_{\max }$ to which the membrane can be charged. This voltage can be found by taking the derivative $d V / d Q$ and setting it equal to zero. This procedure gives



$$

V_{\max }=\sqrt{\frac{k d_{0}^{3}}{\epsilon A}}\left(\frac{2}{3}\right)^{3 / 2}

$$



The corresponding charge at the maximum voltage is given by



$$

Q_{\mathrm{V} \max }^{2}=\frac{2}{3} \epsilon A k d_{0}

$$



For the ion pumps to turn off, we must have $V_{\max }>V_{\mathrm{th}}$. Otherwise the pumps will continue to move charge across the membrane until it collapses. Setting $V_{\max }>V_{\text {th }}$ and solving for $k$ gives



$$

k>\left(\frac{3}{2}\right)^{3} \frac{V_{\mathrm{th}}^{2} \epsilon A}{d_{0}^{3}}

$$

Final answer: $\left(\frac{3}{2}\right)^{3} \frac{V_{\mathrm{th}}^{2} \epsilon A}{d_{0}^{3}}$