Subfield: Electromagnetism

$$

\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:

Electric Slide



Two large parallel plates of area $A$ are placed at $x=0$ and $x=d \ll \sqrt{A}$ in a semiconductor medium. The plate at $x=0$ is grounded, and the plate at $x=d$ is at a fixed potential $-V_{0}$, where $V_{0}>0$. Particles of positive charge $q$ flow between the two plates. You may neglect any dielectric effects of the medium.
Context question:
a. For large $V_{0}$, the velocity of the positive charges is determined by a strong drag force, so that



$$

v=\mu E

$$



where $E$ is the local electric field and $\mu$ is the charge mobility.

i. In the steady state, there is a nonzero but time-independent density of charges between the two plates. Let the charge density at position $x$ be $\rho(x)$. Use charge conservation to find a relationship between $\rho(x), v(x)$, and their derivatives.
Context answer:
$v \frac{\mathrm{d} \rho}{\mathrm{d} x}+\rho \frac{\mathrm{d} v}{\mathrm{~d} x}=0$


Extra Supplementary Reading Materials:

a. For large $V_{0}$, the velocity of the positive charges is determined by a strong drag force, so that



$$

v=\mu E

$$



where $E$ is the local electric field and $\mu$ is the charge mobility.
Context question:
ii. Let $V(x)$ be the electric potential at $x$. Derive an expression relating $\rho(x), V(x)$, and their derivatives. (Hint: start by using Gauss's law to relate the charge density $\rho(x)$ to the derivative of the electric field $E(x)$.)
Context answer:
$$

\frac{d^{2} V}{d x^{2}}=-\frac{\rho}{\epsilon_{0}}

$$


Question: iii. Suppose that in the steady state, conditions have been established so that $V(x)$ is proportional to $x^{b}$, where $b$ is an exponent you must find, and the current is nonzero. Derive an expression for the current in terms of $V_{0}$ and the other given parameters.

Solution:
We have that



$$

\rho \frac{d v}{d x}+v \frac{d \rho}{d x}=0

$$



and now that $v=-\mu \frac{d V}{d x}$, so substituting in Poisson's equation gives us that



$$

\left(\frac{d^{2} V}{d x^{2}}\right)^{2}+\frac{d V}{d x}\left(\frac{d^{3} V}{d x^{3}}\right)=0

$$



Using $V(x)=-V_{0}(x / d)^{b}$ gives



$$

b(b-1) b(b-1)=-b b(b-1)(b-2)

$$



The solution with $b=0$ cannot satisfy the boundary conditions, while $b=1$ has zero current. Assuming $b$ is neither of these values, we have $b-1=-(b-2)$, so $b=3 / 2$. Substituting gives



$$

v=-\frac{3 V_{0} \mu x^{1 / 2}}{2 d^{3 / 2}}

$$



and



$$

\rho=-\frac{3 V_{0} \epsilon_{0}}{4 d^{3 / 2} x^{1 / 2}}

$$



so



$$

I=\rho A v=\frac{9 \epsilon_{0} \mu A V_{0}^{2}}{8 d^{3}}

$$



with the current flowing from left to right.

Final answer: $I=\frac{9 \epsilon_{0} \mu A V_{0}^{2}}{8 d^{3}}$