267 lines
11 KiB
Markdown
267 lines
11 KiB
Markdown
# Homework 1 - Aidan Sharpe
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## 1.
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If $\vec{A} = 4\hat{x} + 4\hat{y} - 2\hat{z}$ and $\vec{B} = 3\hat{x} - 1.5\hat{y} + \hat{z}$, find the acute angle between $\vec{A}$ and $\vec{B}$.
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Definition of dot product:
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$$\vec{A} \cdot \vec{B} = (A_x B_x)+(A_y B_y) + (A_z B_z) =\lVert \vec{A} \rVert \lVert \vec{B} \rVert \cos(\varphi)$$
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Solve the dot product:
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$$\vec{A} \cdot \vec{B} = (4 \cdot 3) + (4 \cdot (-1.5)) + ((-2) \cdot 1) = 4$$
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Magnitudes of the vectors:
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$$\lVert \vec{A} \rVert = \sqrt{4^2 + 4^2 + (-2)^2} = \sqrt{36} = 6$$
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$$\lVert \vec{B} \rVert = \sqrt{3^2 + \left(-\frac{3}{2}\right)^2 + 1^2} = \sqrt{\frac{49}{4}} = \frac{7}{2}$$
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By the definition of the dot product:
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$$4 = 6 \cdot \frac{7}{2} \cos(\varphi)$$
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$$\therefore \cos(\varphi) = \frac{4}{21}$$
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$$\therefore \varphi = \arccos\left(\frac{4}{21}\right) = 1.379 = 79.02^\circ$$
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## 2
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If $\vec{A} = \frac{10}{\rho}\hat{\rho} + 5\hat{\varphi} + 2\hat{z}$ and $\vec{B} = 5\hat{\rho} + \cos(\varphi)\hat{\varphi} + \rho\hat{z}$, find (a) $\vec{A} \cdot \vec{B}$ and (b) $\vec{A} \times \vec{B}$ at $x=1$, $y=1$, $z=1$.
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Convert from cartesian to cylidrical:
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$$\rho = \sqrt{x^2 + y^2} = \sqrt{2}$$
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$$\varphi = \arctan\left(\frac{y}{x}\right) = \frac{\pi}{4}$$
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$$z = z = 1$$
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Find $\vec{A}$ and $\vec{B}$ at $x=1$, $y=1$, $z=1$:
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$$\vec{A} = \frac{10}{\sqrt{2}}\hat{\rho}+5\hat{\varphi}+2\hat{z} = 5\sqrt{2}\hat{\rho}+5\hat{\varphi}+2\hat{z}$$
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$$\vec{B} = 5\hat{\rho}+\cos\left(\frac{\pi}{4}\right)\hat{\varphi}+\sqrt{2}\hat{z}=5\hat{\rho}+\frac{\sqrt{2}}{2}\hat{\varphi}+\sqrt{2}\hat{z}$$
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### a) Find $\vec{A} \cdot \vec{B}$
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$$\vec{A} \cdot \vec{B} = \left(5\sqrt{2} \cdot 5\right) + \left(5 \cdot \frac{\sqrt{2}}{2}\right) + \left(2 \cdot \sqrt{2}\right) = \frac{59\sqrt{2}}{2}$$
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### b) Find $\vec{A} \times \vec{B}$
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$$\vec{A} \times \vec{B} = \begin{vmatrix} \hat{\rho} & \hat{\varphi} & \hat{z} \\ 5\sqrt{2} & 5 & 2 \\ 5 & \frac{\sqrt{2}}{2} & \sqrt{2} \end{vmatrix} = 4\sqrt{2}\hat{\rho} - 20\hat{z}$$
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## 3
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Two point charges have mass $0.2$g. Two insulating threads of length $1$m are used to suspend the charges from a common point. The gravitational force is $980 \times 10^{-5}\text{N/g}$.
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Define $\varphi$ as the angle between the threads, $\alpha$ as half of that angle, and $\beta$ as $\frac{\pi}{2} - \alpha$. This way, $\alpha$ and $\beta$ make a right triangle with the leg adjacent to $\alpha$ making a perpendicular bisector of the distance between the charges, $r$.
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Find the distance, $r$:
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$$\frac{r}{2} = (1\text{m})\cos(\beta)$$
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$$\therefore r = 2\cos(\beta)$$
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By Coulomb's Law, the electric field force on each charge, $\vec{F}_e$, is:
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$$\vec{F}_e = \frac{q^2}{4\pi\varepsilon_0 r^2}\hat{x}$$
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Pluggin in for $r$,
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$$\vec{F}_e = \frac{q^2}{16\pi\varepsilon_0\cos^2(\beta)}\hat{x}$$
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The force due to gravity on each charge, $\vec{F}_g$, is:
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$$\vec{F}_g = 0.2 \cdot 980\times 10^{-5} = 0.00196(-\hat{y})\text{N}$$
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Assuming static equilibrium:
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$$\sum F_x = 0 = F_{e_x} + F_{T_x} + F_{g_x}$$
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$$\sum F_y = 0 = F_{e_y} + F_{T_y} + F_{g_y}$$
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Where:
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$F_{T_x}$ is the x-component of the force due to tension in each thread, $F_T$.
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$F_{T_y}$ is the y-component of $F_T$.
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Since $F_g$ only acts in the $\hat{y}$ direction, and $F_e$ only acts in the $\hat{x}$ direction:
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$$F_{e_x} + F_{T_x} = 0$$
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$$F_{T_y} + F_{g_y} = 0$$
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Define the components of the tension force $F_T$:
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$$F_{T_x} = F_T\cos\left(\frac{\pi}{2} + \alpha\right) = -F_T\cos(\beta)$$
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$$F_{T_y} = F_T\sin\left(\frac{\pi}{2} + \alpha\right) = F_T\sin(\beta)$$
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Solve for $F_T$ using equilibrium in the $\hat{y}$ direction:
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$$F_T\sin(\beta) - 0.00196 = 0$$
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$$\therefore F_T = \frac{0.00196}{\sin(\beta)}$$
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Plug in $F_T$ to solve equilibrium in the $\hat{x}$ direction:
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$$F_e + \left( \frac{0.00196}{\sin(\beta)} \right)(-\cos(\beta)) = 0$$
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$$\therefore F_e = 0.00196 \cot(\beta)$$
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### a) When $\varphi = 45^\circ$, solve for the charge, $q$:
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Since $\varphi = \frac{\pi}{4}$, $\alpha = \frac{\pi}{8}$, and $\beta = \frac{3\pi}{8}$.
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Known:
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$$F_e = \frac{q^2}{16\pi\varepsilon_0\cos^2(\beta)}$$
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$$F_e = 0.00196\frac{\cos(\beta)}{\sin(\beta)}$$
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Set equal:
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$$\frac{q^2}{16\pi\varepsilon_0\cos^2(\beta)} = 0.00196\cot(\beta)$$
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$$\therefore q^2 = 16\pi\varepsilon_0\frac{\cos^3(\beta)}{\sin(\beta)}$$
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Plug in for $\beta$ and solve for $q$:
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$$q = \pm \sqrt{16\pi\varepsilon_0(0.00196)\frac{\cos^3\left(\frac{3\pi}{8}\right)}{\sin\left(\frac{3\pi}{8}\right)}} = \pm 2.300\times10^{-7}\text{C}$$
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### b) When $q=0.5\mu\text{C}$, solve for the angle, $\varphi$:
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Known:
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$$F_e = \frac{q^2}{16\pi\varepsilon_0\cos^2(\beta)}$$
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$$F_e = 0.00196\frac{\cos(\beta)}{\sin(\beta)}$$
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Set equal:
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$$\frac{q^2}{16\pi\varepsilon_0\cos^2(\beta)} = 0.00196\cot(\beta)$$
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Plug in for $q$:
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$$\frac{(0.5\times 10^{-6})^2}{16\pi\varepsilon_0(0.00196)} = \frac{\cos^3(\beta)}{\sin(\beta)}$$
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$$\therefore \frac{\cos^3(\beta)}{\sin(\beta)} = 0.287$$
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$$\therefore \beta = 0.915$$
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Using $\beta$ to solve for $\varphi$:
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$$\alpha = \frac{\pi}{2} - \beta = 0.656$$
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$$\varphi = 2\alpha = 1.314 = 75.257^\circ$$
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## 4
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The magnetic field, $\vec{B} = B_0(\hat{x} + 2\hat{y} - 4\hat{z})$ exists at a point. Find the electric field at that point if the force experienced by a test charge with velocity, $\vec{v} = v_0(3\hat{x} - \hat{y} + 2\hat{z})$ is $0$.
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Lorentz Force Law:
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$$\vec{F} = q\vec{E} + q\vec{v} \times \vec{B}$$
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Since $\vec{F}$ is $0$:
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$$0 = q\vec{E} + q\vec{v} \times \vec{B}$$
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$$\therefore q\vec{E} = -q\vec{v} \times \vec{B}$$
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$$\therefore \vec{E} = -\vec{v} \times \vec{B}$$
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By the definition of the cross product:
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$$-\vec{v} \times \vec{B} = \vec{B} \times \vec{v}$$
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In terms of $\vec{E}$:
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$$\vec{E} = \vec{B} \times \vec{v} = \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ 1 & 2 & -4 \\ 3 & -1 & 2 \end{vmatrix} = -14\hat{y} - 7\hat{z}$$
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## 5
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A circular loop with radius, $a$, exists in the x-y plane. If the loop is uniformly charged and has total charge, $Q$, determine the $\vec{E}$-field intensity at some point along the axis normal to the loop.
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By Coulomb's Law:
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$$\vec{E} = \int \frac{dq}{4\pi\varepsilon_0 r^2}\hat{r}$$
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Calling the distance along the normal axis $z$:
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$$r^2 = a^2 + z^2$$
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Along a uniformly charged line, the charge, $dq$, at any given point is given by the equation:
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$$dq = \lambda dl$$
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Where:
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$\lambda$ is the linear charge density
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The linear charge density, $\lambda$ is defined as:
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$$\lambda = \frac{Q}{L}$$
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Where:
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$Q$ is the total charge
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$L$ is the total length
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For a circular loop with radius, $a$, and total charge, $Q$:
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$$\lambda = \frac{Q}{2\pi a}$$
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$$dl = a d\varphi$$
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$$\therefore dq = \frac{Qd\varphi}{2\pi}$$
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Plugging into Coulomb's Law:
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$$\vec{E} = \int\limits_{0}^{2\pi} \frac{Qd\varphi}{8\pi^2\varepsilon_0(a^2 + z^2)} \hat{r}$$
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$$\therefore \vec{E} = \frac{Q}{8\pi^2\varepsilon_0(a^2 + z^2)} \int\limits_{0}^{2\pi}\hat{r}d\varphi$$
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Find $\hat{r}$ in terms of $\hat{x}$, $\hat{y}$, and $\hat{z}$:
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$$\hat{r} = \frac{\vec{a} + \vec{z}}{\sqrt{a^2 + z^2}}$$
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$$\vec{z} = z\hat{z}$$
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$$\vec{a} = a\cos{\varphi}\hat{x} + a\sin{\varphi}\hat{y}$$
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Back to Coulomb's Law:
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$$\vec{E} = \frac{Q}{8\pi^2\varepsilon_0(a^2 + z^2)^{(3/2)}} \int\limits_{0}^{2\pi} (a\cos(\varphi)\hat{x} + a\sin(\varphi)\hat{y} + z\hat{z})d\varphi$$
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By components:
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$$E_x = \frac{aQ}{8\pi^2\varepsilon_0(a^2 + z^2)^{3/2}} \int\limits_{0}^{2\pi} \cos(\varphi)d\varphi = 0$$
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$$E_y = \frac{aQ}{8\pi^2\varepsilon_0(a^2 + z^2)^{3/2}} \int\limits_{0}^{2\pi} \sin(\varphi)d\varphi = 0$$
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$$E_z = \frac{zQ}{8\pi^2\varepsilon_0(a^2 + z^2)^{3/2}} \int\limits_{0}^{2\pi} d\varphi = \frac{zQ}{4\pi\varepsilon_0(a^2 + z^2)^{3/2}}$$
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Recombining:
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$$\vec{E} = \frac{zQ}{4\pi\varepsilon_0(a^2 + z^2)^{3/2}}\hat{z}$$
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## 6
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Consider a circular ring in the x-y plane with inner radius, $a$, outer radius, $b$, and uniform charge density, $\rho_s$. Find an expression for the $\vec{E}$-field at a point at distance, $z$, along the axis normal to the ring.
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By Coulomb's Law:
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$$d\vec{E} = \frac{dq}{4\pi\varepsilon_0 r^2}\hat{r}$$
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Calling the radial distance, $\rho$:
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$$r^2 = \rho^2 + z^2$$
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For surface charge densities:
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$$dq = \rho_s ds$$
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In cylindrical coordinates:
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$$ds = \rho d\rho d\varphi$$
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Plugging into Coulomb's Law:
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$$\vec{E} = \iint \frac{\rho_s \rho}{4\pi\varepsilon_0(\rho^2 + z^2)}\hat{r}d\rho d\varphi$$
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$\hat{r}$ is defined as:
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$$\hat{r} = \frac{\vec{\rho} + \vec{z}}{\sqrt{\rho^2 + z^2}}$$
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Where:
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$\vec{z} = z\hat{z}$
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$\vec{\rho} = \rho\cos(\varphi)\hat{x} + \rho\sin(\varphi)\hat{y}$
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Splitting $\hat{r}$ by components:
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$$\hat{r} = \frac{\rho\cos(\varphi)\hat{x} + \rho\sin(\varphi)\hat{y} + z\hat{z}}{\sqrt{\rho^2 + z^2}}$$
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Back to Coulomb's Law:
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$$\vec{E} = \frac{\rho_s}{4\pi\varepsilon_0}\int\limits_{0}^{2\pi}\int\limits_{a}^{b} \frac{\rho^2\cos(\varphi)\hat{x} + \rho^2\sin(\varphi)\hat{y} + \rho z\hat{z}}{(\rho^2 + z^2)^{3/2}} d\rho d\varphi$$
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By components:
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$$E_x = \frac{\rho_s}{4\pi\varepsilon_0}\int\limits_{0}^{2\pi}\int\limits_{a}^{b} \frac{\rho^2\cos(\varphi)}{(\rho^2 + z^2)^{3/2}} d\rho d\varphi = 0$$
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$$E_y = \frac{\rho_s}{4\pi\varepsilon_0}\int\limits_{0}^{2\pi}\int\limits_{a}^{b} \frac{\rho^2\sin(\varphi)}{(\rho^2 + z^2)^{3/2}} d\rho d\varphi = 0$$
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$$E_z = \frac{\rho_s}{4\pi\varepsilon_0}\int\limits_{0}^{2\pi}\int\limits_{a}^{b} \frac{\rho z}{(\rho^2 + z^2)^{3/2}} d\rho d\varphi = \frac{\rho_s z}{2\varepsilon_0} \left(\frac{1}{\sqrt{z^2 + a^2}} - \frac{1}{\sqrt{z^2 + b^2}}\right)$$
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Recombining:
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$$\vec{E} = \frac{\rho_s z}{2 \varepsilon_0} \left( \frac{1}{\sqrt{z^2 + a^2}} - \frac{1}{\sqrt{z^2 + b^2}} \right) \hat{z}$$
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## 7
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Consider two concentric cylindrical surfaces. The inner having radius, $a$, and charge density $\rho_s$, and the outer having radius, $b$, and charge density $-\rho_s$.
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By Gauss's Law:
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$$\Phi_E = \frac{Q_{enc}}{\varepsilon_0} = \oint_S \vec{E} \cdot d\vec{A}$$
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The $\vec{E}$-field for a cylindrical surface with radius, $\rho$, and length, $l$, is given by the equation:
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$$2\pi\rho l E = \frac{Q_{enc}}{\varepsilon_0}$$
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### a) For $\rho < a$:
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$$Q_{enc} = 0$$
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$$\therefore E = 0$$
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### b) For $a < \rho < b$:
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$$Q_{enc} = 2\pi a l\rho_s$$
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Where:
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$l$ is the length of the section of the cylender.
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$$2\pi\rho lE = \frac{2\pi a l \rho_s}{\varepsilon_0}$$
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$$\therefore E = \frac{a\rho_s}{\rho \varepsilon_0}$$
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### c) For $\rho > b$:
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$$Q_{enc} = 2\pi l\rho_s(a-b)$$
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$$2\pi \rho l E = \frac{2\pi l \rho_s(a-b)}{\varepsilon_0}$$
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$$\therefore E = \frac{\rho_s (a-b)}{\rho \varepsilon_0}$$
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## 8
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Consider an infinite slab of thickness, $d$, centered on the origin ($x=0$, $y=0$, $z=0$).
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By Gauss's Law:
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$$\Phi_E = \frac{Q_{enc}}{\varepsilon_0} = \oint_S \vec{E} \cdot d\vec{A}$$
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### a) Find the strength of the electric field inside the slab ($\lvert z \rvert < d/2$):
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$$Q_{enc} = \rho_v lwh$$
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$$\oint_S \vec{E} \cdot d\vec{A} = E(2lw + 2lh + 2wh)$$
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$$E = \frac{\rho_v lwh}{2\varepsilon_0(lw + lh + wh)}$$
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Where:
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$l$ is the size of the x dimension of a Gaussian rectangular prism centered on the origin,
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$w$ is the size of the y dimension of that rectangular prism,
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$h$ is the size of the z dimension of that rectangular prism
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### b) Find the strength of the electric field inside the slab ($\lvert z \rvert > d/2$):
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$$Q_{enc} = \frac{\rho_v lwd}{2}$$
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$$\oint_S \vec{E} \cdot d\vec{A} = E(2lw + 2lh + 2wh)$$
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$$E = \frac{\rho_v lwd}{4\varepsilon_0 (lw + lh + wh)}$$
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Where:
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$l$ is the size of the x dimension of a Gaussian rectangular prism centered on the origin,
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$w$ is the size of the y dimension of that rectangular prism,
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$h$ is the size of the z dimension of that rectangular prism
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