Rowan-Classes/5th-Semester-Fall-2023/EEMAGS/Homework/Homework-01.md

Homework 1 - Aidan Sharpe

1.

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}.

Definition of dot product:

\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)

Solve the dot product:

\vec{A} \cdot \vec{B} = (4 \cdot 3) + (4 \cdot (-1.5)) + ((-2) \cdot 1) = 4

Magnitudes of the vectors:

\lVert \vec{A} \rVert = \sqrt{4^2 + 4^2 + (-2)^2} = \sqrt{36} = 6

\lVert \vec{B} \rVert = \sqrt{3^2 + \left(-\frac{3}{2}\right)^2 + 1^2} = \sqrt{\frac{49}{4}} = \frac{7}{2}

By the definition of the dot product:

4 = 6 \cdot \frac{7}{2} \cos(\varphi)

\therefore \cos(\varphi) = \frac{4}{21}

\therefore \varphi = \arccos\left(\frac{4}{21}\right) = 1.379 = 79.02^\circ

2

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.

Convert from cartesian to cylidrical:

\rho = \sqrt{x^2 + y^2} = \sqrt{2}

\varphi = \arctan\left(\frac{y}{x}\right) = \frac{\pi}{4}

z = z = 1

Find \vec{A} and \vec{B} at x=1, y=1, z=1:

\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}

\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}

a) Find \vec{A} \cdot \vec{B}

\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}

b) Find \vec{A} \times \vec{B}

\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}

3

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}.

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.

Find the distance, r:

\frac{r}{2} = (1\text{m})\cos(\beta)

\therefore r = 2\cos(\beta)

By Coulomb's Law, the electric field force on each charge, \vec{F}_e, is:

\vec{F}_e = \frac{q^2}{4\pi\varepsilon_0 r^2}\hat{x}

Pluggin in for r,

\vec{F}_e = \frac{q^2}{16\pi\varepsilon_0\cos^2(\beta)}\hat{x}

The force due to gravity on each charge, \vec{F}_g, is:

\vec{F}_g = 0.2 \cdot 980\times 10^{-5} = 0.00196(-\hat{y})\text{N}

Assuming static equilibrium:

\sum F_x = 0 = F_{e_x} + F_{T_x} + F_{g_x}

\sum F_y = 0 = F_{e_y} + F_{T_y} + F_{g_y}

Where:

F_{T_x} is the x-component of the force due to tension in each thread, F_T.

F_{T_y} is the y-component of F_T.

Since F_g only acts in the \hat{y} direction, and F_e only acts in the \hat{x} direction:

F_{e_x} + F_{T_x} = 0

F_{T_y} + F_{g_y} = 0

Define the components of the tension force F_T:

F_{T_x} = F_T\cos\left(\frac{\pi}{2} + \alpha\right) = -F_T\cos(\beta)

F_{T_y} = F_T\sin\left(\frac{\pi}{2} + \alpha\right) = F_T\sin(\beta)

Solve for F_T using equilibrium in the \hat{y} direction:

F_T\sin(\beta) - 0.00196 = 0

\therefore F_T = \frac{0.00196}{\sin(\beta)}

Plug in F_T to solve equilibrium in the \hat{x} direction:

F_e + \left( \frac{0.00196}{\sin(\beta)} \right)(-\cos(\beta)) = 0

\therefore F_e = 0.00196 \cot(\beta)

a) When \varphi = 45^\circ, solve for the charge, q:

Since \varphi = \frac{\pi}{4}, \alpha = \frac{\pi}{8}, and \beta = \frac{3\pi}{8}.

Known:

F_e = \frac{q^2}{16\pi\varepsilon_0\cos^2(\beta)}

F_e = 0.00196\frac{\cos(\beta)}{\sin(\beta)}

Set equal:

\frac{q^2}{16\pi\varepsilon_0\cos^2(\beta)} = 0.00196\cot(\beta)

\therefore q^2 = 16\pi\varepsilon_0\frac{\cos^3(\beta)}{\sin(\beta)}

Plug in for \beta and solve for q:

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}

b) When q=0.5\mu\text{C}, solve for the angle, \varphi:

Known:

F_e = \frac{q^2}{16\pi\varepsilon_0\cos^2(\beta)}

F_e = 0.00196\frac{\cos(\beta)}{\sin(\beta)}

Set equal:

\frac{q^2}{16\pi\varepsilon_0\cos^2(\beta)} = 0.00196\cot(\beta)

Plug in for q:

\frac{(0.5\times 10^{-6})^2}{16\pi\varepsilon_0(0.00196)} = \frac{\cos^3(\beta)}{\sin(\beta)}

\therefore \frac{\cos^3(\beta)}{\sin(\beta)} = 0.287

\therefore \beta = 0.915

Using \beta to solve for \varphi:

\alpha = \frac{\pi}{2} - \beta = 0.656

\varphi = 2\alpha = 1.314 = 75.257^\circ

4

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.

Lorentz Force Law:

\vec{F} = q\vec{E} + q\vec{v} \times \vec{B}

Since \vec{F} is 0:

0 = q\vec{E} + q\vec{v} \times \vec{B}

\therefore q\vec{E} = -q\vec{v} \times \vec{B}

\therefore \vec{E} = -\vec{v} \times \vec{B}

By the definition of the cross product:

-\vec{v} \times \vec{B} = \vec{B} \times \vec{v}

In terms of \vec{E}:

\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}

5

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.

By Coulomb's Law:

\vec{E} = \int \frac{dq}{4\pi\varepsilon_0 r^2}\hat{r}

Calling the distance along the normal axis z:

r^2 = a^2 + z^2

Along a uniformly charged line, the charge, dq, at any given point is given by the equation:

dq = \lambda dl

Where:

\lambda is the linear charge density

The linear charge density, \lambda is defined as:

\lambda = \frac{Q}{L}

Where:

Q is the total charge

L is the total length

For a circular loop with radius, a, and total charge, Q:

\lambda = \frac{Q}{2\pi a}

dl = a d\varphi

\therefore dq = \frac{Qd\varphi}{2\pi}

Plugging into Coulomb's Law:

\vec{E} = \int\limits_{0}^{2\pi} \frac{Qd\varphi}{8\pi^2\varepsilon_0(a^2 + z^2)} \hat{r}

\therefore \vec{E} = \frac{Q}{8\pi^2\varepsilon_0(a^2 + z^2)} \int\limits_{0}^{2\pi}\hat{r}d\varphi

Find \hat{r} in terms of \hat{x}, \hat{y}, and \hat{z}:

\hat{r} = \frac{\vec{a} + \vec{z}}{\sqrt{a^2 + z^2}}

\vec{z} = z\hat{z}

\vec{a} = a\cos{\varphi}\hat{x} + a\sin{\varphi}\hat{y}

Back to Coulomb's Law:

\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

By components:

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

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

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}}

Recombining:

\vec{E} = \frac{zQ}{4\pi\varepsilon_0(a^2 + z^2)^{3/2}}\hat{z}

6

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.

By Coulomb's Law:

d\vec{E} = \frac{dq}{4\pi\varepsilon_0 r^2}\hat{r}

Calling the radial distance, \rho:

r^2 = \rho^2 + z^2

For surface charge densities:

dq = \rho_s ds

In cylindrical coordinates:

ds = \rho d\rho d\varphi

Plugging into Coulomb's Law:

\vec{E} = \iint \frac{\rho_s \rho}{4\pi\varepsilon_0(\rho^2 + z^2)}\hat{r}d\rho d\varphi

\hat{r} is defined as:

\hat{r} = \frac{\vec{\rho} + \vec{z}}{\sqrt{\rho^2 + z^2}}

Where:

\vec{z} = z\hat{z}

\vec{\rho} = \rho\cos(\varphi)\hat{x} + \rho\sin(\varphi)\hat{y}

Splitting \hat{r} by components:

\hat{r} = \frac{\rho\cos(\varphi)\hat{x} + \rho\sin(\varphi)\hat{y} + z\hat{z}}{\sqrt{\rho^2 + z^2}}

Back to Coulomb's Law:

\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

By components:

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

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

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)

Recombining:

\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}

7

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.

By Gauss's Law:

\Phi_E = \frac{Q_{enc}}{\varepsilon_0} = \oint_S \vec{E} \cdot d\vec{A}

The $\vec{E}$-field for a cylindrical surface with radius, \rho, and length, l, is given by the equation:

2\pi\rho l E = \frac{Q_{enc}}{\varepsilon_0}

a) For \rho < a:

Q_{enc} = 0

\therefore E = 0

b) For a < \rho < b:

Q_{enc} = 2\pi a l\rho_s

Where:

l is the length of the section of the cylender.

2\pi\rho lE = \frac{2\pi a l \rho_s}{\varepsilon_0}

\therefore E = \frac{a\rho_s}{\rho \varepsilon_0}

c) For \rho > b:

Q_{enc} = 2\pi l\rho_s(a-b)

2\pi \rho l E = \frac{2\pi l \rho_s(a-b)}{\varepsilon_0}

\therefore E = \frac{\rho_s (a-b)}{\rho \varepsilon_0}

8

Consider an infinite slab of thickness, d, centered on the origin (x=0, y=0, z=0).

By Gauss's Law:

\Phi_E = \frac{Q_{enc}}{\varepsilon_0} = \oint_S \vec{E} \cdot d\vec{A}

a) Find the strength of the electric field inside the slab (\lvert z \rvert < d/2):

Q_{enc} = \rho_v lwh

\oint_S \vec{E} \cdot d\vec{A} = E(2lw + 2lh + 2wh)

E = \frac{\rho_v lwh}{2\varepsilon_0(lw + lh + wh)}

Where:

l is the size of the x dimension of a Gaussian rectangular prism centered on the origin,

w is the size of the y dimension of that rectangular prism,

h is the size of the z dimension of that rectangular prism

b) Find the strength of the electric field inside the slab (\lvert z \rvert > d/2):

Q_{enc} = \frac{\rho_v lwd}{2}

\oint_S \vec{E} \cdot d\vec{A} = E(2lw + 2lh + 2wh)

E = \frac{\rho_v lwd}{4\varepsilon_0 (lw + lh + wh)}

Where:

l is the size of the x dimension of a Gaussian rectangular prism centered on the origin,

w is the size of the y dimension of that rectangular prism,

h is the size of the z dimension of that rectangular prism