Find the rectangular coordinates of the point where the angle \(\frac{5\pi}{3}\) meets the unit circle. If this were a point in the complex plane, what would be the rectangular and exponential forms of the complex number? (See figure.)

Write \(\vec{dr}\) in rectangular, cylindrical, and spherical coordinates.

1. Rectangular: \begin{equation} \vec{dr}= \end{equation}
2. Cylindrical: \begin{equation} \vec{dr}= \end{equation}
3. Spherical: \begin{equation} \vec{dr}= \end{equation}

Start with \(d\vec{r}\) in rectangular, cylindrical, and spherical coordinates. Use these expressions to write the scalar area elements \(dA\) (for different coordinate equals constant surfaces) and the volume element \(d\tau\). It might help you to think of the following surfaces: The various sides of a rectangular box, a finite cylinder with a top and a bottom, a half cylinder, and a hemisphere with both a curved and a flat side, and a cone.

1. Rectangular: \begin{align} dA&=\\ d\tau&= \end{align}
2. Cylindrical: \begin{align} dA&=\\ d\tau&= \end{align}
3. Spherical: \begin{align} dA&=\\ d\tau&= \end{align}

The electrostatic potential due to a point charge at the origin is given by: \begin{equation} V=\frac{1}{4\pi\epsilon_0} \frac{q}{r} \end{equation}

1. Find the electric field due to a point charge at the origin as a gradient in rectangular coordinates.
2. Find the electric field due to a point charge at the origin as a gradient in spherical coordinates.
3. Find the electric field due to a point charge at the origin as a gradient in cylindrical coordinates.

Consider the fields at a point \(\vec{r}\) due to a point charge located at \(\vec{r}'\).

1. Write down an expression for the electrostatic potential \(V(\vec{r})\) at a point \(\vec{r}\) due to a point charge located at \(\vec{r}'\). (There is nothing to calculate here.)
2. Write down an expression for the electric field \(\vec{E}(\vec{r})\) at a point \(\vec{r}\) due to a point charge located at \(\vec{r}'\). (There is nothing to calculate here.)
3. Working in rectangular coordinates, compute the gradient of \(V\).
4. Write several sentences comparing your answers to the last two questions.

(Straightforward algebra) Purpose: Discover the change of variables that allows you to go from the solution to the reduced mass system back to the original system. Practice solving systems of two linear equations.

For systems of particles, we used the formulas \begin{align} \vec{R}_{cm}&=\frac{1}{M}\left(m_1\vec{r}_1+m_2\vec{r}_2\right) \nonumber\\ \vec{r}&=\vec{r}_2-\vec{r}_1 \label{cm} \end{align} to switch from a rectangular coordinate system that is unrelated to the system to coordinates adapted to the center-of-mass. After you have solved the equations of motion in the center-of-mass coordinates, you may want to transform back to the original coordinate system. Find the inverse transformation, i.e. solve for: \begin{align} \vec{r}_1&=\\ \vec{r}_2&= \end{align} Hint: The system of equations (\ref{cm}) is linear, i.e. each variable is to the first power, even though the variables are vectors. In this case, you can use all of the methods you learned for solving systems of equations while keeping the variables vector valued, i.e. you can safely ignore the fact that the \(\vec{r}\)s are vectors while you are doing the algebra as long as you don't divide by a vector.

You have a charge distribution on the \(x\)-axis composed of two point charges: one with charge \(+3q\) located at \(x=-d\) and the other with charge \(-q\) located at \(x=+d\).

1. Sketch the charge distribution.
2. Write an expression for the volume charge density \(\rho (\vec{r})\) everywhere in space.

Sketch each of the vector fields below.

1. \(\boldsymbol{\vec F} =-y\,\boldsymbol{\hat x} + x\,\boldsymbol{\hat y}\)
2. \(\boldsymbol{\vec G} = x\,\boldsymbol{\hat x} + y\,\boldsymbol{\hat y}\)
3. \(\boldsymbol{\vec H} = y\,\boldsymbol{\hat x} + x\,\boldsymbol{\hat y}\)

The distance \(\left\vert\vec r -\vec r\,{}'\right\vert\) between the point \(\vec r\) and the point \(\vec r'\) is a coordinate-independent, physical and geometric quantity. But, in practice, you will need to know how to express this quantity in different coordinate systems.

1. Find the distance \(\left\vert\vec r -\vec r\,{}'\right\vert\) between the point \(\vec r\) and the point \(\vec r'\) in rectangular coordinates.

   

2. Show that this same distance written in cylindrical coordinates is: \begin{equation} \left|\vec r -\vec r\,{}'\right| =\sqrt{s^2+s\,{}'^2-2ss\,{}'\cos(\phi-\phi\,{}') +(z-z\,{}')^2} \end{equation}
3. Show that this same distance written in spherical coordinates is: \begin{equation} \left\vert\vec r -\vec r\,{}'\right\vert =\sqrt{r'^2+r\,{}^2-2rr\,{}' \left[\sin\theta\sin\theta\,{}'\cos(\phi-\phi\,{}') +\cos\theta\cos\theta\,{}'\right]} \end{equation}
4. Now assume that \(\vec r\,{}'\) and \(\vec r\) are in the \(x\)-\(y\) plane. Simplify the previous two formulas.