\documentclass[11pt]{scrartcl} % scrreprt for longer papers
\usepackage{custompackage}

\begin{document}

    \title{AP Physics C: Electricity and Magnetism FRQ Solution Manual}
    %\date{}
    \author{}

    \maketitle


    \begin{enumerate}
    	\ii A solid plastic sphere of radius $a$ and a conducting spherical shell of inner radius $b$ and outer radius $c$ are shown in the figure below. The shell has an unknown charge. The solid plastic sphere has a charge per unit volume given by $\rho(r) = \beta r$, where $\beta$ is a positive constant and $r$ is the distance from the center of the sphere. Express your answers to parts (a), (b), and (c) in terms of $\beta, r, a$, and physical constants, as appropriate.
    	\placeimg{0.4}{1-1}
    	\begin{enumerate}
    		\ii Consider a Gaussian sphere of radius $r$ concentric with the plastic sphere. Derive an expression for the charge enclosed by the Gaussian sphere for the following regions.
    		\begin{enumerate}
    			\ii $r < a$
    			\begin{answer*}
    				$$\boxed{Q\enc = \beta \pi r^4}$$
    			\end{answer*}
    			\begin{soln}
    				We have that the total charge is the integral of the charge density over a differential volume:
    				$$Q\enc = \int_{0}^{r} \rho(R)\dd{V}$$
    				The goal is to write $\dd{V}$ in terms of $\dd{R}$ so we can integrate the expression succssfully. For a sphere of radius $R$, $V = \frac{4}{3}\pi R^3$, so $\dd{V} = 4\pi R^2 \dd{R}$. Thus
    				$$Q\enc = \int_0^r 4\pi R^2 \rho(R)\dd{R}$$
    				Substituting in $\rho$, we get
    				$$Q\enc = \int_0^r 4\beta \pi R^3 \dd{R}$$
    				This evaluates to
    				$$Q\enc = \beta \pi r^4$$
    				as desired.
    			\end{soln}
    			\ii $a < r < b$
    			\begin{answer*}
    				$$\boxed{Q\enc = \beta \pi a^4}$$
    			\end{answer*}
    			\begin{soln}
    				Here, the analysis is exactly the same, except in the spherical region between $a$ and $b$ there is no charge. If we really wanted to we could break it into two integrals, but just evaluating the above integral at $a$ works fine as well since the second one vanishes. So our answer is just
    				$$Q\enc = \beta \pi a^4$$
    				as desired.
    			\end{soln}
    		\end{enumerate}
    		\ii Use Gauss's Law to derive an expression for the magnitude of the magnetic field in the following regions.
    		\begin{enumerate}
    			\ii $r < a$
    			\begin{answer*}
    				$$\boxed{\vb{E} = \frac{\beta r^2}{4 \eps_0}\vu{r}}$$
    			\end{answer*}
    			\begin{soln}
    				Gauss's law in integral form (the differential form yields a one-line solution but using it is not in the spirit of the question) states that
    				$$\oint \vb{E} \vdot \dd{\vb{a}} = \frac{Q\enc}{\eps_0}$$
    				Since we have spherical symmetry, the left hand side evaluates to $EA = E(4\pi r^2)$. Using our expression for enclosed charge ($Q\enc = \beta \pi r^4$), we see that
    				$$E = \frac{\beta r^2}{4 \eps_0}$$
    				and indeed, due to radial symmetry,
    				$$\vb{E} = \frac{\beta r^2}{4 \eps_0}\vu{r}$$
    				as desired.
    			\end{soln}
    			\ii $a < r < b$
    			\begin{answer*}
    				$$\boxed{\vb{E} = \frac{\beta a^4}{4\eps_0 r^2}\vu{r}}$$
    			\end{answer*}
    			\begin{soln}
    				We use the same solution method as above. We have that $Q\enc = \beta \pi a^4$, so
    				$$E(4\pi r^2) = \beta \pi a^4$$
    				$$E = \frac{\beta a^4}{4\eps_0 r^2}$$
    				and because of radial symmetry,
    				$$\vb{E} = \frac{\beta a^4}{4\eps_0 r^2}\vu{r}$$
    				as desired.
    			\end{soln}
    		\end{enumerate}
    		\ii At any point outside of the conducting shell, it is observed that the magnitude of the electric field is zero.
    		\begin{enumerate}
    			\ii Determine the charge on the inner surface of the conducting shell. Justify your answer.
    			\begin{answer*}
    				$$\boxed{Q_{\mathrm{inner}} = -\beta \pi a^4}$$
    			\end{answer*}
    			\begin{soln}
    				The electric field on the inside of a conductor is zero, because if there were a field, then charges would be on the inside of it and not only on the surface. By drawing a concentric Gaussian sphere at $b + \epsilon$ for some $\epsilon < (c - b)$, we see that the enclosed charge is zero because the field is zero, and so the inner surface of the conductor must have equal and opposite charge to the plastic sphere.
    			\end{soln}
    			\ii Determine the charge on the outer surface of the conducting shell.
    			\begin{answer*}
    				$$\boxed{Q_{\mathrm{outer}} = \SI{0}{\coulomb}}$$
    			\end{answer*}
    			\begin{soln}
    				Outside of the conductor the electric field is zero, and so the enclosed charge is zero. If the charge on the inner surface of the conductor is equal and opposite to the charge on the sphere, then any additional charge would upset the balance. Thus the charge on the outer surface on the conductor has to be zero.
    			\end{soln}
    		\end{enumerate}
    		\ii See the below questions.
    		\begin{enumerate}
    			\ii On the axes below, sketch the electric field $E$ as a fucntion of distance $r$ from the center of the sphere. Sketch the graph for the range $r = 0$ at the center of the sphere to $r = c$ at the outside of the conducting shell.
    			\begin{answer*}
    				Here is a sample response:
    				\placeimg{0.5}{1-3}
    			\end{answer*}
    			\begin{soln}
    				Note that from $0 < r < a$ the electric field varies with the square of $r$, and from $a < r < b$ the electric field varies with the inverse of the square of $r$. From $b < r < c$ the field is zero as at that point we are inside a conductor.
    			\end{soln}
    			\ii The figure below shows the sphere and shell with four points labeled W, X, Y, and Z. Point W is at the center of the sphere, point X is on the surface of the sphere, and points Y and Z are on the inner and outer surface of the shell. Rank the points according to the electric potential at that point, with 1 indicating the largest electric potential. If two points have the same electric potential, give them the same numerical ranking.
    			\placeimg{0.5}{1-4}
    			\begin{answer*}
    				$$\framebox{W > X > Y = Z}$$
    			\end{answer*}
    			\begin{soln}
    				We define the potential at radius $r$ to be
    				$$V(r) = -\int_{\infty}^{r}\vb{E}(R) \vdot \dd{\vb{R}} = -\int_{b}^{r}\vb{E}(R) \vdot \dd{\vb{R}}$$
    				since the electric field at $r > b$ is zero for all $r$. Naturally, this makes the potentials at Y and Z equal and zero. Further, as we move inwards we work against the field, increasing the potential. Thus it is clear that W has higher potential than X, which has greater potential than Y and Z, as desired.
    			\end{soln}
    		\end{enumerate}
    	\end{enumerate}
    	\newpage

    	\ii An experiment is designed to measure the dielectric constant of paper that has an area $A = \SI{0.060}{\meter\squared}$. Using aluminum foil, two parallel plates are created with the same area as the paper. Five hundred sheets of paper are placed between the aluminum foil plates to create a parallel plate capacitor, as shown in the figure below.
    	\placeimg{0.5}{2-1}
    	Using a multimeter, the capacitance $C$ of the capacitor is measured. The number of sheets and the total thickness $d$ of the stack of paper are recorded. The experiment is repeated, reducing the number of sheets of paper each time. The data are recorded in the table below.
		\begin{table}[H]
			\centering
			\begin{tabular}{|c|c|c|c|}
				\hline
				Sheets of Paper & $d$ (\si{\meter}) & $C$ (\si{\farad})    & {\color[HTML]{FD6864} $\frac{1}{d}$ $(\si{\per\meter})$} \\ \hline
				500             & 0.045             & $6.5 \times 10^{-11}$  & {\color[HTML]{FD6864} 22.2222}                           \\ \hline
				400             & 0.036             & $7.4 \times 10^{-11}$  & {\color[HTML]{FD6864} 27.7778}                           \\ \hline
				300             & 0.027             & $8.9 \times 10^{-11}$  & {\color[HTML]{FD6864} 37.0370}                           \\ \hline
				200             & 0.018             & $11.9 \times 10^{-11}$ & {\color[HTML]{FD6864} 55.5556}                           \\ \hline
				100             & 0.010             & $21.0 \times 10^{-11}$ & {\color[HTML]{FD6864} 100}                               \\ \hline
			\end{tabular}
		\end{table}
		\begin{enumerate}
			\ii Indicate below which quantities should be graphed to yield a straight line whose slope could be used to calculate a numerical value for the dielectric constant of the paper. Use the remaining columns in the table above, as needed, to record any quantities that you indicated that are not given. Label each column you use and include units.
			\begin{answer*}
				Vertical: \framebox{capacitance}

				Horizontal: \framebox{inverse of distance}
			\end{answer*}
			\begin{soln}
				We use the equation
				$$C = \frac{\kappa \eps_0 A}{d}$$
				When we hold $A$ constant, we can measure $\kappa$ by varying $\frac{1}{d}$ and seeing how $C$ responds.

				See the rightmost column of the data table for extra calculated values.
			\end{soln}
			\ii Plot the data points for the quantities indicated in part (a) on the graph below. Clearly scale and label all axes, including units if appropriate. Draw a straight line that best represents the data.
			\begin{answer*}
				A graph is below.
				\placeimg{0.6}{2-2}
			\end{answer*}
			\ii Using the straight line, calculate a dielectric constant for the paper.
			\begin{answer*}
				$$\boxed{\kappa \approx 3.5152133}$$
			\end{answer*}
			\begin{soln}
				We graph and get the regression line through calculator to be $y = (1.8674219292158 \times 10^{-12})x + (2.0795454545455 \times 10^{-11})$. Thus, from the equation
				$$C = \frac{\kappa \eps_0 A}{d}$$
				we see that $\kappa \eps_0 A\lambda = 1.8674219292158 \times 10^{-12}$, where $\lambda$ is some constant that makes the units cancel. Thus,
				$$\kappa = \frac{1.8674219292158 \times 10^{-12}}{\eps_0 A} = \frac{1.8674219292158 \times 10^{-12}}{(8.854 \times 10^{-12})(0.060)} = 3.5152133$$
				as desired.
			\end{soln}
			The student now makes a capacitor using the same aluminum foil plates and just one sheet of paper. Using the experimentally determined dielectric constant, the student calculates the capacitance to be $\SI{18}{\nano\farad}$. The student uses this uncharged capacitor to build a circuit using wire, a $\SI{36}{\volt}$ battery, $3$ identical $\SI{80}{\ohm}$ resistors, and an open switch, as shown in the figure below.
			\placeimg{0.6}{2-3}
			\ii Calculate the current in the battery immediately after the switch is closed.
			\begin{answer*}
				$$\boxed{i\ini = \SI{0.3}{\ampere}}$$
			\end{answer*}
			\begin{soln}
				Let $R = \SI{80}{\ohm}$ be the resistance of every resistor in the circuit.
				Immediately after the switch is closed, the capacitor acts like a wire (since it has no charge, and thus no voltage). The equivalent resistance of the two resistors in parallel is
				$$R\eq = \frac{1}{\frac{1}{R} + \frac{1}{R}} = \frac{R}{2} = \frac{\SI{80}{\ohm}}{2} = \SI{40}{\ohm}$$
				This equivalent resistor and the resistor on the bottom of the circuit add in series, giving a total resistance of
				$$R\tot = R\eq + R =\SI{40}{\ohm} + \SI{80}{\ohm}  = \SI{120}{\ohm}$$
				Thus, by Ohm's Law, $V = iR\tot$ and so
				$$i\ini = \frac{V}{R\tot} = \frac{\SI{36}{\volt}}{\SI{120}{\ohm}} = \SI{0.3}{\ampere}$$
				as desired.
			\end{soln}
			\ii Determine the time constant for this circuit.
			\begin{answer*}
				$$\boxed{\tau = \SI{2.16e-6}{\second}}$$
			\end{answer*}
			\begin{soln}
				Collapsing the $\SI{80}{\ohm}$ resistors into one $\SI{120}{\ohm}$ resistor as before, we get, through an application of Kirchhoff's Voltage Law, the differential equation
				$$V - \dv{q}{t}R\tot - q\frac{1}{C} = 0$$
				Solving this differential equation with the initial condition $q(0) = \SI{0}{\coulomb}$ gives the equation
				$$q(t) = CV\left(1- \exp\left(-\frac{t}{R\tot C}\right)\right)$$
				Taking time derivatives yields
				$$i(t) = \frac{V}{R\tot}\exp\left(-\frac{t}{R\tot C}\right)$$
				From this we know that $R\tot C$ is the time constant $\tau$, which can be calculated to be
				$$\tau = R\tot C = (\SI{120}{\ohm})(\SI{18}{\nano\farad}) = \SI{2.16e-6}{\second}$$
				as desired.
			\end{soln}
			\ii Students A and B measure the time it takes after the switch is closed for the voltage across the capacitor to reach half its maximum value and find that it is longer than expected.
			\begin{enumerate}
				\ii Student A assumes that the capacitance value is correct. Would Student A conclude that the resistance value is larger or smaller than measured? Explain experimentally what could account for this.
				\begin{answer*}
					\framebox{Resistance is larger than measured.}
				\end{answer*}
				\begin{soln}
					We know that, since it takes longer to reach a given percentage of final value, that the time constant $\tau = RC$ is greater than it theoretically should be. Since the capacitance should be correct, the resistance should be higher. Many different reasons for this are possible; a plausible one is that the circuit wires themselves have additional resistance.
				\end{soln}
				\ii Student B assumes that the resistance value is correct. Would Student B conclude that the capacitance value is larger or smaller than measured? Explain experimentally what could account for this.
				\begin{answer*}
					\framebox{Capacitance is larger than measured.}
				\end{answer*}
				\begin{soln}
					We know that, since it takes longer to reach a given percentage of final value, that the time constant $\tau = RC$ is greater than it theoretically should be. Since the resistance should be correct, the capacitance should be higher. Many different reasons for this are possible; a plausible one is that the capacitor used has a calculated capacitance that underestimates the true value due to loss of precision in the calculation.
				\end{soln}
			\end{enumerate}
		\end{enumerate}

		\newpage

		\ii The figures below represent different views of two long, straight, horizontal wires, $1$ and $2$, carrying currents $I_1 = I$ and $I_2 = 2I$, respectively, in the directions shown. The wires are held in place. In Figure 1, the current in wire $1$ is directed out of the page, and wire $1$ is a distance $d$ above wire 2. Point P is a horizontal distance $d$ from wire $1$ and a distance $d$ directly above wire 2. Express your answers to parts (a) and (b) in terms of $I$, $d$, and physical constants, as appropriate.
		\placeimg{0.6}{3-1}
		\begin{enumerate}
			\ii Use Ampere's law to derive an expression for the magnitude of the magnetic field at point P due to wire $1$.
			\begin{answer*}
				$$\boxed{\vb{B}_1 = \frac{\mu_0 I}{2\pi d}\vu{k}}$$
			\end{answer*}
			\begin{soln}
				Note that the $xz$ plane is depicted in the Side View and the $xy$ plane is depicted in the Top View.

				Ampere's law states that
				$$\oint \vb{B} \vdot \dd{\vb{l}} = \mu_0 I\enc$$
				Drawing an Amperian loop of radius $d$ around wire 1 yields the expression
				$$B_1(2\pi d) = \mu_0 I_1 = \mu_0 I$$
				Thus,
				$$B_1 = \frac{\mu_0 I}{2\pi d}$$
				Using the right hand rule (pointing the thumb in direction of current and wrapping the hand around, examining where the fingers point) gives that the field is pointing in the $+\vu{k}$ direction. Thus,
				$$\vb{B}_1 = \frac{\mu_0 I}{2\pi d}\vu{k}$$
				as desired.
			\end{soln}
			\ii Derive an expression for the magnitude of the net magnetic field at point P.
			\begin{answer*}
				$$\boxed{\vb{B} = \frac{\mu_0 I}{2\pi d}\left(2\vu{j} + \vu{k}\right)}$$
			\end{answer*}
			\begin{soln}
				We know from the prior part that $\dst \vb{B}_1 = \frac{\mu_0 I}{2\pi d}\vu{k}$. Drawing an Amperian loop of radius $d$ around wire 2 yields the expression
				$$B_2 (2\pi d) = \mu_0 I_2 = 2\mu_0 I$$
				Thus,
				$$B_2 = \frac{\mu_0 I}{\pi d}$$
				Using the right hand rule as before gives that the field is pointing in the $+\vu{j}$ direction. Thus,
				$$\vb{B}_2 = \frac{\mu_0 I}{\pi d}\vu{j}$$
				Summing them by superposition yields
				$$\vb{B} = \frac{\mu_0 I}{2\pi d}\left(2\vu{j} + \vu{k}\right)$$
				as desired.
			\end{soln}
			\ii Calculate the numerical value of the angle to the horizontal for the direction of the net magnetic field at point P.
			\begin{answer*}
				$$\boxed{\theta = \arctan\left(\frac{1}{2}\right) \approx \SI{0.463647609}{\radian}}$$
			\end{answer*}
			\begin{soln}
				The angle $\theta$ from the horizontal is just
				$$\theta = \arctan\left(\frac{B_1}{B_2}\right)$$
				(draw it out on the $xy$ plane). This evaluates to
				$$\theta = \arctan\left(\frac{1}{2}\right) \approx \SI{0.463647609}{\radian}$$
				as desired.
			\end{soln}
			\ii Wire $1$ is now released. Which of the following best describes the initial motion of wire $1$ due to the magnetic field of wire $2$? Assume gravitational effects are negligible. Justify your answer.
			\begin{answer*}
				\framebox{Wire 1 does not move}.
			\end{answer*}
			\begin{soln}
				The magnetic field from wire $2$ is
				$$\vb{B}_2 = \frac{\mu_0 I}{\pi d}\vu{j}$$
				The differential length element $\dd{\vb*{\ell}_1} = -\dd{y}\vu{j}$ points in the $-\vu{j}$ direction. The force on wire $1$ from the field generated from wire 2 is
				$$\vb{F}_{12} = \int I \dd{\vb*{\ell}} \cross \vb{B}_2 = -I \int \left(\dd{y} \vu{j}\right) \cross \left(\frac{\mu_0 I}{\pi d}\vu{j}\right) = \vb{0}\si{\newton}$$
				since the cross product of a vector (specifically the $\vu{j}$ unit vector) with itself is zero. Thus there is no force and no acceleration on wire 1.
			\end{soln}
			Wire $1$ is now replaced by a conducting rectangular loop of length $\ell$, width $w$, and resistance $R$. The loop is placed a distance $d$ from wire $2$, as shown. The loop, wire, and distance $d$ are all in the plane of the page. The long side of the loop is parallel to the wire. The current $I_2$ for wire $2$ is decreasing linearly as a function of time $t$ according to the equation $I_2 = 2I_0(1 - kt)$, where $k$ is a positive constant with units of $\si{\per\second}$.
			\placeimg{0.6}{3-2}
			\ii Find the integration that will give an expression for the flux $\Phi$ as a function of time $t$.
			\begin{answer*}
				$$\boxed{\Phi = \int\limits_{r=d}^{r = d + w} \frac{\mu_0  I_0(1- kt)}{\pi r}\ell \dd{r}}$$
			\end{answer*}
			\begin{soln}
				We know that the flux $\Phi$ is given by the equation
				$$\Phi = \int \vb{B} \vdot \dd{\vb{A}} = \int \vb{B} \vdot \left(\dd{\vb*{\ell}} \cross \dd{\vb{w}}\right) = \int \ell \vb{B} \vdot (\vu{i} \cross \dd{\vb{r}}) = \int \ell \vb{B} \vdot (\vu{i} \cross \vu{k})\dd{r} = \int \ell \vb{B} \vdot \vu{j} \dd{r}$$
				using the change of variables $\dd{\vb{w}} = \dd{\vb{r}}$.

				Calculating $\vb{B}$ is done through Ampere's law. Drawing an Amperian loop of radius $r$, where $d < r < (d + w)$, yields the equation
				$$\oint \vb{B} \vdot \dd{\vb{l}} = \mu_0 I_2 = \mu_0 2I_0(1-kt)$$
				Evaluating the left hand side through the Amperian loop yields
				$$B(2\pi r) = \mu_0 I_2 = \mu_0 2I_0(1-kt)$$
				and so
				$$B = \frac{\mu_0 I_0(1-kt)}{\pi r}$$
				Application of the right hand rule in a similar way to as before yields that the magnetic field is pointing in the $+\vu{j}$ direction, so
				$$\vb{B} = \frac{\mu_0 I_0(1-kt)}{\pi r}\vu{j}$$
				Plugging this back into the flux equation, we get that $\vu{j} \vdot \vu{j} = 1$, so it simplifies to
				$$\Phi = \int \frac{\mu_0 I_0 (1 - kt)\ell }{\pi r} \dd{r}$$
				Clearly, our bounds of integration are from $r = d$ to $r = d + w$, so our final equation for flux is
				$$\Phi = \int\limits_{r = d}^{r = d + w} \frac{\mu_0 I_0 (1 - kt)}{\pi r} \ell \dd{r}$$
				as desired.
			\end{soln}
			\ii Given that the flux through the rectangular loop as a function of time $t$ is given by the equation $\dst \Phi = \frac{\mu_0 I_0 \ell (1 - kt)}{\pi}\ln\left(\frac{d+w}{d}\right)$, derive an expression for the magnitude of the current, if any, induced in the loop. Express your answers in terms of $I_0$, $d$, $r$, $R$, $w$, $k$, $\ell$, and physical constants, as appropriate.
			\begin{answer*}
				$$\boxed{i = -\frac{\mu_0 I_0 \ell k}{\pi r}\ln\left(\frac{d+w}{d}\right)}$$
			\end{answer*}
			\begin{soln}
				We know that by Faraday's law (again, the differential form is overpowered, so we'll use the integral form for now),
				$$\mathcal{E} = -\dv{\Phi}{t}$$
				and by Ohm's law
				$$i = \frac{\mathcal{E}}{R} = -\frac{1}{R}\dv{\Phi}{t} = -\frac{1}{R}\dv{\left(\frac{\mu_0 I_0 \ell (1 - kt)}{\pi}\ln\left(\frac{d+w}{d}\right)\right)}{t} = -\frac{\mu_0 I_0 \ell k}{\pi r}\ln\left(\frac{d+w}{d}\right)$$
				as desired (where the negative denotes the reversed direction of current).
			\end{soln}
			\ii What is the direction of the current, if any, induced in the loop as seen in Figure 3? Justify your answer.
			\begin{answer*}
				\framebox{Clockwise}.
			\end{answer*}
			\begin{soln}
				The magnetic field going in the $+\vu{j}$ direction (into the page) is decreasing as a function of time, so the flux is as well. By Lenz's law, a back electromotive force is created to counteract this decrease in flux, so the induced electromotive force creates flux to oppose the loss of magnetic field. This back pseudo-magnetic field also points in the $+\vu{j}$ direction. By the right hand rule (sticking the thumb in the direction of the pseudo-magnetic field and curling the fingers around to indicate the direction of the current), the induced current must be going clockwise.
			\end{soln}
		\end{enumerate}
    \end{enumerate}

\end{document}