Exam 3 Solutions

1. A hollow spherical shell carries charge density
   
   $$\rho = \frac{k}{r^2}$$
   
   
   in the region $a\le r\le b$.
   1. Find the electric field in the three regions (i) $r{<}a$ (ii) $a{<} r{<} b$ (iii) $r{>}b$.
      
      

      (i) $r{<}a$

      We use Gauss' law, for a spherically symmetric radial electric field. Thus

      
      

      $$\int \vec{E} \cdot d\vec{a} = \frac{Q_{\rm enclosed}}{\ep... ...uad \Rightarrow \quad\quad \vec{E} = 0, \mbox{for} r < a .$$
      
      

      (ii) $a{<} r{<} b$

      Here Gauss' law, for a spherically symmetric radial electric field, gives

      $$\begin{eqnarray*} \int \vec{E} \cdot d\vec{a}& =& \frac{Q_{\rm enclosed}}{\e... ...r^2} 4\pi r^2dr = \ \frac{4\pi k}{\epsilon_o} (r-a) . \end{eqnarray*}$$
      
      
      
      
      
      Thus
      
      $$\vec{E} = \frac{k}{\epsilon_o r^2} (r-a) \hat{r}, \mbox{for} a< r <b .$$
      
      

      (iii) $r{>}b$

      As the only charge lies between $a\le r\le b$, Gauss' law gives

      $$\begin{eqnarray*} \int \vec{E} \cdot d\vec{a}& =& \frac{Q_{\rm enclosed}}{\e... ...r^2} 4\pi r^2dr = \ \frac{4\pi k}{\epsilon_o} (b-a) . \end{eqnarray*}$$
      
      
      
      
      
      Thus
      
      $$\vec{E} = \frac{k}{\epsilon_o r^2} (b-a) \hat{r}, \mbox{for} b< r .$$
      
      

   2. Find the electric potential in the three regions (i) $r{<}a$ (ii) $a{<} r{<} b$ (iii) $r{>}b$.
      
      

      Setting $r\rightarrow\infty$ as the zero of potential, we'll work our way backwards in finding the potential:

      (iii) $r{>}b$

      $$\begin{eqnarray*} V&=& - \int_\infty^r \vec{E} \cdot d\vec{l} = - \... ... \frac{dr}{r^2} = \ \frac{k(b-a)}{\epsilon_o r } . \end{eqnarray*}$$
      
      
      
      
      

      (ii) $a{\le} r{\le} b$

      $$\begin{eqnarray*} V&=& - \int_\infty^r E dr = \ - \int_\infty^b \f... ...) - \frac{k}{\epsilon_o } \ln \left( \frac{r}{b} \right) . \end{eqnarray*}$$
      
      
      
      
      

      (i) $r{<}a$

      
      

      $$V = - \int_\infty^r E dr = - \int_\infty^a E dr... ... - \frac{k}{\epsilon_o } \ln \left( \frac{a}{b} \right) ,$$
      
      
      using our results from the integrals in part (ii) at $r=a$.

2. A parallel plate capacitor has plates at $x = \pm a$ and carries charge $Q$. A conducting plate of thickness $2\alpha$ is inserted midway between the capacitor plates.
   1. Display quantitatively the charge distribution and electric field for this configuration. (That is, display, as a function of $x$, exactly where and how much charge is carried, and display both the magnitude and direction of $\vec{E} (x)$. Neglect finite size effects in determining $\vec{E}$, and don't show a derivation.)

      

      The capacitor has, of course, charge $+Q$ and $-Q$ on its plates. The conductor has $\vec{E} = 0$ inside, and thus must have screening charges $-Q$ and $+Q$ on its outer surfaces as shown. The electric field in the capacitor, for $\alpha < \vert x\vert < a$, is its usual value $\sigma/\epsilon_o$ for a parallel plate capacitor; the two additional conductor surfaces at $\pm \alpha$ make cancelling contributions to the field here.

   2. By what factor is the capacitance changed, when the conducting plate is inserted? (Show all your work.)

      
      
      

      $C = Q/V$ and the separated charge $Q$ is the same in both cases. $V$ is given by
      

      $$V= - \int_-^+ \vec{E} \cdot d\vec{l} = \int_+^- {E} dx .$$
      
      
      Before $\vec{E} = \sigma/\epsilon_o \hat{x}$ for the entire domain $-a <x <a$ between the capacitor plates, so
      
      $$V = \int_{-a}^a \frac{\sigma}{\epsilon_o} dx = \frac{2a\sigma}{\epsilon_o} .$$
      
      
      After $\vec{E} = \sigma/\epsilon_o \hat{x}$ is nonzero only for $\alpha < \vert x\vert < a$, so
      
      $$V' = \int_{-a}^{-\alpha} \frac{\sigma}{\epsilon_o} dx + \... ...silon_o} dx = \frac{2(a-\alpha)\sigma}{\epsilon_o} .$$
      
      
      Thus the new capacitance is
      
      $$C' = \frac{Q}{V'} = \frac{Q}{V} \frac{V}{V'} = C \frac{V}{V'} = C \cdot \frac{a}{a-\alpha } .$$
      
      
      The capacitance increases, as we store the same charge with a smaller potential difference (voltage).

3. 
   

   $\parbox{2.3in}{Two infinite grounded metal plates lie parallel to th... ...trip insulated from the two plates, carrying constant charge density $\sigma$.}$

   NOTE: The symmetry of this situation implies that $V (x,y) = V (-x, y)$.

   1. What are the boundary conditions for $V(x,y)$, at
      1. $y=0$ $V(y=0) = 0$
      2. $y=a$ $V(y=a) = 0$

         (This is the meaning of the term ``grounded.'')

         
         
         

      3. $x \rightarrow \pm \infty$ $V(x \rightarrow \pm \infty) \rightarrow 0$

         Our finite charge distribution at $x=0$ must have negligible impact on the potential infinitely far away.

         
         
         

      4. $x=0$

         The surface charge $\sigma$ at $x=0$ implies the boundary condition
         

         $$\vec{E}_{\rm right} - \vec{E}_{\rm left} = \frac{\sigma}{\epsilon_o} \hat{x}$$
         
         
         or
         
         $$E_{\rm x, right} - E_{\rm x, left} = \frac{\sigma}{\epsil... ...V}{\partial x} \right\vert _{\rm left} \mbox{at $x=0$} .$$
         
         

      5. Use your result for (iv) and the symmetry of the problem to state the boundary condition for $x=0$, solely in terms of values approaching zero from the right (positive values of $x$).
         
         

         The symmetry of the problem tells us that $V (x,y) = V (-x, y)$ so $E_x (x,y) = -E_x (-x, y)$ (the electric field points away from $x=0$ in both cases.) In particular at $x=0$, $E_{\rm x, right} = - E_{\rm x, left}$ so
         

         $$2 E_{\rm x, right} = \frac{\sigma}{\epsilon_o} = -2 \left.... ...{\partial x} \right\vert _{\rm right} \mbox{at $x=0$} .$$
         
         

   2. Solve Laplace's equation for $V(x,y)$ in the region $x\ge 0, 0 \le y \le a$. Show all of your reasoning.

      
      
      

      Separating variables, $V(x,y) = X(x) Y(y)$ gives
      

      $$\nabla^2 V = 0\quad\quad \Rightarrow \quad\quad \frac{1}{X} ... ...l x^2} + \frac{1}{Y} \frac{\partial^2 Y}{\partial y^2} = 0 ,$$
      
      
      which means that each term must be an equal and opposite constant, so as to cancel. We choose the constant
      
      $$\frac{1}{X} \frac{\partial^2 X}{\partial x^2} = k^2 = -\ \frac{1}{Y} \frac{\partial^2 Y}{\partial y^2} ,$$
      
      
      because this gives exponential solutions for $x$ (which can be made to vanish as $x\rightarrow \infty$) and sinusoids for $y$ (which can be made to vanish at $y = 0, a$). This gives the solutions
      
      $$X = \left\{ \begin{array}{l} e^{kx}\ e^{-kx}\end{array} \right. ;$$
      
      
      we discard the $e^{kx}$ solution as it violates our $x\rightarrow \infty$ boundary condition. For $y$ it gives
      
      $$Y = \left\{ \begin{array}{l} \sin{ky}\ \cos{ky}\end{array} \right. ;$$
      
      
      we discard the $\cos{ky}$ solution as it violates our $y=0$ boundary condition. To obey our $y=a$ boundary condition we fix $k = n\pi/a$, so that $Y(y)$ has a zero at $y=a$. Thus far, then, we have imposed boundary conditions (i) -- (iii) and have as our most general solution obeying these boundary conditions
      
      $$V(x,y) = \sum_n a_n \sin \frac{n\pi y}{a} e^{ - \frac{n\pi x}{a}} .$$
      
      
      This has x-derivative
      
      $$\frac{\partial V}{\partial x} = \sum_n \left(- \frac{a_n ... ...right) \sin \frac{n\pi y}{a} e^{ - \frac{n\pi x}{a}} ,$$
      
      
      or, at $x=0$,
      
      $$\left.\frac{\partial V}{\partial x} \right\vert _{\rm x=0, r... ...t) \sin \frac{n\pi y}{a} = - \frac{\sigma}{2\epsilon_o} ,$$
      
      
      imposing our boundary condition (v) in the final step. We recognize this as a Fourier sin series for the constant function $-{\sigma}/{2\epsilon_o}$, with coefficients
      $$\begin{eqnarray*} \left(- \frac{a_n n\pi}{a} \right) &=& \frac{2}{a} \int_0... ...{for $n$ even}\ 2& \mbox{for $n$ odd}\end{array} \right. \end{eqnarray*}$$
      
      
      
      
      
      Thus
      
      $$a_n = \left\{ \begin{array}{cl} 0&\mbox{for $n$ even}\ \f... ...\pi)^2 \epsilon_o}& \mbox{for $n$ odd}\end{array} \right.$$
      
      
      giving
      
      $$V(x, y) = \frac{2a\sigma}{ \epsilon_o} \sum_n \frac{1}{(n\pi)^2} \sin \frac{n\pi y}{a} e^{ - \frac{n\pi x}{a}} .$$
      
      

4. The potential at the surface of an infinitely long metal pipe, of radius $R$, is given by
   
   $$V_o = k\cos 3\phi .$$
   
   
   1. Find the potential inside and outside the metal pipe.

      
      
      

      Outside the infinitely long pipe we have Laplace's equation (since there's no charge density), in cylindrical coordinates with no z-dependence. We've found the general solution for such coordinates to be
      

      $$V(s,\phi) = a + b\ln s + \sum_{n=1}^\infty (A_ns^{-n} + B_ns^n) (C_n\cos n\phi + D_n\sin n\phi ) .$$
      
      
      In the region $s\le R$ our solution must be finite as $s\rightarrow 0$, implying that $b$ and $A_n$ must vanish. This leaves solution
      
      $$V(s,\phi) = a + \sum_{n=1}^\infty s^n (C_n\cos n\phi + D_n\sin n\phi ) ,$$
      
      
      renormalizing the coefficients $C_n$ and $D_n$ for convenience.

      In the region $s\ge R$ our solution must be finite as $s\rightarrow \infty$, implying that $b$ and $B_n$ must vanish. This leaves solution
      

      $$V(s,\phi) = a' + \sum_{n=1}^\infty s^{-n} (C_n'\cos n\phi + D_n'\sin n\phi ) .$$
      
      

      Solutions from both domains must give the stated $V_o = k\cos 3\phi$ on the boundary, where $s=R$. This means that of all the allowed angular dependences, only the $\cos 3\phi$ term is present, with
      

      $$C_n R^n = C_n' R^{-n} = k .$$
      
      
      Thus our solution is
      
      $$V(s,\phi) = k\cos 3\phi \left\{ \begin{array}{ll}s^3/R^3 & s <R\\ R^3/s^3& s>R \end{array} \right. .$$
      
      

   2. What is the charge density $\sigma (\phi)$ on the pipe? (Assume there's no charge inside or outside the pipe.)
      
      

      The charge density on the pipe is fixed by the boundary condition
      

      $$- \frac{\sigma}{\epsilon_o} = \left. \left( \frac{\partial ... ...tial V_{\rm in}}{\partial s} \right) \right\vert _{s=R} .$$
      
      
      Thus
      
      $$\sigma (\phi) = -\epsilon_o k\cos3\phi \left. \left( -3R^... ... \right\vert _{s=R} = \frac{6\epsilon_o k\cos3\phi}{R} .$$
      
      

5. 
   

   $\parbox{2.5in}{ Current flows through a hollow conducting pipe of ra... ...on, $\vec{B}$points in the $+\hat{\phi}$ direction. \par \end{enumerate}\par }$

   1. Taking the current density $J$ to be constant for $R-a \le s \le R$, find the induced magnetic field in three regions: (i) $s{<} R-a$, (ii) $R-a {<} s {<} R$, (iii) $s {>} R$.

      
      
      We use Ampere's law to find the induced magnetic field,
      

      $$\oint \vec{B} \cdot d\vec{l} = \mu_o I_{\rm enclosed} ,$$
      
      
      where we take a circular path at fixed $s$, with $\vec{B} = B(s) \hat{\phi}$ by symmetry. Thus

      (i) for $s{<} R-a$

      Here $I_{\rm enclosed} = 0$, so $\vec{B} = 0$.

      (ii) for $R-a \le s \le R$

      Ampere's law here gives

      $$\begin{eqnarray*} 2\pi s B &=& \mu_o \int^s J(s) da\\ &=& \mu_o \int_{R... ...J_o 2\pi s ds = \mu_o\pi J_o ( s^2 - (R-a)^2 ) , \end{eqnarray*}$$
      
      
      
      
      
      so that
      
      $$\vec{B} = \frac{\mu_oJ_o}{2s} ( s^2 - (R-a)^2 ) \hat{\phi} .$$
      
      

      (iii) for $s {>} R$

      Here Ampere's law gives
      

      $$2\pi s B = \mu_o I_{\rm total} = \mu_o\pi J_o ( R^2 - (R-a)^2 ) ,$$
      
      
      using our integral above to find $I_{\rm total}$. Thus
      
      $$\vec{B} = \frac{\mu_o I_{\rm total}}{2\pi s} \hat{\phi} = \frac{\mu_oJ_o}{2s} ( R^2 - (R-a)^2 ) \hat{\phi} .$$
      
      

   2. A rectangular current loop of height $l$, width $a$, and resistance $R$ is placed to the right of the conducting pipe, with its left leg a distance $s {>} R$ from the center of the pipe. The current through the pipe increases at a constant rate, $dI/dt = k$.
      1. What current is induced in the loop, and in what direction does it flow?
         
         

         $\parbox{2.4in}{We have the geometry pictured, with $\vec{B}$ inside... ...lux enclosed by the current loop. Thus by Faraday's law we have an induced emf}$

         $$\begin{eqnarray*} \varepsilon &=& - \frac{d\Phi}{dt} =\ - \frac{d }{dt} \l... ...{dI }{dt} =\ - \frac{kl\mu_o }{2\pi } \ln \frac{s+a}{s} . \end{eqnarray*}$$
         
         
         
         
         
         The current induced in the loop is thus $I= \varepsilon/R$ or
         
         $$I = - \frac{kl\mu_o }{2\pi R} \ln \frac{s+a}{s} .$$
         
         
         The minus sign means that positive current flows in the negative, or left-handed, sense about $\vec{B}$, which is counterclockwise in the loop as drawn.

      2. Explain why the direction of the induced current agrees with Lenz' law.

         
         
         

         The induced current induces its own secondary induced magnetic field, whose direction is given by the right hand rule. For counterclockwise current in the loop, the secondary induced $\vec{B}$ points out of the page, in the area inside the current loop. This secondary induced $\vec{B}$ field opposes the increasing $\vec{B}$ field into the page, due to the increasing current through the conducting pipe. It acts to oppose the change in magnetic flux, in agreement with Lenz' law.

6. A solenoid carries the alternating current $I = I_o \cos \omega t$.

   NOTE: Questions (a) and (b) are just a special case of homework 11, assigned problem 5 -- feel free to cite your result from there.

   1. What is the magnetic field due to the solenoid current?
      
      

      Citing the result from homework 11, assigned problem 5,
      

      $$\vec{B} = \left\{ \begin{array}{ll} \mu_o nI(t) \hat{z} =... ...mbox{inside} \ [5pt] 0&\mbox{outside}\end{array} \right. \ ,$$
      
      
      where $\hat{z}$ is the axis that the solenoid's current flows counterclockwise about.

   2. What, if any, electric field does this magnetic field induce?
      
      

      Citing the result from homework 11, assigned problem 5, Faraday's law determines an induced electric field, due to the changing magnetic flux, of
      

      $$\vec{E} = \left\{ \begin{array}{ll} - \frac{ \mu_o n s}{2}\... ...\hat{\phi}&\mbox{outside the solenoid}\end{array} \right. .$$
      
      

   3. What, if any, $\vec{B}$ field is induced outside the solenoid by the changing $\vec{E}$ field you calculated in (b)?
      
      

      $\parbox{2.6in}{ Outside the solenoid, we have a tangential $\vec{E}$... ...dius $s$, $\vec{B}$ must point upwards, in the $\hat{z}$-direction, as shown.}$

      
      
      

      $\parbox{2.4in}{ We have the Ampere-like displacement current law \be... ...e plane of the loop. $\vec{B}$ points up in the z-direction, as argued above.}$

      For the left hand side, since $\vec{B} = B(s) \hat{z}$, the loop integral gets contributions only on the vertical legs, of
      

      $$\oint \vec{B} \cdot d\vec{l} = \int_0^l B(s_o) dz - \int_0^l\ B(s) dz = l (B(s_o) - B(s)) .$$
      
      

      For the right hand side, since ${\partial \vec{E}}/{\partial t}$ is perpendicular to the loop surface, we have

      $$\begin{eqnarray*} \mu_o\epsilon_o \int\ \frac{\partial \vec{E}}{\partial t} \c... ...psilon_o n a^2l}{2} I_o\omega^2 \cos\omega t \ln(s/s_o) , \end{eqnarray*}$$
      
      
      
      
      
      substituting our result of (b) for $\vec{E}$ outside the solenoid.

      Putting both results together,
      

      $$\vec{B} = B(s) \hat{z} = \left( - \frac{ \mu_o^2\epsilon_o... ...} I_o\omega^2 \cos\omega t \ln s + k \right) \hat{z} \ ,$$
      
      
      where $k$ is an $s$-independent constant. Note the sign: this induced $\vec{B}$ is out of phase with the solenoid-induced $\vec{B}$ of part (a). This makes sense: this secondary induced magnetic field is opposing the change in flux caused by the changing solenoid field.