slides_lecture02_main.tex

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\title[PHYS 201 / Lecture \thislecture]
{
  PHYS 201 / Lecture \thislecture \\
    {\it Electric flux; Gauss' law}\\
}

\input{slides_author.tex}

\begin{frame}[plain]
  \titlepage
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\input{slides_lecture01_summary.tex}

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\begin{frame}{Plan for Lecture \thislecture}

\begin{itemize}
{\small
\item {\bf Electric flux}
  \begin{itemize}
  {\scriptsize
     \item The "flow" of an electric field (i.e. the number of electric field lines) through a surface.
  }
  \end{itemize}

\item {\bf Gauss' law}
  \begin{itemize}
  {\scriptsize
     \item Relates the flux through a closed surface with the net charge contained in it.
     \item Our first Maxwell equation! Will study both its `differential' and `integral' forms.
  }
  \end{itemize}
}
\end{itemize}

\end{frame}

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\begin{frame}{Electric flux}

The flux $\Phi_E$ of the electric field $\vec{E}$ through a surface S is the {\em number of field lines}
flowing through S.
\begin{itemize}
 \item
   The number of field lines, and thus $\Phi_E$ is proportional to $|\vec{E}|$ and, obviously, proportional to S.
 \item
   The flux decreases if the surface is not perpendicular to the field, and becomes zero when the surface
   is parallel to the field.
\end{itemize}

\begin{columns}
  \begin{column}{0.33\textwidth}
   \begin{center}
    \includegraphics[width=0.98\textwidth]{./images/schematics/electric_flux_surface_perpendicular.png}\\
   \end{center}
  \end{column}
  \begin{column}{0.33\textwidth}
   \begin{center}
    \includegraphics[width=0.98\textwidth]{./images/schematics/electric_flux_surface_angle.png}\\
   \end{center}
  \end{column}
  \begin{column}{0.33\textwidth}
   \begin{center}
    \includegraphics[width=0.98\textwidth]{./images/schematics/electric_flux_surface_parallel.png}\\
   \end{center}
  \end{column}
\end{columns}

\begin{center}
  We can write the electric flux as $\Phi_E = |\vec{E}| \; S \; cos\theta$.
\end{center}

\end{frame}


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\begin{frame}{Electric flux}

\begin{columns}
  \begin{column}{0.33\textwidth}
   \begin{center}
    \includegraphics[width=0.98\textwidth]{./images/schematics/electric_flux_surface.png}\\
    \includegraphics[width=0.98\textwidth]{./images/schematics/surface_normal.png}\\
    {\scriptsize
     Note that if the surface is closed, it is customary
     to define the direction of each element dS to point outwards.\\
   }
   \end{center}
  \end{column}
  \begin{column}{0.67\textwidth}
  {\small
    It is useful to express a surface S as a vector:
    The surface vector $\vec{S}$ is perpendicular to the surface, and has a size which is proportional to its area.
    The vector can point to either side of S. \\
    Then the electric flux $\Phi_E$ can be expressed as the dot product of $\vec{E}$ and $\vec{S}$:
    \begin{equation*}
       \Phi_{E} = \vec{E} \cdot \vec{S} =  |\vec{E}| \; |\vec{S}| \; cos\theta
    \end{equation*}

    Of course, S may not be flat, or $\vec{E}$ might change along S.
    But we can always divide S in infinitesimally small areas dS that are approximately
    flat and within which $\vec{E}$ remains constant.\\
    Then the flux $d\Phi_{E}$ through each element dS is given by $d\Phi_{E} = \vec{E} \cdot d\vec{S}$
    and the total flux $\Phi_E$ by the integral:
    \begin{equation*}
       \Phi_{E} = \int_{S} \vec{E} \cdot d\vec{S}
    \end{equation*}
  }
  \end{column}
\end{columns}

\end{frame}

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{
\problemslide

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% Halliday-Resnick, Question 1, page 621
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\begin{frame}{Worked example: Calculation of electric flux (Easy)}

  \begin{blockexmplque}{Question}
    A surface has the area vector $\vec{A}$ = (2$\hat{x}$+3$\hat{y}$) m$^2$.
    What is the flux of a uniform electric field through the area if the field is
    \begin{itemize}
       \item $\vec{E}$ = 4$\hat{x}$ N/C, and
    	 \item $\vec{E}$ = 4$\hat{z}$ N/C?
    \end{itemize}
  \end{blockexmplque}
  \vspace{0.1cm}

  The flux of a uniform electric field $\vec{E}$ through a planar surface
  $\vec{A}$ is given by
  \begin{equation*}
  	\Phi = \vec{E} \cdot \vec{A}
  \end{equation*}

  Therefore, for the 2 cases:
  \begin{itemize}
    \item $\Phi = \Big\{ 4\hat{x} \cdot (2\hat{x} + 3\hat{y}) \Big\} \; \frac{N\cdot m^2}{C}
  	            = \Big\{ 8 (\hat{x} \cdot \hat{x}) + 12 (\hat{x} \cdot \hat{y}) \Big\} \; \frac{N\cdot m^2}{C}
  	            = 8 \; \frac{N\cdot m^2}{C}$
  	\item $\Phi = \Big\{ 4\hat{z} \cdot (2\hat{x} + 3\hat{y}) \Big\} \; \frac{N\cdot m^2}{C}
               	= \Big\{ 8 (\hat{z} \cdot \hat{x}) + 12 (\hat{z} \cdot \hat{y}) \Big\} \; \frac{N\cdot m^2}{C}
  	            = 0 \; \frac{N\cdot m^2}{C}$
  \end{itemize}

\end{frame}


} % Example
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{
\problemslide

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\begin{frame}{Worked example: Calculation of electric flux}

  \begin{blockexmplque}{Question}
    An electric field $\vec{E}$ is given by:
    \begin{equation*}
     \vec{E} =
      \big\{ 2x \; \frac{N}{m \cdot C} \big\} \hat{x} +
      \big\{ 2 \; \frac{N}{C} - 2y \frac{N}{m \cdot C} \big\} \hat{y} +
      \big\{ 4z \; \frac{N}{m \cdot C} \big\} \hat{z}
    \end{equation*}
    \vspace{-0.3cm}
    \begin{columns}
      \begin{column}{0.47\textwidth}
        \begin{center}
          \includegraphics[width=0.99\textwidth]{./images/problems/lect02_electric_flux_area_on_cylinder.png}\\
        \end{center}
      \end{column}
      \begin{column}{0.53\textwidth}
        Determine the electric flux crossing \\
        a 1 mm $\times$ 1 mm area on the surface \\
        of the cylindrical shell at:
        \begin{equation*}
          r = 10 \; m, \;\; z = 2 \; m, \; \phi = 53.2^{o}
        \end{equation*}
     \end{column}
   \end{columns}
  \end{blockexmplque}
  \vspace{0.4cm}

\end{frame}

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\begin{frame}{Worked example: Calculation of electric flux}

  At the point P:
  \begin{equation*}
     x = r  cos\phi = (10 \; m)  cos53.2^o = (10 \; m)  0.6 = 6 \; m
  \end{equation*}
  \begin{equation*}
     y = r  sin\phi = (10 \; m)  sin53.2^o = (10 \; m)  0.8 = 8 \; m
  \end{equation*}

  \vspace{0.2cm}

  The electric field $\vec{E}$ varies very little over the small given area.
  Taking it to be constant is a good approximation.\\
  Therefore, I will just substitute x = 6 m, y = 8 m, and z = 2 m
  in the given expression for $\vec{E}$:
  \begin{equation*}
   \vec{E} =
     \big\{ 2 \cdot (6\; m) \; \frac{N}{m \cdot C} \big\} \hat{x} +
     \big\{ 2 \; \frac{N}{C} - 2 \cdot (8 \; m) \frac{N}{m \cdot C} \big\} \hat{y} +
     \big\{ 4 \cdot (2\;m) \; \frac{N}{m \cdot C} \big\} \hat{z} \Rightarrow
  \end{equation*}
  \begin{equation*}
   \vec{E} =
   \big\{ 12 \hat{x} - 14 \hat{y} + 8 \hat{z} \big\} \frac{N}{C}
 \end{equation*}

\end{frame}

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\begin{frame}{Worked example: Calculation of electric flux}

Now, I need to express the area $d\vec{S}$ in vector form.\\
\vspace{0.2cm}
Because the area is very small ($|d\vec{S}|$ = 1 mm$^2$) in comparison to the
size of the cylinder (r = 10 m), we can consider the area $d\vec{S}$ to be {\em planar}.\\
\vspace{0.1cm}
Also, as one can easily see, the vector $d\vec{S}$ lies on the xy plane.\\
\vspace{0.3cm}
Therefore:
\begin{equation*}
  d\vec{S} = dS_x \hat{x} + dS_y \hat{y}
\end{equation*}
where:
\begin{equation*}
   dS_x = |d\vec{S}| cos\phi = (10^{-6} \; m^2) \; cos53.2^o = 0.6 \times 10^{-6} \; m^2
\end{equation*}
\begin{equation*}
   dS_y = |d\vec{S}| sin\phi = (10^{-6} \; m^2) \; sin53.2^o = 0.8 \times 10^{-6} \; m^2
\end{equation*}
\vspace{0.2cm}
Putting everything together:
\begin{equation*}
  d\vec{S} = \big\{ 0.6 \hat{x} + 0.8 \hat{y} \big\} \times 10^{-6} \; m^2
\end{equation*}

\end{frame}

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\begin{frame}{Worked example: Calculation of electric flux}

We're now ready to calculate the flux d$\Phi$ of the electric field $\vec{E}$
through the small patch $d\vec{S}$:\\
\vspace{0.2cm}
\begin{equation*}
  d\Phi = \vec{E} \cdot d\vec{S} \Rightarrow
\end{equation*}
\vspace{0.1cm}
\begin{equation*}
  d\Phi =
  \big\{ 12 \hat{x} - 14 \hat{y} + 8 \hat{z} \big\} \frac{N}{C}
   \cdot \big\{ 0.6 \hat{x} + 0.8 \hat{y} \big\} \times 10^{-6} \; m^2 \Rightarrow
\end{equation*}
\vspace{0.1cm}
\begin{equation*}
  d\Phi =
  \big\{ 12 \cdot 0.6 - 14 \cdot 0.8 + 8 \cdot 0 \big\} \times 10^{-6} \frac{N \cdot m^2}{C} =
  \big\{ 7.2 - 11.2 + 0 \big\} \times 10^{-6} \frac{N \cdot m^2}{C}
  \Rightarrow
\end{equation*}
\vspace{0.1cm}
\begin{equation*}
  d\Phi = -4 \times 10^{-6} \frac{N \cdot m^2}{C}
\end{equation*}
\vspace{0.1cm}
Q: What is the significance of the minus sign?
\end{frame}

} % Example
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\begin{frame}{Electric flux through a closed surface}

\begin{center}
 What is the electric flux through the closed surface shown?\\
\end{center}
\vspace{0.2cm}

\begin{columns}
  \begin{column}{0.38\textwidth}
   {\small
     No charge is contained in the closed surface.\\
     \vspace{0.2cm}
     Since flux lines start from positive charges and end in negative charges,
     every line that enters the closed surface eventually exits.\\
     \vspace{0.2cm}
     Field lines that enter and field lines that exit contribute with different signs in $\oint \vec{E} d\vec{S}$
     (recall that the surface vector points outwards).\\
     \vspace{0.2cm}
     The flux is 0.
   }
   \begin{center}
   \end{center}
  \end{column}
  \begin{column}{0.62\textwidth}
    \includegraphics[width=0.90\textwidth]{./images/schematics/electric_dipole_field_lines_closed_surf_noq.png}\\
  \end{column}
\end{columns}
\end{frame}


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\begin{frame}{Electric flux through a closed surface}

\begin{center}
 What is the electric flux through the closed surface shown?\\
\end{center}
\vspace{0.2cm}

\begin{columns}
  \begin{column}{0.38\textwidth}
   {\small
     Both a positive and an equal negative charge are contained within the closed surface.
     Certain lines originating from the positive charge can reach the negative charge without exiting the closed surface.
     Lines that do exit though the closed surface, re-enter to terminate on the negative charge. \\
     \vspace{0.2cm}
     The flux is 0 again.
   }
   \begin{center}
   \end{center}
  \end{column}
  \begin{column}{0.62\textwidth}
    \includegraphics[width=0.90\textwidth]{./images/schematics/electric_dipole_field_lines_closed_surf_bothq.png}\\
  \end{column}
\end{columns}
\end{frame}


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\begin{frame}{Electric flux through a closed surface}

\begin{center}
 What is the electric flux through the closed surface shown?\\
\end{center}
\vspace{0.2cm}

\begin{columns}
  \begin{column}{0.38\textwidth}
   {\small
     All field lines are exiting.
     The flux is non-zero and positive.
   }
   \begin{center}
   \end{center}
  \end{column}
  \begin{column}{0.62\textwidth}
    \includegraphics[width=0.90\textwidth]{./images/schematics/electric_dipole_field_lines_closed_surf_posq.png}
  \end{column}
\end{columns}
\end{frame}


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\begin{frame}{Electric flux through a closed surface}

\begin{center}
 What is the electric flux through the closed surface shown?\\
\end{center}
\vspace{0.2cm}

\begin{columns}
  \begin{column}{0.38\textwidth}
   {\small
     Now all field lines are entering.
     The flux is non-zero. It has the same value as in the previous page, but now it is negative.
   }
   \begin{center}
   \end{center}
  \end{column}
  \begin{column}{0.62\textwidth}
    \includegraphics[width=0.90\textwidth]{./images/schematics/electric_dipole_field_lines_closed_surf_negq.png}\\
  \end{column}
\end{columns}
\end{frame}


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\begin{frame}{Gauss' law}

In the previous few slides, the {\em essence} of {\bf Gauss' law} was illustrated.\\
\vspace{0.3cm}

\begin{columns}
  \begin{column}{0.33\textwidth}
   \begin{center}
    \includegraphics[width=0.90\textwidth]{./images/people/gauss.jpg}\\
    \vspace{0.2cm}
    {\scriptsize
      Carl Friedrich Gauss\\ (1777-1855)\\ German mathematician\\
    }
   \end{center}
  \end{column}
  \begin{column}{0.67\textwidth}
      {\em \bf
       The electric flux $\Phi_{E} = \oint \vec{E} d\vec{S}$ through a closed surface
       is related to the net charge $Q_{enc}$ enclosed within the surface.\\
      }
      \vspace{0.4cm}
      As it turns out, this relation is simply:
      \begin{equation*}
         {\color{red}
          \Phi_{E} = \frac{Q_{enc}}{\epsilon_0}
         }
      \end{equation*}

      It doesn't matter how elaborate the shape of the surface or the enclosed
      charge distribution (and the resulting field) may be:\\
      The above simple relation {\bf always holds}!\\
  \end{column}
\end{columns}
\end{frame}

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% Worked example
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{
\problemslide

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\begin{frame}{Worked example: Finding $\Phi$ from $Q_{enc}$ (Easy)}

  \begin{blockexmplque}{Question}
    \begin{columns}
      \begin{column}{0.40\textwidth}
        \begin{center}
          \includegraphics[width=0.99\textwidth]{./images/problems/lect02_enclosed_charges.png}\\
        \end{center}
      \end{column}
      \begin{column}{0.60\textwidth}
        The figure on the left shows 5 charged lumps of a material.
        The cross-section of the surface $S$ is indicated.\\
        What is the {\em net electric flux} through the surface if:\\
        \begin{itemize}
         \item $q_1$ = $q_4$ = 3.1 nC,
         \item $q_2$ = $q_5$ = -5.9 nC, and
         \item $q_3$ = -3.1 nC?
        \end{itemize}
     \end{column}
   \end{columns}
  \end{blockexmplque}
  \vspace{0.2cm}
  \begin{equation*}
  \Phi =
    \frac{Q_{enc}}{\epsilon_0} =
    \frac{q_1+q_2+q_3}{\epsilon_0} =
    \frac{(3.1 - 5.9 - 3.1) \times 10^{-9} \; C}{8.85 \times 10^{-12} \; C^2/(N \cdot m^2)} =
    -0.67 \times 10^{3} \frac{N \cdot m^2}{C}
  \end{equation*}

\end{frame}

} % Example
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% Worked example
% Halliday-Resnick, Sec. 23.4, Problem 5, page 622
%

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{
\problemslide

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\begin{frame}{Worked example: Finding $\Phi$ from $Q_{enc}$}

  \begin{blockexmplque}{Question}
       A proton is a distance $d$/2 directly above the center of a square of side $d$.
    	 What is the magnitude of the electric flux through the square?
       \begin{center}
           \includegraphics[width=0.70\textwidth]{./images/problems/lect02_charge_above_surface.png}
       \end{center}
  \end{blockexmplque}

\end{frame}

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\begin{frame}{Worked example: Finding $\Phi$ from $Q_{enc}$}

  Consider a closed Gaussian surface in the shape of a cube of edge length $d$,
  with the proton (of charge $q$) positioned in the centre of the cube.
  Gauss's law tells us that the flux $\Phi$ through that surface is given by:
  \begin{equation*}
  	\Phi = \frac{q}{\epsilon_0}
  \end{equation*}

  Because of the symmetry of the configuration, the same electric flux
  $\Phi_{face}$ passes through each of the 6 faces of the cube, so:
  \begin{equation*}
  	\Phi = 6 \Phi_{face}
  \end{equation*}

  Therefore:
  \begin{equation*}
  	\Phi_{face} = \frac{q}{6 \epsilon_0} \Rightarrow
  \end{equation*}
  \begin{equation*}
  	\Phi_{face} =
  	 \frac{1.6 \times 10^{-19} \; C}
  	      {6 (8.85 \times 10^{-12} \; \frac{C^2}{N \cdot m^2})} =
          3 \times 10^{-9} \; \frac{N \cdot m^2}{C}
  \end{equation*}

\end{frame}

} % Example
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% Worked example
% Halliday-Resnick, Sec. 23.4, Problem 10, page 622
%

{
\problemslide

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\begin{frame}{Worked example: Finding $Q_{enc}$ from $\Phi$}

  \begin{blockexmplque}{Question}
    The figure below shows a closed surface in the shape of a cube of
    edge length 2 m. It lies in a region where the non-uniform electric field
    $\vec{E}$ is given by the following:
    \begin{equation*}
      \vec{E} = (3x+4)\hat{x} + 6\hat{y} + 7\hat{z}.
    \end{equation*}
    In the above expression, $|\vec{E}|$ has units of N/C if $x$ is given in m.
    What is the net charge contained by the cube?
    \begin{center}
        \includegraphics[width=0.30\textwidth]{./images/problems/lect02_cube.png}
    \end{center}
  \end{blockexmplque}

\end{frame}

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\begin{frame}{Worked example: Finding $Q_{enc}$ from $\Phi$}

  It is easy to see that the constant terms of $\vec{E}$ make no contribution
  to the total flux, therefore we can consider only the non-constant component.\\
  \vspace{0.3cm}

  Let $\vec{E}_{nc}$ be that non-constant component of the field,
  which is given by:
  \begin{equation*}
  	\vec{E}_{nc} = (3x \frac{N \cdot m}{C}) \hat{x}
  \end{equation*}

  Let $\vec{S}_{x+}$ ($\vec{S}_{x-}$) be the area vector of the front (back)
  face of the cube lying on the $yz$ plane. That vector points towards
  positive (negative) $x$.\\
  \vspace{0.3cm}

  For a given $x$, $\vec{E}_{nc}$ is constant along the $yz$ plane.
  Since the field is uniform and the surfaces are planar,
  the flux through the 2 cube faces lying on $yz$ planes can be written as:
  \begin{equation*}
  	\Phi_{x} =
  	     \vec{S}_{x+} \cdot \vec{E}_{nc}(x = 0 \; m)
  		 + \vec{S}_{x-} \cdot \vec{E}_{nc}(x = -2 \; m)
  \end{equation*}

\end{frame}

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\begin{frame}{Worked example: Finding $Q_{enc}$ from $\Phi$}

  Because the magnitude of $\vec{E}_{nc}$ is proportional
  to $x$ and the first term is evaluated at $x$=0, the first term is zero.
  What remains is:
  \begin{equation*}
  	\Phi_{x} = \vec{S}_{x-} \cdot \vec{E}_{nc}(x = -2 \; m)
      	     = (4 \; m^2) (-\hat{x}) \cdot (-6 \; N/C) \hat{x}
  					 = 24 \; N \cdot m^2/C
  \end{equation*}

  Since $\vec{E}_{nc}$ points along $\hat{x}$,
  the flux $\Phi_{y}$ ($\Phi_{z}$) through the cube faces lying on $xz$ ($xy$)
  planes is zero.
  Therefore, the total flux through all 6 faces of the cube is:
  \begin{equation*}
  	\Phi =  24 \; N \cdot m^2/C
  \end{equation*}

  Finally, from Gauss's law:
  \begin{equation*}
  	\Phi = \frac{q_{enc}}{\epsilon_0} \Rightarrow
  	q_{enc} = \Phi \epsilon_0 \Rightarrow
  \end{equation*}
  \begin{equation*}
  	q_{enc} = (24 \; N \cdot m^2/C) (8.85 \times 10^{-12}
     \; \frac{C^2}{N \cdot m^2}) \Rightarrow
  \end{equation*}
  \begin{equation*}
  	q_{enc} = 2.124  \times 10^{-10} \; C
  \end{equation*}

\end{frame}

} % Example
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\begin{frame}{Gauss' law - Derivation for a simple case}

%\begin{center}
We will derive Gauss' law for a very simple case: A single charge Q in the centre of a sphere of radius R.
%\end{center}

\begin{columns}
  \begin{column}{0.34\textwidth}
   \begin{center}
    \includegraphics[width=0.85\textwidth]{./images/schematics/gauss_law_S.png}
    \begin{equation*}
        \vec{E}(\vec{r}) = \frac{Q}{4\pi\epsilon_0 r^{2}} \hat{r}
    \end{equation*}
    \begin{equation*}
        d\vec{S} = dS \hat{r}
    \end{equation*}
   \end{center}
  \end{column}
  \begin{column}{0.66\textwidth}
    \begin{equation*}
        \Phi_E = \oint_{sphere} \vec{E} \cdot d\vec{S} \Rightarrow
    \end{equation*}
    \begin{equation*}
        \Phi_E = \oint_{sphere} \big( \frac{Q}{4\pi\epsilon_0 R^{2}} \hat{r} \big) \cdot \big( dS \hat{r} \big) \xRightarrow {\hat{r}\cdot\hat{r}=1}
    \end{equation*}
    \begin{equation*}
        \Phi_E = \frac{Q}{4\pi\epsilon_0 R^{2}} \oint_{sphere} dS \Rightarrow
    \end{equation*}
    \begin{equation*}
        \Phi_E = \frac{Q}{4\pi\epsilon_0 R^{2}} 4\pi R^{2} \Rightarrow
    \end{equation*}
    \begin{equation*}
    {\bf \color{magenta}
        \Phi_E = \frac{Q}{\epsilon_0}
    }
    \end{equation*}
  \end{column}
\end{columns}
\end{frame}


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\begin{frame}{Gauss' law - Generalisation for any surface}

Although the law was derived for a very simple case, it is a {\bf general result}.\\
\vspace{0.2cm}

\begin{columns}
  \begin{column}{0.34\textwidth}
   \begin{center}
    \includegraphics[width=0.98\textwidth]{./images/schematics/gauss_law_generalisation_geom_0.png}\\
   \end{center}
  \end{column}
  \begin{column}{0.66\textwidth}
    The same result would have been obtained
    \begin{itemize}
     \item if the charge was not in the centre but in any other position within the sphere, or
     \item if I had any other closed surface shape instead of a sphere.
    \end{itemize}
  \end{column}
\end{columns}

Only one thing matters for the calculation of $\Phi_E$: The {\bf net charge within the surface}.

\end{frame}


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\begin{frame}{Gauss' law - Generalisation for any surface}

\begin{columns}
  \begin{column}{0.34\textwidth}
   \begin{center}
    \includegraphics[width=0.90\textwidth]{./images/schematics/gauss_law_generalisation_geom_1.png}\\
    {\scriptsize
     Consider element $dS$ on a general surface $S$ and the corresponding element $dS^{\prime}$ on a sphere $S^{\prime}$,
     both covering the same solid angle $d\Omega$
    }
   \end{center}
  \end{column}
  \begin{column}{0.66\textwidth}
     By definition, on the spherical surface $S^{\prime}$:
     \begin{equation*}
       d\Omega = \frac{dS^{\prime}}{r^{2}} \Rightarrow dS^{\prime} = r^{2} d\Omega
     \end{equation*}
     By construction:
     \begin{equation*}
       dS^{\prime} = dS cos\theta \Rightarrow dS = \frac{dS^{\prime}}{cos\theta} \Rightarrow dS = \frac{r^{2} d\Omega}{cos\theta}
     \end{equation*}
     So, the flux $\Phi_E$ through the general surface $S$ is
     \begin{equation*}
       \Phi_E = \int_{S} \vec{E} \cdot d\vec{S} = \int_{S} E dS cos\theta
          = \int_{\Omega} E \frac{r^{2} d\Omega}{\cancel{cos\theta}} \cancel{cos\theta} =
     \end{equation*}
     \begin{equation*}
        \int_{\Omega} E r^{2} d\Omega = \int_{S^{\prime}} E dS^{\prime}
        = \int_{S^{\prime}} \vec{E} d\pvec{S}' = \Phi_E^{sphere} = \frac{Q}{\epsilon_0}
     \end{equation*}
  \end{column}
\end{columns}
\end{frame}


%
%
%

\begin{frame}{Gauss' law - Generalization for any charge distribution}


Our result for a {\em single charge} enclosed by an arbitrary closed surface S is:
\begin{equation*}
   \oint_{S} \vec{E} \cdot d\vec{S} = \frac{Q}{\epsilon_0}
\end{equation*}

What is the {\bf form of Gauss' law if the surface S encloses an array of charges} (or a continuous charge distribution)?
The result can be obtained via a straightforward application of the superposition principle.\\

\begin{equation*}
  \left.
     \begin{array}{l}
      \oint_{S} \vec{E_{1}} \cdot d\vec{S} = \frac{Q_{1}}{\epsilon_0}\\
      \\
      \oint_{S} \vec{E_{2}} \cdot d\vec{S} = \frac{Q_{2}}{\epsilon_0}\\
      \\
      ...                                                      \\
      \\
      \oint_{S} \vec{E_{n}} \cdot d\vec{S} = \frac{Q_{n}}{\epsilon_0}
     \end{array}
  \right\}
%  \oint_{S} (\vec{E_{1}}+\vec{E_{2}}+...+\vec{E_{n}}) d\vec{S} = \frac{Q_{1}+Q_{2}+...+Q_{n}}{\epsilon_0} \Rightarrow
  \oint_{S} (\sum_{i=1}^{n}\vec{E_{i}}) \cdot d\vec{S} = \frac{\sum_{i=1}^{n}Q_{i}}{\epsilon_0} \Rightarrow
  \oint_{S} \vec{E} \cdot d\vec{S} = \frac{Q_{enc}}{\epsilon_0}
\end{equation*}

\end{frame}


%
%
%

\begin{frame}{Integral form of Gauss' law}

Our result for any array of charges (or continuous charge distribution), with net charge $Q_{enc}$,
enclosed by an arbitrary closed surface S is:
\begin{equation*}
  \oint \vec{E} \cdot d\vec{S} = \frac{Q_{enc}}{\epsilon_0}
\end{equation*}

This is known as the {\bf \underline{integral form} of Gauss' law}.\\
\vspace{0.2cm}

This form is useful in cases where the problem at hand has a symmetry.\\
The symmetry can be exploited to simplify the calculation of the integral.\\

\end{frame}



%
% Worked example
%

{
\problemslide

%
% Another
%

\begin{frame}{Worked example (Planar symmetry)}

  \begin{blockexmplque}{Question}
    The figure on the left shows two large, parallel, non-conducting sheets,
    each with a fixed uniform charge on one side.\\
    \begin{columns}
      \begin{column}{0.20\textwidth}
        \begin{center}
          \includegraphics[width=0.99\textwidth]{./images/problems/lect02_2_charged_planes.png}\\
        \end{center}
      \end{column}
      \begin{column}{0.80\textwidth}
        The magnitudes of the surface charge densities are\\
        $\sigma_{(+)}$ = 6.8 $\mu$C/m$^2$ for the positively charged sheet, and
        $\sigma_{(-)}$ = - 4.3 $\mu$C/m$^2$ for the negatively charged sheet.\\
        Find the electric field $\vec{E}$
        (a) to the left of the sheets,
        (b) between the sheets, and
        (c) to the right of the sheets.
     \end{column}
   \end{columns}
  \end{blockexmplque}
  \vspace{0.4cm}

  Note: We will revisit this example when we study the {\bf parallel plate capacitor}
  in Lecture 4.\\

\end{frame}

%
%
%

\begin{frame}{Worked example (Planar symmetry)}

  Consider an infinite positively charged plane with surface charge density
  $\sigma$ and the (appropriately chosen)
  cylindrical {\em Gaussian} surface shown below.\\

  \begin{columns}
   \begin{column}{0.60\textwidth}
    The symmetry of the problem indicates that $\vec{E}$ is
    {\bf perpendicular to the sheet}: There is no
    flux through the curved surface but only through the 2 circular end caps.\\
    \vspace{0.2cm}
    Additionally, because the plane is positively charged, the field $\vec{E}$
    is directed outwards.\\
   \end{column}
   \begin{column}{0.40\textwidth}
    \begin{center}
      \includegraphics[width=0.70\textwidth]{./images/problems/lect02_charged_plane_3d.png}\\
    \end{center}
   \end{column}
  \end{columns}
  The surface charge enclosed by the Gaussian surface is $Q_{enc} = \sigma A$.\\
  \vspace{0.2cm}
  Therefore, applying Gauss' law, we get:
  \begin{equation*}
    \oint \vec{E} \cdot d\vec{S} = \frac{Q_{enc}}{\epsilon_0} \Rightarrow
    EA + EA = \frac{\sigma A}{\epsilon_0} \Rightarrow
    {\color{red}
      E = \frac{\sigma}{2\epsilon_0}
    }
  \end{equation*}
  The electric field stays the same for any point, irrespectively of its distance from the plane.

\end{frame}

%
%
%

\begin{frame}{Worked example (Planar symmetry)}

Repeating the exercise for a negatively charged plane yield of the same magnitude.
with the difference that it would point inwards.\\
\vspace{0.2cm}

For the case of the 2 parallel plates, the electric field $\vec{E}$ anywhere in
space is given by the superposition of 2 fields:\\
\vspace{0.2cm}
\begin{itemize}
  \item The field $\vec{E}_{(+)}$, due to the positively charged plane, with direction
        away from that plane and magnitude:
        \begin{equation*}
            E_{(+)} =
              \frac{\sigma_{(+)}}{2\epsilon_0} =
              \frac{6.8 \times 10^{-6} \; C/m^2}{2(8.85 \times 10^{-12} \; C/(N \cdot m^2))} =
              3.84 \times 10^{5} N/C
        \end{equation*}
  \item The field $\vec{E}_{(-)}$, due to the positively charged plane, with direction
        away from that plane and magnitude:
        \begin{equation*}
            E_{(-)} =
              \frac{\sigma_{(-)}}{2\epsilon_0} =
              \frac{4.3 \times 10^{-6} \; C/m^2}{2(8.85 \times 10^{-12} \; C/(N \cdot m^2))} =
              2.43 \times 10^{5} N/C
        \end{equation*}
\end{itemize}

\end{frame}

%
%
%

\begin{frame}{Worked example (Planar symmetry)}

  \begin{columns}
   \begin{column}{0.50\textwidth}
     We can now work out the electric field on the left of the planes,
     between the planes, and on the right of the planes.
   \end{column}
   \begin{column}{0.50\textwidth}
    \begin{center}
      \includegraphics[width=0.90\textwidth]{./images/problems/lect02_2_charged_planes_solution.png}\\
    \end{center}
   \end{column}
  \end{columns}

     \begin{itemize}
        \item
          On the {\bf left} of the planes:\\
          $E_{L} = E_{(+)} - E_{(-)} = 1.4 \times 10^{5} \; N/C$
          directed to the left.\\
          \vspace{0.2cm}
        \item
          {\bf Between} the planes:\\
          $E_{B} = E_{(+)} + E_{(-)} = 6.3 \times 10^{5} \; N/C$
          directed to the right.\\
          \vspace{0.2cm}
        \item
          On the {\bf right} of the planes:\\
          $E_{R} = E_{(+)} - E_{(-)} = 1.4 \times 10^{5} \; N/C$
          directed to the right.\\
      \end{itemize}

\end{frame}


%
% Another
%

\begin{frame}{Worked example (Cylindrical symmetry)}

\begin{blockexmplque}{Question}
A long, straight, thin plastic rod is uniformly charged with +10 nC/m.
Calculate the magnitude of the electric field at 10 cm away from the rod.
What is the direction of the electric field at any point away from the rod?
\end{blockexmplque}
\vspace{0.4cm}

Consider a cylindrical surface with the rod running along its axis.\\
\vspace{0.2cm}

Gauss' law in integral form is:
\begin{equation*}
 \oint \vec{E}(r) \cdot d\vec{S} = \frac{Q_{encl}}{\epsilon_0}
\end{equation*}

Assuming the rod is infinitely long (i.e. ignoring fringe effects)
the electric field is everywhere perpendicular to the rod and points away from it.
Therefore, all the flux is going out of the sides of the cylinder.\\

\end{frame}

%
%
%

\begin{frame}{Worked example (Cylindrical symmetry)}

For a length L and linear charge density $\lambda$:

\begin{columns}
  \begin{column}{0.45\textwidth}
   \begin{center}
     \includegraphics[width=0.65\textwidth]{./images/problems/lect1_charged_rod.png}\\
   \end{center}
  \end{column}
  \begin{column}{0.55\textwidth}

    \begin{equation*}
      \oint \Big( E(r) \hat{r} \Big) \cdot \Big( dS \hat{r} \Big) = \frac{\lambda L}{\epsilon_0} \Rightarrow
    \end{equation*}

    \begin{equation*}
      E(r) 2\pi r L = \frac{\lambda L}{\epsilon_0} \Rightarrow
    \end{equation*}

    \begin{equation*}
      E(r) = \frac{\lambda}{2 \pi \epsilon_0 r}
    \end{equation*}

  \end{column}
\end{columns}

\vspace{0.3cm}

At the given distance away from the rod (10 cm), the magnitude of the electric field is:

\begin{equation*}
 E = 2 \cdot (9.0 \times 10^9 \; \frac{N \cdot m^2}{C^2}) \frac{10 \times 10^{-9} \; C/m}{0.1 \; m} \Rightarrow
 E = 1.8 \; kV/m
\end{equation*}

\end{frame}


%
% Another
%

\begin{frame}{Worked example (Spherical symmetry)}

  \begin{blockexmplque}{Question}
    A solid non-conductive sphere with radius R is uniformly charged and carries total charge Q.
    Calculate the electric field E at radius r from its centre for the following two cases:
    \begin{enumerate}
       \item Inside the sphere (r $<$ R).
       \item On the surface or outside the sphere (r $\geq$ R).
    \end{enumerate}
    How is the result for case 2 above related to that for a point charge Q at
    the origin? \\
  \end{blockexmplque}
  \vspace{0.4cm}

  The charge density is:
  \begin{equation*}
    \rho = \frac{Q}{\tau} = \frac{Q}{4\pi R^3 /3} = \frac{3Q}{4\pi R^3}
  \end{equation*}

\end{frame}

%
%
%

\begin{frame}{Worked example (Spherical symmetry)}

  Exploiting the spherical symmetry of the problem,
  the flux of the electric field at any distance r is:
  \begin{equation*}
    \Phi(r) = 4\pi r^2 E(r)
  \end{equation*}

  From Gauss' law:
  \begin{equation*}
    \Phi(r) = \frac{Q_{encl}(r)}{\epsilon_0}
  \end{equation*}

  Therefore:
  \begin{equation*}
    4\pi r^2 E(r) = \frac{Q_{encl}(r)}{\epsilon_0}
  \end{equation*}
  \begin{equation*}
    E(r) = \frac{Q_{encl}(r)}{4 \pi \epsilon_0 r^2}
  \end{equation*}

\end{frame}

%
%
%

\begin{frame}{Worked example (Spherical symmetry)}

  {\bf For $r < R$:}

  \begin{equation*}
    Q_{encl}(r) =  \rho \tau(r) = \frac{3Q}{4\pi R^3} \cdot \frac{4\pi r^3}{3} = Q \frac{r^3}{R^3}
  \end{equation*}

  \begin{equation*}
    E(r) = \frac{Q \frac{r^3}{R^3}}{4 \pi \epsilon_0 r^2} = \frac{Q}{4 \pi \epsilon_0 R^3} r
  \end{equation*}

  {\bf For $r \ge R$:}

  \begin{equation*}
    Q_{encl}(r) =  Q
  \end{equation*}

  \begin{equation*}
    E(r) = \frac{Q}{4 \pi \epsilon_0 r^2}
  \end{equation*}

  The result for the case of $r \ge R$  is identical to that for point charge Q at the origin.

\end{frame}

} % Example

%
%
%

\begin{frame}{Integral and differential forms of Gauss' law}

Our result for any array of charges (or continuous charge distribution), with net charge $Q_{enc}$,
enclosed by an arbitrary closed surface S is:
\begin{equation*}
  \oint \vec{E} \cdot d\vec{S} = \frac{Q_{enc}}{\epsilon_0}
\end{equation*}

This is known as the {\bf \underline{integral form} of Gauss' law}.\\
\vspace{0.2cm}

This form is useful in cases where the problem at hand has a symmetry.\\
The symmetry can be exploited to simplify the calculation of the integral.\\
Will will see several examples of the above in the workshops.\\
\vspace{0.2cm}

It is very useful to obtain Gauss' law in its {\bf \underline{differential form}}.\\
This is a more practical form for the analytical or numerical solution of problems
that do not have a symmetry that we can exploit.\\
\vspace{0.2cm}

Before we can proceed, we need to continue brushing up our calculus skills.

\end{frame}



% starting reminder
{
\reminderslide

%
%
%

\begin{frame}{Reminder: Gradient, Divergence, Curl and all that}

\begin{itemize}
  \item The $\vec\nabla$ ({\em nabla}) operator
    \begin{equation*}
       \vec{\nabla} = (\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z})
    \end{equation*}
  \item The nabla operator is a kind of a vector
    \begin{itemize}
        \item It is a {\em vector operator} that acts upon the quantity on the right
        \item It appears as a vector as long as we do not make a distinction between "acting upon" and "multiplying"
    \end{itemize}
  \vspace{0.3cm}
  \item Just as we have 3 kinds of vector multiplications (with scalar, dot product, cross product),
        we have 3 ways the nabla operator can act
    \begin{itemize}
       \item $\vec\nabla$ (a scalar function) $\rightarrow$ {\bf gradient}
       \item $\vec\nabla$ $\cdot$ (a vector function) $\rightarrow$ {\bf divergence}
       \item $\vec\nabla$ $\times$ (a vector function) $\rightarrow$ {\bf curl}
    \end{itemize}
\end{itemize}

\end{frame}


%
%
%

\begin{frame}{Reminder: Gradient, Divergence, Curl and all that}

\underline{\bf Gradient}:
This is a {\bf generalization to many dimensions} of the concept of the derivative of a 1-dimensional function.\\
\vspace{0.1cm}
Let $f(x,y)$ be a scalar field in a 2-dimensional space. Its gradient is:
\begin{equation*}
   \vec\nabla f(x,y) =
     \Big(
       \frac{\partial f(x,y)}{\partial x},
       \frac{\partial f(x,y)}{\partial y}
     \Big)
\end{equation*}

\begin{columns}
  \begin{column}{0.50\textwidth}
    Like the usual derivative, $\vec\nabla f(x,y)$ tells us how fast $f(x,y)$ changes as we move in space.\\
    \vspace{0.1cm}
    Notice that $\vec\nabla f(x,y)$ is a vector.
    It tells us, not only how fast $f(x,y)$ changes, but also the {\bf direction of the steepest ascent}.\\
    \vspace{0.1cm}
    Similarly for 3 or more dimensions.
  \end{column}
  \begin{column}{0.50\textwidth}
   \begin{center}
     \includegraphics[width=0.85\textwidth]{./images/schematics/gradient_steepest_ascent.png}\\
   \end{center}
  \end{column}
\end{columns}

\end{frame}

%
%
%

\begin{frame}{Reminder: Gradient, Divergence, Curl and all that}

In the previous slide, we used a {\bf partial derivative}.
It is worth stopping to make sure you all understand what this is.
\vspace{0.1cm}

Consider a function f of several variables x,y,z,... Notice that\\
\begin{center}
{\color{red}
 {\Large
  $\frac{\partial f(x,y,z,...)}{\partial x}$
 }
 {\bf is not the same as}
 {\Large
  $\frac{df(x,y,z,...)}{dx}$\\
 }
}
\end{center}

\vspace{0.2cm}
A partial derivative, with respect to x, of a function of several variables, e.g. f(x,y,z,...),
is the derivative with respect to x {\bf while all other variables are held constant}.\\

\vspace{0.2cm}
On the other hand, the total derivative takes into account, not only the change of the function as x changes,
but also {\bf adds in all the changes due to the indirect dependencies}:
\begin{equation*}
   \frac{df(x,y,z,...)}{dx} =
     \frac{\partial f(x,y,z,...)}{\partial x} +
     \frac{\partial f(x,y,z,...)}{\partial y} \cdot \frac{\partial y}{\partial x} +
     \frac{\partial f(x,y,z,...)}{\partial z} \cdot \frac{\partial z}{\partial x} + ...
\end{equation*}

\end{frame}


%
%
%

\begin{frame}{Reminder: Gradient, Divergence, Curl and all that}

\underline{\bf Divergence}:
Let $\vec{F}(x,y)=\Big(F_{x}(x,y), \; F_{y}(x,y) \Big)$
be a vector field in a 2-dimensional space. Its divergence is given by:
\begin{equation*}
   \vec\nabla \cdot \vec{F}(x,y) =
       \frac{\partial F_{x}(x,y)}{\partial x} +
       \frac{\partial F_{y}(x,y)}{\partial y}
\end{equation*}
Similarly for 3 or more dimensions.\\
\vspace{0.1cm}

\begin{columns}
  \begin{column}{0.40\textwidth}
    Notice that the divergence of a vector field is a scalar.\\
    Its value for a particular point expresses the magnitude of the vector field's source (or sink) at that point.
  \end{column}
  \begin{column}{0.60\textwidth}
   \begin{center}
     \includegraphics[width=0.98\textwidth]{./images/schematics/divergence_examples.png}\\
   \end{center}
  \end{column}
\end{columns}

\vspace{0.2cm}
\underline{\bf Curl}: This won't be introduced till later so let's forget it for now.

\end{frame}


%
%
%

\begin{frame}{Reminder: Gradient, Divergence, Curl and all that}

\begin{columns}
  \begin{column}{0.25\textwidth}
   \begin{center}
     {\bf Confused?}\\
     \vspace{0.3cm}
     {\small
      Here is a piece of advice that comes from Feynman.\\
     }
     \vspace{0.3cm}
     \includegraphics[width=0.99\textwidth]{./images/people/feynman_bongos2.jpg}\\
   \end{center}
  \end{column}
  \begin{column}{0.75\textwidth}
    \begin{itemize}
    {\scriptsize
      \item If you are solving a particular problem where $\vec{\nabla} \vec{E}$ appears,
            {\bf do not just stare at it} if you are not quite sure what it means or if it confuses you.
      \item This is a nice compact notation that is supposed to make our lives easier,
            not more difficult.
      \item But it takes a while to get used to compact notations and be sure that we truly understand what they mean
      \item If you find that $\vec{\nabla} \vec{E}$ confuses you, write it out as
        \begin{equation*}
         \vec{\nabla} \cdot \vec{E} = \frac{\partial E_{x}}{\partial x} + \frac{\partial E_{y}}{\partial y} + \frac{\partial E_{z}}{\partial z}
        \end{equation*}
      \item There is {\bf no shame at writing the components and not using the fancy compact notation}
    }
    \end{itemize}

    \begin{center}
    {\scriptsize
     Marking your scripts, I may excuse you for being confused by $\vec{\nabla} \cdot \vec{E}$.\\
     But not for ignoring Feynman's advice!
    }
    \end{center}

  \end{column}
\end{columns}

\end{frame}


%
%
%

\begin{frame}{Reminder: Gauss' theorem (Divergence theorem)}

\begin{center}
  Gauss' theorem: {\bf Not to be confused with Gauss' law} we saw earlier.
\end{center}

\begin{columns}
  \begin{column}{0.35\textwidth}
   \begin{center}
     \includegraphics[width=0.95\textwidth]{./images/schematics/divergence_theorem_tau.png}\\
   \end{center}
  \end{column}
  \begin{column}{0.65\textwidth}
     \begin{itemize}
       \item Assume a volume $\tau$ whose boundary is the closed surface S
       \item Assume a vector field $\vec{F}$ which is
        \begin{itemize}
           \item defined anywhere in $\tau$
           \item continuous / differentiable anywhere in $\tau$
        \end{itemize}
     \end{itemize}
  \end{column}
\end{columns}

%\begin{block}{}
 The integral of the flux of the field $\vec{F}$ over the closed surface S is
 equal to the integral of the divergence of $\vec{F}$ in the volume $\tau$\\
 \begin{equation*}
   \oint_{S} \vec{F} \cdot d\vec{S} = \int_{\tau} \vec{\nabla} \cdot \vec{F} d\tau
 \end{equation*}
%\end{block}

\end{frame}

} % ending reminder


%
%
%

\begin{frame}{Deriving the differential form of Gauss' law}

Expressing $Q_{enc}$ in terms of the density $\rho$,
the integral form of Gauss' law is:
\begin{equation*}
   \oint_{S} \vec{E} \cdot d\vec{S} = \frac{1}{\epsilon_0} Q_{enc} =  \frac{1}{\epsilon_0} \int \rho d\tau
\end{equation*}

Applying Gauss' theorem to the electric field:
\begin{equation*}
   \oint_{S} \vec{E} \cdot d\vec{S} = \int_{\tau} \vec{\nabla} \cdot \vec{E} d\tau
\end{equation*}

Therefore:
\begin{equation*}
  \int_{\tau} \vec{\nabla} \cdot \vec{E} d\tau = \frac{1}{\epsilon_0} \int \rho d\tau \Rightarrow
  \int_{\tau} \Big( \vec{\nabla} \cdot \vec{E} - \frac{\rho}{\epsilon_0} \Big) d\tau = 0 \Rightarrow
  {\color{magenta} \vec{\nabla} \cdot \vec{E} = \frac{\rho}{\epsilon_0}}
\end{equation*}

The above is the {\bf \underline{differential form} of Gauss' law}.\\
It tells us that the divergence of the electric field at each point is proportional to the local charge density.

\end{frame}


%
%
%

\begin{frame}{The differential form of Gauss' law}

{\bf \underline{Differential form} of Gauss' law}:
The divergence of the electric field at each point is proportional to the local charge density.

\begin{equation*}
  \vec{\nabla} \cdot \vec{E}(\vec{r}) = \frac{\rho(\vec{r})}{\epsilon_0}
\end{equation*}

Recall the geometrical interpretation of the divergence of a vector field.\\
\begin{center}
  \includegraphics[width=0.60\textwidth]{./images/schematics/electric_field_lines_pos_and_neg_point_charges_1.png}\\
\end{center}

Gauss' law tells us, and it is easier to see that in its differential form, that
the {\bf electric charge is the source of the electric field}.

\end{frame}

% ------------------------------------------------------------------------------
% ------------------------------------------------------------------------------

%
% What to remember
%

\renewcommand{\lecturesummarytitle}{Main points to remember }
\input{slides_lecture02_summary.tex}

%
% At the next lecture
%

\begin{frame}{At the next lecture (Lecture \nextlecture)}

\begin{itemize}
  \item {\bf Potential energy of a charge in an electrostatic field}
    \begin{itemize}
       \item we will generalise for discrete and continuous charge distributions
    \end{itemize}

  \vspace{0.4cm}
  \item {\bf Electric potential}

  \vspace{0.4cm}
  \item {\bf Circuital law for electrostatics} in differential and integral forms

  \vspace{0.4cm}
  \item Continue brushing up our relevant calculus skills

\end{itemize}

\end{frame}

%
% Optional reading
%

\input{slides_lecture02_optional.tex}