lectromagnetic Waves: Electric Flux - Exercises and Complete Worked Solutions
LectromagneticWavesElectricFluxExercisesAndCompleteWorkedSolutions
2026-09-16 03:41:50
2026-09-16 03:41:50
Example
Electromagnetic Waves: Electric Flux
electric flux
electric field
area vector
surface normal
dot product
projected area
differential area
surface integral
closed surface
Coulomb field
spherical flux
Gauss law preparation
exercises
worked solutions
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\section*{Electromagnetic Waves, Antennas, and RF: Electric Flux - Exercises and Complete Worked Solutions}
This companion article provides self-study exercises for EM06, \emph{Electric Flux}. All exercises are stated first. Complete worked solutions follow in Part II.
The problems reinforce four ideas developed in EM06:
\begin{enumerate}
\item a surface must be given an orientation through its normal vector;
\item only the component of $\mathbf E$ normal to the surface contributes to flux;
\item a general surface requires a surface integral; and
\item a closed surface uses outward-pointing area vectors.
\end{enumerate}
For a flat surface of area $A$ with unit normal $\hat{\mathbf n}$,
\begin{equation}
\boxed{\mathbf A=A\hat{\mathbf n}.}
\end{equation}
For a small surface patch,
\begin{equation}
\boxed{d\mathbf A=\hat{\mathbf n}\,dA.}
\end{equation}
The differential flux is
\begin{equation}
\boxed{d\Phi_E=\mathbf E\cdot d\mathbf A,}
\end{equation}
and for a uniform field over a flat surface,
\begin{equation}
\boxed{\Phi_E=EA\cos\theta,}
\end{equation}
where $\theta$ is measured between the electric field and the chosen surface normal.
For a general open surface,
\begin{equation}
\boxed{\Phi_E=\int_S\mathbf E\cdot d\mathbf A,}
\end{equation}
while for a closed surface the outward-normal convention gives
\begin{equation}
\boxed{\Phi_E=\oint_S\mathbf E\cdot d\mathbf A.}
\end{equation}
These are the same definitions and conventions used in EM06 \cite{Griffiths2017,OpenStaxV2,PurcellMorin2013,MIT802}.
\section*{How to use this problem set}
Attempt all exercises in Part I before reading Part II. For every flux problem, explicitly identify
\begin{enumerate}
\item the field vector $\mathbf E$;
\item the surface normal $\hat{\mathbf n}$;
\item whether the surface is open or closed;
\item the angle between the field and the normal, not the field and the surface; and
\item the sign of $\mathbf E\cdot d\mathbf A$.
\end{enumerate}
In Exercises 12 and 13, use the centered point-charge spherical result derived directly from Coulomb's field in EM06. Do not invoke the general form of Gauss's law; that theorem is reserved for EM07.
\section*{Part I: Exercises}
\section*{Exercise 1: area vector and orientation}
A flat surface has area
\begin{equation}
A=0.40\,\text{m}^2
\end{equation}
and chosen unit normal
\begin{equation}
\hat{\mathbf n}=\frac{3}{5}\hat{\mathbf x}+\frac{4}{5}\hat{\mathbf y}.
\end{equation}
Find the area vector $\mathbf A$. Then write the area vector if the orientation is reversed.
\section*{Exercise 2: field normal to a surface}
A uniform electric field has magnitude
\begin{equation}
E=250\,\text{N/C}.
\end{equation}
It passes through a flat surface of area
\begin{equation}
A=0.32\,\text{m}^2
\end{equation}
in the same direction as the chosen surface normal. Find the electric flux.
\begin{center}
\includegraphics{EM06E_fig01_flux_angle_geometry.png}
\vspace{0.45em}
\textbf{Figure.}
The flux angle is measured between the electric field and the surface normal. Tilting the surface changes the normal and therefore changes the projected area seen by the field.
\end{center}
\section*{Exercise 3: tilted flat surface}
A uniform field of magnitude
\begin{equation}
E=250\,\text{N/C}
\end{equation}
crosses a flat surface of area
\begin{equation}
A=0.32\,\text{m}^2.
\end{equation}
The angle between $\mathbf E$ and the chosen surface normal is
\begin{equation}
\theta=60^\circ.
\end{equation}
Find the flux.
\section*{Exercise 4: angle given relative to the surface}
A field of magnitude
\begin{equation}
E=180\,\text{N/C}
\end{equation}
crosses a flat surface of area
\begin{equation}
A=0.50\,\text{m}^2.
\end{equation}
The field makes an angle of $30^\circ$ with the \emph{surface itself}. Find the electric flux for the normal that makes an acute angle with the field.
\section*{Exercise 5: reversing orientation}
For one chosen normal, the electric flux through an open surface is
\begin{equation}
\Phi_E=+18\,\text{N}\,\text{m}^2/\text{C}.
\end{equation}
What is the flux if the surface orientation is reversed? Has the electric field changed?
\section*{Exercise 6: compute flux with a vector dot product}
A uniform electric field is
\begin{equation}
\mathbf E=(120\hat{\mathbf x}-50\hat{\mathbf y}+30\hat{\mathbf z})\,\text{N/C}.
\end{equation}
A flat surface has area
\begin{equation}
A=0.40\,\text{m}^2
\end{equation}
and unit normal
\begin{equation}
\hat{\mathbf n}=\frac{3}{5}\hat{\mathbf x}+\frac{4}{5}\hat{\mathbf y}.
\end{equation}
Compute the electric flux using $\Phi_E=\mathbf E\cdot\mathbf A$.
\section*{Exercise 7: zero flux with a nonzero field}
A uniform field is
\begin{equation}
\mathbf E=500\hat{\mathbf x}\,\text{N/C}.
\end{equation}
A rectangular surface lies in the $xy$ plane and has chosen normal $+\hat{\mathbf z}$.
\begin{enumerate}
\item[(a)] Find the flux through the rectangle.
\item[(b)] Explain why the result does not imply that the electric field is zero.
\end{enumerate}
\section*{Exercise 8: a nonuniform field over a plane}
The electric field is
\begin{equation}
\mathbf E(y)=(10+4y)\hat{\mathbf x}\,\text{N/C}.
\end{equation}
A rectangular surface lies in the plane $x=2\,\text{m}$, with chosen normal $+\hat{\mathbf x}$. Its coordinate limits are
\begin{equation}
0\le y\le1\,\text{m},
\qquad
0\le z\le0.50\,\text{m}.
\end{equation}
Evaluate
\begin{equation}
\Phi_E=\int_S\mathbf E\cdot d\mathbf A.
\end{equation}
\begin{center}
\includegraphics{EM06E_fig02_open_closed_surfaces.png}
\vspace{0.45em}
\textbf{Figure.}
An open surface requires a chosen normal. A closed surface instead uses the outward normal everywhere.
\end{center}
\section*{Exercise 9: uniform field through a closed box}
A rectangular box is placed in a uniform field
\begin{equation}
\mathbf E=E_0\hat{\mathbf x}.
\end{equation}
The two faces perpendicular to the $x$ axis each have area $A$. Determine the flux through each of those two faces, the flux through the other four faces, and the total flux through the closed box.
\begin{center}
\includegraphics{EM06E_fig03_uniform_field_box.png}
\vspace{0.45em}
\textbf{Figure.}
For a uniform field, the positive outward flux through one face is canceled by equal negative flux through the opposite face.
\end{center}
\section*{Exercise 10: a nonuniform field through a rectangular box}
Let
\begin{equation}
\mathbf E=\alpha x\hat{\mathbf x},
\end{equation}
with
\begin{equation}
\alpha=5.0\,\text{N}/(\text{C}\,\text{m}).
\end{equation}
A rectangular box spans
\begin{equation}
1\le x\le3\,\text{m},
\qquad
0\le y\le2\,\text{m},
\qquad
0\le z\le1\,\text{m}.
\end{equation}
Compute the flux through all six faces directly and find the net closed-surface flux.
\section*{Exercise 11: flux through a hemisphere in a uniform field}
A hemisphere of radius
\begin{equation}
R=0.20\,\text{m}
\end{equation}
occupies the $+x$ side of a sphere. Its curved surface uses the outward normal. A uniform field
\begin{equation}
\mathbf E=300\hat{\mathbf x}\,\text{N/C}
\end{equation}
passes through it.
Using the projected-area interpretation, find the flux through the curved hemispherical surface.
\section*{Exercise 12: centered point charge and spherical flux}
A point charge
\begin{equation}
q=3.0\,\text{nC}
\end{equation}
is at the center of a spherical surface. Using the Coulomb-field derivation from EM06,
\begin{equation}
\Phi_E=\frac{q}{\epsilon_0},
\end{equation}
find the electric flux through the sphere.
\begin{center}
\includegraphics{EM06E_fig04_centered_charge_spheres.png}
\vspace{0.45em}
\textbf{Figure.}
For centered spherical surfaces, the point-charge field weakens as $1/r^2$ while the spherical area grows as $r^2$. Their product is independent of radius.
\end{center}
\section*{Exercise 13: compare two spherical radii}
The same positive point charge is surrounded by two centered spherical surfaces of radii
\begin{equation}
r_1=0.20\,\text{m},
\qquad
r_2=0.80\,\text{m}.
\end{equation}
Find:
\begin{enumerate}
\item[(a)] the ratio $E(r_1)/E(r_2)$;
\item[(b)] the ratio $A_2/A_1$ of spherical areas;
\item[(c)] the ratio $\Phi_1/\Phi_2$ of total fluxes.
\end{enumerate}
Explain why the result is consistent with the inverse-square field.
\section*{Exercise 14: diagnose conceptual statements}
For each statement, decide whether it is correct. If it is incorrect, rewrite it accurately.
\begin{enumerate}
\item[(a)] ``Electric flux is the number of physical electric-field lines crossing a surface.''
\item[(b)] ``A nonzero electric field always produces nonzero flux through every surface.''
\item[(c)] ``The angle in $EA\cos\theta$ is measured from the field to the surface normal.''
\item[(d)] ``Reversing the normal of an open surface reverses the sign of its flux.''
\item[(e)] ``For a closed surface, outward normals are the standard orientation.''
\item[(f)] ``The result $q/\epsilon_0$ derived in EM06 for a centered point charge and sphere is already the full general statement of Gauss's law.''
\end{enumerate}
\section*{Part II: Complete Worked Solutions}
\section*{Solution 1: area vector and orientation}
The area vector is
\begin{equation}
\mathbf A=A\hat{\mathbf n}.
\end{equation}
Therefore,
\begin{align}
\mathbf A
&=(0.40)
\left(
\frac35\hat{\mathbf x}+\frac45\hat{\mathbf y}
\right)\,\text{m}^2\\
&=\left(0.24\hat{\mathbf x}+0.32\hat{\mathbf y}\right)\,\text{m}^2.
\end{align}
Thus
\begin{equation}
\boxed{\mathbf A=(0.24\hat{\mathbf x}+0.32\hat{\mathbf y})\,\text{m}^2.}
\end{equation}
Reversing the orientation changes $\hat{\mathbf n}$ to $-\hat{\mathbf n}$, so
\begin{equation}
\boxed{\mathbf A_{\text{reversed}}=(-0.24\hat{\mathbf x}-0.32\hat{\mathbf y})\,\text{m}^2.}
\end{equation}
The scalar area has not changed. Only the chosen orientation has changed.
\section*{Solution 2: field normal to a surface}
Because the field points in the same direction as the chosen normal,
\begin{equation}
\theta=0.
\end{equation}
Therefore,
\begin{align}
\Phi_E
&=EA\cos0\\
&=(250)(0.32)\\
&=80\,\text{N}\,\text{m}^2/\text{C}.
\end{align}
Hence
\begin{equation}
\boxed{\Phi_E=80\,\text{N}\,\text{m}^2/\text{C}.}
\end{equation}
\section*{Solution 3: tilted flat surface}
Use
\begin{equation}
\Phi_E=EA\cos\theta.
\end{equation}
Then
\begin{align}
\Phi_E
&=(250)(0.32)\cos60^\circ\\
&=80\left(\frac12\right)\\
&=40\,\text{N}\,\text{m}^2/\text{C}.
\end{align}
Thus
\begin{equation}
\boxed{\Phi_E=40\,\text{N}\,\text{m}^2/\text{C}.}
\end{equation}
The tilt reduces the effective projected area by the factor $\cos60^\circ=1/2$.
\section*{Solution 4: angle given relative to the surface}
The given $30^\circ$ is measured from the field to the surface, not to the normal. The normal is perpendicular to the surface, so the angle to the normal is
\begin{equation}
\theta=90^\circ-30^\circ=60^\circ.
\end{equation}
Hence
\begin{align}
\Phi_E
&=EA\cos60^\circ\\
&=(180)(0.50)\left(\frac12\right)\\
&=45\,\text{N}\,\text{m}^2/\text{C}.
\end{align}
Therefore,
\begin{equation}
\boxed{\Phi_E=45\,\text{N}\,\text{m}^2/\text{C}.}
\end{equation}
This exercise illustrates why the angle convention must be checked before applying the cosine formula.
\section*{Solution 5: reversing orientation}
Reversing the normal sends
\begin{equation}
d\mathbf A\rightarrow-d\mathbf A.
\end{equation}
Therefore,
\begin{equation}
\Phi_E\rightarrow-\Phi_E.
\end{equation}
Thus
\begin{equation}
\boxed{\Phi_E=-18\,\text{N}\,\text{m}^2/\text{C}.}
\end{equation}
The electric field has not changed. Only the bookkeeping orientation of the open surface changed.
\section*{Solution 6: compute flux with a vector dot product}
First construct the area vector:
\begin{align}
\mathbf A
&=A\hat{\mathbf n}\\
&=0.40
\left(
\frac35\hat{\mathbf x}+\frac45\hat{\mathbf y}
\right)\,\text{m}^2\\
&=\left(0.24\hat{\mathbf x}+0.32\hat{\mathbf y}\right)\,\text{m}^2.
\end{align}
Now take the dot product:
\begin{align}
\Phi_E
&=\mathbf E\cdot\mathbf A\\
&=(120)(0.24)+(-50)(0.32)+(30)(0)\\
&=28.8-16.0\\
&=12.8\,\text{N}\,\text{m}^2/\text{C}.
\end{align}
Therefore,
\begin{equation}
\boxed{\Phi_E=12.8\,\text{N}\,\text{m}^2/\text{C}.}
\end{equation}
The $z$ component of the field contributes nothing because the chosen surface normal has no $z$ component.
\section*{Solution 7: zero flux with a nonzero field}
The field points along $+\hat{\mathbf x}$ while the surface normal points along $+\hat{\mathbf z}$. These directions are perpendicular, so
\begin{equation}
\theta=90^\circ.
\end{equation}
Thus
\begin{equation}
\boxed{\Phi_E=EA\cos90^\circ=0.}
\end{equation}
The field is still
\begin{equation}
\mathbf E=500\hat{\mathbf x}\,\text{N/C},
\end{equation}
which is nonzero. The flux vanishes because the field runs parallel to the surface instead of through it.
\section*{Solution 8: a nonuniform field over a plane}
The rectangle lies in the plane $x=2\,\text{m}$ and has normal $+\hat{\mathbf x}$, so
\begin{equation}
d\mathbf A=\hat{\mathbf x}\,dy\,dz.
\end{equation}
Therefore,
\begin{equation}
\mathbf E\cdot d\mathbf A=(10+4y)\,dy\,dz.
\end{equation}
The total flux is
\begin{align}
\Phi_E
&=\int_0^{0.50}\int_0^1(10+4y)\,dy\,dz\\
&=\int_0^{0.50}\left[10y+2y^2\right]_0^1 dz\\
&=\int_0^{0.50}12\,dz\\
&=6.0\,\text{N}\,\text{m}^2/\text{C}.
\end{align}
Hence
\begin{equation}
\boxed{\Phi_E=6.0\,\text{N}\,\text{m}^2/\text{C}.}
\end{equation}
The simple product $EA$ is not sufficient here because the field magnitude changes with $y$ across the surface.
\section*{Solution 9: uniform field through a closed box}
For the face whose outward normal is $+\hat{\mathbf x}$,
\begin{equation}
\boxed{\Phi_R=+E_0A.}
\end{equation}
For the opposite face, the outward normal is $-\hat{\mathbf x}$, so
\begin{equation}
\boxed{\Phi_L=-E_0A.}
\end{equation}
On the remaining four faces, the normals are perpendicular to $\mathbf E$, so
\begin{equation}
\boxed{\Phi=0}
\end{equation}
for each of those faces.
Adding all six contributions,
\begin{equation}
\Phi_{\text{closed}}=E_0A-E_0A=0.
\end{equation}
Therefore,
\begin{equation}
\boxed{\Phi_{\text{closed}}=0.}
\end{equation}
The field is nonzero, but equal flux enters and leaves the box.
\section*{Solution 10: a nonuniform field through a rectangular box}
The field is
\begin{equation}
\mathbf E=\alpha x\hat{\mathbf x}.
\end{equation}
Only the two faces perpendicular to the $x$ axis contribute. Their area is
\begin{equation}
A=(2\,\text{m})(1\,\text{m})=2\,\text{m}^2.
\end{equation}
At the right face, $x=3\,\text{m}$ and the outward normal is $+\hat{\mathbf x}$. The field magnitude there is
\begin{equation}
E_R=\alpha x=(5.0)(3)=15\,\text{N/C}.
\end{equation}
Therefore,
\begin{equation}
\Phi_R=(15)(2)=30\,\text{N}\,\text{m}^2/\text{C}.
\end{equation}
At the left face, $x=1\,\text{m}$ and the outward normal is $-\hat{\mathbf x}$. The field magnitude there is
\begin{equation}
E_L=(5.0)(1)=5.0\,\text{N/C}.
\end{equation}
Thus,
\begin{equation}
\Phi_L=-(5.0)(2)=-10\,\text{N}\,\text{m}^2/\text{C}.
\end{equation}
The other four faces have normals perpendicular to the field, so their flux is zero. Hence
\begin{align}
\Phi_{\text{closed}}
&=30-10\\
&=20\,\text{N}\,\text{m}^2/\text{C}.
\end{align}
Therefore,
\begin{equation}
\boxed{\Phi_{\text{closed}}=20\,\text{N}\,\text{m}^2/\text{C}.}
\end{equation}
Unlike the uniform-field case, the field is stronger on the right side than on the left, so the outward and inward contributions do not cancel.
\section*{Solution 11: flux through a hemisphere in a uniform field}
For a uniform field, the flux through the curved hemisphere equals the field magnitude times the area projected onto a plane perpendicular to the field.
The projection of the hemisphere onto the $yz$ plane is a disk of area
\begin{equation}
A_\perp=\pi R^2.
\end{equation}
Therefore,
\begin{align}
\Phi_E
&=E\pi R^2\\
&=(300)\pi(0.20)^2\\
&=12\pi\,\text{N}\,\text{m}^2/\text{C}\\
&\approx37.7\,\text{N}\,\text{m}^2/\text{C}.
\end{align}
Hence
\begin{equation}
\boxed{\Phi_E\approx37.7\,\text{N}\,\text{m}^2/\text{C}.}
\end{equation}
The flux is positive because the outward normals on the $+x$ hemisphere have positive $x$ components.
\section*{Solution 12: centered point charge and spherical flux}
EM06 derived, directly from Coulomb's field for a centered point charge,
\begin{equation}
\Phi_E=\frac{q}{\epsilon_0}.
\end{equation}
Using
\begin{equation}
q=3.0\times10^{-9}\,\text{C}
\end{equation}
and
\begin{equation}
\epsilon_0\approx8.854\times10^{-12}\,\text{C}^2/(\text{N}\,\text{m}^2),
\end{equation}
we obtain
\begin{align}
\Phi_E
&=\frac{3.0\times10^{-9}}{8.854\times10^{-12}}\\
&\approx3.39\times10^2\,\text{N}\,\text{m}^2/\text{C}.
\end{align}
Thus
\begin{equation}
\boxed{\Phi_E\approx339\,\text{N}\,\text{m}^2/\text{C}.}
\end{equation}
This result is independent of the sphere radius for the centered point-charge geometry.
\section*{Solution 13: compare two spherical radii}
For a point charge,
\begin{equation}
E(r)\propto\frac{1}{r^2}.
\end{equation}
Therefore,
\begin{align}
\frac{E(r_1)}{E(r_2)}
&=\left(\frac{r_2}{r_1}\right)^2\\
&=\left(\frac{0.80}{0.20}\right)^2\\
&=16.
\end{align}
Hence
\begin{equation}
\boxed{\frac{E(r_1)}{E(r_2)}=16.}
\end{equation}
Spherical area scales as $r^2$, so
\begin{align}
\frac{A_2}{A_1}
&=\left(\frac{r_2}{r_1}\right)^2\\
&=16.
\end{align}
Thus
\begin{equation}
\boxed{\frac{A_2}{A_1}=16.}
\end{equation}
Flux is the product of field magnitude and spherical area in this centered geometry. The factor of $16$ decrease in field magnitude from $r_1$ to $r_2$ is exactly canceled by the factor of $16$ increase in area. Therefore,
\begin{equation}
\boxed{\frac{\Phi_1}{\Phi_2}=1.}
\end{equation}
\section*{Solution 14: diagnose conceptual statements}
\begin{enumerate}
\item[(a)] Incorrect. Electric-field lines are a visualization convention. Electric flux is defined mathematically by
\begin{equation}
\Phi_E=\int_S\mathbf E\cdot d\mathbf A.
\end{equation}
\item[(b)] Incorrect. A nonzero field can give zero flux through a surface if the field is tangent to the surface everywhere, so that $\mathbf E\cdot d\mathbf A=0$.
\item[(c)] Correct. The angle in
\begin{equation}
\Phi_E=EA\cos\theta
\end{equation}
is measured between $\mathbf E$ and the chosen surface normal.
\item[(d)] Correct. Reversing the normal changes $d\mathbf A$ to $-d\mathbf A$, so the flux changes sign.
\item[(e)] Correct. Closed surfaces conventionally use outward-pointing area vectors.
\item[(f)] Incorrect. EM06 derived $q/\epsilon_0$ only for a sphere centered on a point charge, using Coulomb's field. EM07 will state and analyze the general law for arbitrary closed surfaces and enclosed charge distributions.
\end{enumerate}
\section*{Common mistakes}
\begin{itemize}
\item \textbf{Using the angle to the surface rather than the normal.} Convert to the complementary angle before using $EA\cos\theta$.
\item \textbf{Treating area as a scalar when orientation matters.} Flux uses $d\mathbf A=\hat{\mathbf n}\,dA$.
\item \textbf{Assuming a strong field guarantees large flux.} Tangential field contributes zero normal flux.
\item \textbf{Forgetting the outward-normal convention on a closed surface.}
\item \textbf{Using $EA$ for a field that varies over the surface.} In that case, evaluate the surface integral.
\item \textbf{Calling field lines physical objects.} They are only a visualization of the continuous vector field.
\item \textbf{Invoking general Gauss's law too early.} EM06 and EM06E use only flux definitions plus the special centered-sphere result derived from Coulomb's field.
\end{itemize}
\section*{What EM06E reinforces}
The fundamental local statement is
\begin{equation}
\boxed{d\Phi_E=\mathbf E\cdot d\mathbf A.}
\end{equation}
For a uniform field over a flat surface,
\begin{equation}
\boxed{\Phi_E=EA\cos\theta.}
\end{equation}
For a general surface,
\begin{equation}
\boxed{\Phi_E=\int_S\mathbf E\cdot d\mathbf A,}
\end{equation}
and for a closed surface,
\begin{equation}
\boxed{\Phi_E=\oint_S\mathbf E\cdot d\mathbf A.}
\end{equation}
The centered point-charge sphere provides the important preview
\begin{equation}
\boxed{\Phi_E=\frac{q}{\epsilon_0},}
\end{equation}
but the general relationship between closed-surface flux and enclosed charge belongs to EM07, \emph{Gauss's Law}.
\begin{thebibliography}{9}
\bibitem{Griffiths2017}
David J. Griffiths,
\emph{Introduction to Electrodynamics},
4th ed., Cambridge University Press, 2017.
\bibitem{OpenStaxV2}
Samuel J. Ling, Jeff Sanny, and William Moebs,
\emph{University Physics, Volume 2},
OpenStax, 2016,
sections on electric flux and Gauss's law.
\bibitem{PurcellMorin2013}
Edward M. Purcell and David J. Morin,
\emph{Electricity and Magnetism},
3rd ed., Cambridge University Press, 2013.
\bibitem{MIT802}
Massachusetts Institute of Technology,
\emph{8.02 Physics II: Electricity and Magnetism},
MIT OpenCourseWare,
materials on electric fields, flux, and Gauss's law.
\bibitem{FeynmanV2}
Richard P. Feynman, Robert B. Leighton, and Matthew Sands,
\emph{The Feynman Lectures on Physics, Volume II},
Addison-Wesley, 1964,
chapters on electrostatics and electric fields.
\end{thebibliography}