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\begin{bf}
\noindent
Institutt for fysikk, NTNU

\noindent
TFY4155/FY1003 Elektrisitet og magnetisme

\noindent
V{\aa}r 2005

\end{bf}

\begin{center}
\begin{large}
Solution to {\o}ving 9
\end{large}
\end{center}

\noindent
Guidance week 9 and 10.

\vspace{1cm}
\noindent

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Question &\hspace{0.3cm} A \hspace{0.3cm} &\hspace{0.3cm} B \hspace{0.3cm} &\hspace{0.3cm} C \hspace{0.3cm} &
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\vspace{1.5cm}
\noindent
$1)$ The charge $Q$ is distributed over the surface of the conductor and results in
some kind of surface charge density $\sigma$ (which will in general vary from place
to place on the conductor's surface, unless it is spherical). The twice as large
charge $2Q$ will distribute itself in the same manner and result in a twice as large
surface charge density everywhere, $2\sigma$. The electric field in point $P$ can
then be computed with Coulomb's law. With charge $Q$:
$$\bold{E}_1(P) = \int_S \frac{\sigma \; dA}{4\pi\varepsilon_0 r^2} \hat{r}$$
With charge $2Q$:
$$\bold{E}_2(P) = \int_S \frac{2\sigma \; dA}{4\pi\varepsilon_0 r^2} \hat{r} = 2\bold{E}_1(P)$$
Here, the integrals are taken over the (closed) surface $S$ of the conductor, $r$ is the
distance from the surface element $dA$ to $P$, and $\hat{r}$ is a unit vector along the
direction from $dA$ to $P$.

\noindent
Since the field everywhere (i.e., everywhere {\em outside} the conductor!) has become
twice as large, the potential (relative to zero, chosen at infinity)
$$V(P) = -\int_\infty^P \bold{E}\cdot d\bold{l}$$
must also have become twice as large.


\vspace{0.4cm}
\noindent
$2)$ Since $E=0$ everywhere inside the metal sphere, the net charge $Q$ must (because
of Gauss' law) distribute itself over the surface of the sphere, and from symmetry
reasons, uniformly over the surface. Then Gauss' law yields, with a spherical Gaussian
surface with a radius $r$ larger than the radius of the metal sphere,
$$E(r) = \frac{Q}{4\pi\varepsilon_0 r^2}$$
i.e., as for a point charge $Q$ at the center of the sphere. The potential $V(r)$
must therefore also become as for a point charge,
$$V(r) = \frac{Q}{4\pi\varepsilon_0 r}$$
Inside the metal sphere and on the surface of the sphere, the potential is
constant and equal to $Q/4\pi\varepsilon_0 R$, where $R$ is the radius of the sphere.
                                                                                                                                            
\noindent
If the distance from the center of the sphere to $B$ is twice the distance to $A$, the field
is reduced by a factor of 4, while the potential is reduced by a factor of 2.


\vspace{0.4cm}
\noindent
$3)$ The potential on a metal sphere with radius $R$ and charge $Q$ is (see the former
question)
$$V(R) = \frac{Q}{4\pi\varepsilon_0 R}$$
Let us e.g. put $V_0 = V(a) = Q/4\pi\varepsilon_0 a$, so that $V_1=V_0/2=15V_0/30$,
$V_2=V_0/3=10V_0/30$ and $V_3=V_0/5=6V_0/30$. Then we see that $V_1:V_2:V_3=15:10:6$.

\noindent
It is intuitively clear that the potential is largest for the smallest sphere: Imagine
starting with a neutral metal sphere and adding charge until the total charge is $Q$.
Clearly, it must be harder to add this charge the smaller the sphere is, due to
repulsion between the charges added. So, you must perform a larger amount of work
in order to put the charge on the smaller sphere. In other words, the smallest sphere
ends up with the largest potential energy, and therefore also the largest electric potential.

\noindent
[Had the charge been negative, the smallest sphere would have had the {\em smallest}
electric potential, i.e., the {\em most negative} potential relative to $V(\infty)=0$. But
still the smallest sphere would have had the largest potential energy, since
$\Delta U = Q \Delta V = -|Q| \Delta V$ when $Q<0$.]

\vspace{0.4cm}
\noindent
$4)$ Let us find the direction of the electric field in the vicinity of the
two charges, knowing that the electric field points towards a negative charge.
Just to the left of $x=-a$, the field must point right, i.e., $E(x)>0$.
All curves are consistent with this. But just to the right of $x=-a$,
the field must point left, i.e., $E(x)<0$. Curve 1 is not consistent with this.
Just to the left of $x=a$, the field must point right, i.e., $E(x)>0$. Curve 4
is not consistent with this. And finally, just to the right of $x=a$,
the field must point left, i.e., $E(x)<0$. Curve 2 is not consistent with this.
We are left with curve 3, which is correct. We see that $E(0)=0$ in curve 3,
which obviously must be correct.


\vspace{0.4cm}
\noindent
$5)$ $E=0$ inside the metal sphere, thus A and C are out of the question. The field
lines in B correspond, as we shall see in a few weeks, to the magnetic field around
a current carrying wire perpendicular to the paper plane.

\vspace{0.4cm}
\noindent
$6)$ First, let's agree what happens here: The charge $q$ is uniformly distributed on
the surface of the metal sphere and creates an electric field $E_0(r)=q/4\pi\varepsilon_0 r^2$
outside (i.e. $r>R$; $E=0$ inside the metal sphere). Electric dipoles in the plastic are aligned
because of $E_0$, and the net effect of the polarization is an induced negative charge
$-q_i$ on the inner surface of the plastic layer and a positive charge $q_i$ on the outer
surface of the plastic layer. The induced charge $-q_i$ creates an electric field
$E_i(r)=-q_i/4\pi\varepsilon_0 r^2$ inside the plastic, directed opposite to $E_0$, i.e.
radially inwards, so that the total electric field $E=E_0+E_i$ becomes weaker in the plastic
than if we had only air. On the outside of the plastic, the field strength is "reestablished"
by the induced charge $q_i$ on the outer surface. Alternatively, with Gauss' law: Total charge
inside a spherical Gaussian surface with radius $r>2R$ is $q-q_i+q_i=q$, so the electric field
here is $E_0(r)=q/4\pi\varepsilon_0 r^2$. Total charge inside a spherical Gaussian surface with radius
$R<r<2R$ is $q-q_i$, so here, i.e., in the plastic layer, the electric field is
$E(r)=(q-q_i)/4\pi\varepsilon_0 r^2$.

\noindent
Polarization $\bold{P}$ is, by definition, electric dipole moment pr unit volume. We only have
dipoles within the dielectric plastic layer, so $P$ can be nonzero only here. Field lines for $E$
would be as in the figure, but in addition we would have new field lines starting on the outer surface
of the plastic layer.

\noindent
We are left with electric displacement $D$, and we remember that $D$ is determined by free
charge: Gauss' law for $D$ reads
$$\oint \bold{D} \cdot d\bold{A} = Q_{\rm free}$$
And by {\em free} charge, we simply mean all charge except induced (bound) charges associated with
the polarization of dielectrics that are present in our system. Here, the charge $q$ on the metal 
sphere represents the free charge, whereas $\pm q_i$ represent induced, bound charge.

\noindent
Gauss' law for $D$ immediately yields $D(r)=q/4\pi r^2$ everywhere outside the metal
sphere, and the field lines are consistent with this.

\vspace{0.4cm}
\noindent
$7)$ Here, we may assume that we have approximately infinitely large planes. Then we know
that the electric field from one plane equals the charge pr unit area on the plane
divided by $2\varepsilon_0$, i.e., independent of the distance from the plane. Therefore,
we may immediately exclude C, since the field outside must be zero (an equally large, but oppositely
directed field contribution from the two planes when we are on the outside). Further, A must
be wrong: There is no chance that the electric field is stronger inside the dielectric than
in the layer with vacuum.

\noindent
But why isn't B correct? Didn't we learn that the field would become weaker if we
inserted a dielectric? Well, yes, but: Imagine starting with vacuum everywhere.
Then the charges $\pm Q$ must be uniformly distributed on the two plates. Next, we
insert the dielectric in the left half. Because of the field from the metal plates,
the dipoles in the dielectric will be aligned with the external field. The net effect
of this alignment is an induced surface charge on the dielectric, in our case positive
on top and negative at the bottom. If nothing more than this happened, we would no
longer have electrostatic equilibrium: It is no longer "advantageous" to have the free
charge on the metal plates uniformly distributed. The induced positive charge on top
of the dielectric will pull some of the free electrons from the right half of the upper
metal plate to the left half, and at the bottom, the opposite thing will take place.
And when do we have electrostatic equilibrium? When the potential everywhere on the
upper metal plate has the same constant value $V_-$ and the potential everywhere
on the lower metal plate has the same constant value $V_+$. (Remember: A metal plate
is an equipotential in equilibrium!) In equilibrium, we have the same {\em total}
charge density on the left and the right side. On the right side, we have
$\sigma = \sigma_f^0$ and on the left side we have $\sigma = \sigma_f^1 - \sigma_i$.
Here, subscript $f$ denotes free charge on the metal plates and $i$ denotes induced
charge. With plate area $A$, these charge densities must of course fulfill the
following: $\pm Q = \pm (\sigma_f^0 A/2 + \sigma_f^1 A/2)$ = total free charge
on the metal plates. From symmetry reasons, the electric field must still be directed
perpendicularly to the plates, and since the potential is constant on a given plate,
the field must be uniform, $E = \Delta V/d = (V_+ - V_-)/d$,
where $d$ is the distance between the plates.


\vspace{0.4cm}
\noindent
$8)$ Zero field inside the metal eliminates A and C. Polarization in the plastic
reduces the field with a factor $\varepsilon_1/\varepsilon_0 = 10$ in comparison to
what we would have with vacuum or air present. The electric field decays as
$1/r^2$, but this gives only a "reduction factor" of
$(5/4)^2/(5/2)^2 = 1/4$ when we compare positions B and D. Thus, the field will be
largest in position D.


\vspace{0.4cm}
\noindent
$9)$ In the point $(x,0)$ ($x>a$), the charges contribute with pairwise
equally large field (in absolute value), and with directions as shown in the figure:
                                                                                                                                            
%%FIGURE
\vspace{0.3cm}
\centerline{\epsfxsize=3.0in \epsfbox{los8_fig1.eps}}
\vspace{0.3cm}
%%END FIGURE
                                                                                                                                            
\noindent
The vector sum of the four thin vectors results in a total electric field pointing
in the positive $y$ direction.


\vspace{0.4cm}
\noindent
$10)$ The potential from a point charge $q$ is
$$V(r) = \frac{q}{4\pi\varepsilon_0 r}$$
where $r$ is the distance from the point charge. The total potential is the sum of
the potentials from each point charge (remember, the superposition principle). Here, we
have pairs of charges with opposite sign but equal distance. Hence, the sum must be zero.
Thinking a little, we realize that the whole $xz$ plane is in fact an equipotential surface
with potential $V=0$. Obviously, the same must also be the case for the $yz$ plane.


\vspace{0.4cm}
\noindent
$11)$ In the lectures, we used a parallel plate capacitor as an example and started with
$$U = \int_0^Q v(q)\;dq$$
and then showed that with an electric field $\bold{E}$, we have a potential energy pr
unit volume
$$u(E) = \frac{1}{2}\varepsilon_0 E^2$$
Here, $v(q)$ represented the potential difference between the capacitor plates, so that
$v(q)\;dq$ represented the work required to increase the charge on the capacitor plates
from $\pm q$ to $\pm (q+dq)$. Hence, $U$ becomes the total potential energy stored
in the capacitor when we have charged it from 0 to final charge $\pm Q$.
                                                                                                                                            
\noindent
This means that we have two alternative ways of determining the potential energy:
We may associate $U$ with the charge $Q$ and use the first formula, of course provided
we know $v(q)$. Alternatively, we may associate $U$ with the electric field
and use the formula
$$U = \int_V u(E)\; dV = \int_V \frac{1}{2}\varepsilon_0 E^2 \; dV$$
provided we know what the field looks like in the volume $V$.
                                                                                                                                            
\noindent
In this question, the system is quite simple, and we know both the potential of
the metal sphere when it has a charge $q$,
$$v(q) = \frac{q}{4\pi\varepsilon_0 R}$$
and the electric field in all the space surrounding the sphere,
$$E(r) = \frac{Q}{4\pi\varepsilon_0 r^2}$$
when is is "fully" charged.
Of course, it is sufficient to calculate the potential energy $U$ one way, but
let us do it both ways here, in order to check that we obtain the same answer.

\noindent
$U$ associated with the charge $Q$:
$$U = \int_0^Q v(q)\;dq = \int_0^Q \frac{q}{4\pi\varepsilon_0 R}\;dq = \frac{Q^2}{8\pi\varepsilon_0 R}$$
                                                                                                                                            
\vspace{0.3cm}
\noindent
$U$ associated with the field $E$:
$$U = \int_V u(E)\; dV = \frac{1}{2}\varepsilon_0 \int_R^\infty E^2 \cdot 4\pi r^2 \;dr =
\frac{1}{2}\varepsilon_0 \left(\frac{Q}{4\pi\varepsilon_0}\right)^2 \cdot 4\pi
\int_R^\infty \frac{r^2 \;dr}{r^4} = \frac{Q^2}{8\pi\varepsilon_0 R}$$
Here, we used spherical coordinates, but since $E$ only depends upon $r$, we could
do the two angular integrations "directly" and use the volume element
$dV = 4\pi r^2 \;dr$ = volume of a spherical shell with radius $r$ and thickness $dr$.
                                                                                                                                            
\noindent
We see that the answers are equal. In conclusion, we may choose if we like to associate
the potential energy with the charges or with the electric field created by the charges.


\vspace{0.4cm}
\noindent
$12)$ The total charge on the two capacitors coupled in parallel is
$$Q = Q_1 + Q_2$$
Since $C_1 = Q_1/\Delta V$ and $C_2 = Q_2/\Delta V$, we may write
$$Q = C_1 \Delta V + C_2 \Delta V = (C_1 + C_2) \Delta V$$
The total capacitance is (pr definition)
$$C = \frac{Q}{\Delta V}$$
so we find
$$C = C_1 + C_2$$
In other words: The capacitance of two capacitors coupled in parallel equals
the sum of the capacitances of each capacitor. Here, we have shown that this is
true for two capacitors. It is straightforward to generalize to an arbitrary number
of capacitors coupled in parallel:
$$C = \sum_i C_i$$
where the sum over $i$ runs from 1 to $N$ = the number of capacitors coupled in parallel.


\vspace{0.4cm}
\noindent
$13)$ The total voltage drop across the two capacitors coupled in series is
$$\Delta V = \Delta V_1 + \Delta V_2$$
Since $C_1 = Q/\Delta V_1$ and $C_2 = Q/\Delta V_2$, we may write
$$\Delta V = \frac{Q}{C_1} + \frac{Q}{C_2} = Q \left( \frac{1}{C_1} + \frac{1}{C_2} \right)$$
The total capacitance is
$$C = \frac{Q}{\Delta V}$$
so we obtain
$$C = \left( \frac{1}{C_1} + \frac{1}{C_2} \right)^{-1}$$
In other words: The inverse capacitance of two capacitors coupled in series equals
the sum of the inverse capacitances of each capacitor. Here, we have shown that this
is true with two capacitors. It is straightforward to generalize to an arbitrary number
of capacitors:
$$\frac{1}{C} = \sum_i \frac{1}{C_i}$$
where the sum over $i$ runs from 1 to $N$ = the number of capacitors coupled in series.

\vspace{0.4cm}
\noindent
$14)$ The electric field from the two capacitor plates will induce charge on the upper
and lower surface of the inserted metal sheet. In equilibrium, we must have zero
electric field inside the metal sheet. This is achieved if the induced charge is
$-Q$ on the upper surface and $Q$ on the lower surface of the metal sheet. Why
exactly $-Q$ and $Q$? Well, because the electric field from an infinitely large
charged plane is independent of the distance to the plane, and given by
the surface charge density $\sigma$:
$$E_0 = \sigma/2\varepsilon_0$$
In our case, $\sigma = Q/A$, where $A$ is the plate area. The direction of
the field from a charged plane is {\em away from} the plane if it is positively
charged and {\em towards} the plane if it is negatively charged. Our system then
becomes as shown in the figure. With e.g. the $y$ axis upwards, the different contributions
to the total electric field in the different regions of space then are either
$(-\sigma/2\varepsilon_0)\hat{y}$ or $(\sigma/2\varepsilon_0)\hat{y}$, see the figure.
                                                                                                                                            
%%FIGURE
\vspace{0.3cm}
\centerline{\epsfxsize=4.0in \epsfbox{los8_fig2.eps}}
\vspace{0.3cm}
%%END FIGURE
                                                                                                                                            
\noindent
So we have essentially 4 infinitely large planes, two with charge density
$\sigma$ (1 and 3) and two with charge density $-\sigma$ (2 and 4). In the figure,
we have drawn the contributions from each of the four planes in all the five
"different" regions. The total electric field is simply the vector sum in each
region, so
$$E=0$$
on the outside, and inside the metallic sheet. In the two regions between the
metallic sheet and the capacitor plates we see that the field becomes
$$\bold{E} = -\frac{\sigma}{\varepsilon_0} \hat{y}$$
i.e., the same as before we inserted the metal sheet.


\noindent
Finally, to what the question was all about, namely the potential difference between
the capacitor plates. Without the metallic sheet present, the potential difference becomes
$$\Delta V = -\int_{(-)}^{(+)} \bold{E} \cdot d\bold{l} = \frac{\sigma d}{\varepsilon_0}$$
because then $\bold{E} = (-\sigma/\varepsilon_0)\hat{y}$ everywhere between the plates.
(We choose $d\bold{l} = dy\;\hat{y}$.)

\noindent
With the metal sheet present, the potential difference becomes
$$\Delta V = -\int_{(-)}^{(+)} \bold{E} \cdot d\bold{l} = \frac{\sigma (d-h)}{\varepsilon_0}
= \frac{\sigma d}{3\varepsilon_0}$$
because now $\bold{E} = (-\sigma/\varepsilon_0)\hat{y}$ only in the two regions between
the metal sheet and the capacitor plates. These two regions have a total extent
$d-h=d-2d/3=d/3$; inside the metal slab we have $E=0$.

\noindent
In conclusion: The potential difference becomes three times smaller.

\noindent
Comment: This was a very lengthy solution. Nevertheless, I personally find the approach
suggested above a nice way of thinking in questions like these, with infinitely large
charged planes. The only thing we need to know is that the field from one plane is
$\sigma/2\varepsilon_0$, with direction away from or towards the plane, with positive or
negative charge, respectively. In addition, we must of course know that $E=0$ inside a metal.
Finally, we need to know that the superposition principle applies for the electric field.
                                                                                                                                            

\vspace{0.4cm}
\noindent
$15)$ An infinitely large uniformly charged plane with charge pr unit area $\sigma$
creates a constant electric field $E=\sigma/2\varepsilon_0$. If $\sigma$ is positive,
$E$ is directed away from the plane, and opposite if $\sigma$ is negative. Further, we know
that the potential {\em decreases} if we move {\em with} the electric field. Remember:
Decreasing potential when we move away from a positive charge, since a positive test
charge must have decreasing potential energy when it moves away from a positive charge.

\noindent
Here, we have a positively charged plane, so the potential will decrease when we move
away from the plane. Is has been chosen the value $V=-20$ V on the plane, so $V$ must
here be negative everywhere.

\noindent
(If the plane had been negatively charged, with charge density $\sigma = -4$ nC/m$^2$,
we would have had $V=0$ in a distance $d = \Delta V/E = \Delta V \cdot 2\varepsilon_0/\sigma =
20 \cdot 2 \cdot 8.85 \cdot 10^{-12}/4\cdot 10^{-9} \simeq 0.09$ m.)

\vspace{0.4cm}
\noindent
$16)$ If the electric field is $E_0$ between the metal plates before we insert the two
dielectric layers, the field strength will be reduced by a factor $1/\varepsilon_{rj}$ inside
dielectric number $j$ ($j=1,2$). Here, $\varepsilon_{rj}$ is the relative permittivity of
dielectric $j$, i.e., $\varepsilon_{r1}=4$ and $\varepsilon_{r2}=2$. The field strength
is not changed in the vacuum layer between $x=2a$ and $x=3a$. Hence:
\begin{eqnarray}
E &=& \frac{1}{4} E_0 \;\;\;\;\; 0 < x < 2a \nonumber \\
E &=& E_0 \;\;\;\;\;\;\; 2a < x < 3a \nonumber \\
E &=& \frac{1}{2} E_0 \;\;\;\;\; 3a < x < 5a \nonumber
\end{eqnarray}
This means that the potential will decrease most slowly between 0 and $2a$,
fastest (more precisely: four times faster) between $2a$ and $3a$, and "intermediately"
fast (more precisely: twice as fast as between 0 and $2a$) between $3a$ and $5a$. 
Only curve 3 is consistent with this.


\vspace{0.4cm}
\noindent
$17)$ See question 7: There, we agreed that the electric field $\bold{E}$ must be
constant throughout the space between the two plates. Next, we must have
$\bold{P} = 0$ in the right half: No electric dipoles to be aligned in vacuum!
However, in the dielectric we have such electric dipoles that are aligned, resulting
in a polarization $\bold{P}$ with the same direction as $\bold{E}$. But then it should
be clear that the electric displacement
$\bold{D} \equiv \varepsilon_0 \bold{E} + \bold{P}$ becomes larger in the left half.
This is also consistent with what we said, that $\bold{D}$ can be associated with
free charge: In question 7, we concluded that we had the largest amount
of free charge on the left half of the metal plates, where we have the dielectric.

\vspace{0.4cm}
\noindent
$18)$ Starting from questions 7 and 17, we have agreed that the electric field
is constant in the whole region between the plates, and also that the density of
free charge on the metal plates is biggest on the side where we have the dielectric
present. Hence, we may write
$$E_1 = \frac{\sigma^{(1)}}{\varepsilon_0} = \frac{\sigma_f^{(1)} - \sigma_i^{(1)}}{\varepsilon_0}$$
for the field in region 1, to the left, and
$$E_2 = \frac{\sigma^{(2)}}{\varepsilon_0} = \frac{\sigma_f^{(2)}}{\varepsilon_0}$$
for the field in region 2, to the right. Here, $\sigma^{(1)}$ and $\sigma^{(2)}$ are
{\em total} charge densities on the left and right side, respectively,
$\sigma_f^{(1)}$ and $\sigma_f^{(2)}$ are {\em free} charge density (on the metal plates)
on the left and right side, respectively, and $\sigma_i^{(1)}$ is {\em induced} charge
density (on the surface of the dielectric) on the left side. These fields are supposed
to be equal, $E_1 = E_2 = E$, and the potential difference between the plates is determined
by this field strength:
$$\Delta V = Ed$$
Let us briefly repeat the various physical quantities that we introduced in the lectures
when we talked about polarization in linear media. We assumed that the polarization
is proportional to the electric field:
$$\bold{P} = \chi_e \varepsilon_0 \bold{E}$$
where $\chi_e$ = the {\em susceptibility} of the medium. With the {\em definition}
$$\bold{D} \equiv \varepsilon_0 \bold{E} + \bold{P}$$
we could then write
\begin{eqnarray}
\bold{D} &=& \varepsilon_0 \bold{E} + \bold{P} \nonumber \\
&=& (1 + \chi_e) \varepsilon_0 \bold{E} \nonumber \\
&=& \varepsilon_r \varepsilon_0 \bold{E} \nonumber \\
&=& \varepsilon \bold{E} \nonumber
\end{eqnarray}
where we had introduced $\varepsilon_r \equiv 1 + \chi_e$ = the {\em relative}
permittivity and $\varepsilon \equiv \varepsilon_r \varepsilon_0$ = the
permittivity of the medium.

\noindent
The electric dipole moment of the dielectric in region 1 is
$$p_1 = (\sigma_i^{(1)}A/2)d$$
so the polarization here becomes
$$P_1 = \frac{p_1}{(Ad)/2} = \sigma_i^{(1)}$$
since the volume of region 1 is $Ad/2$. Then, the electric displacement in
region 1 becomes
$$D_1 = \varepsilon_0 E_1 + P_1 = \sigma_f^{(1)} - \sigma_i^{(1)} + \sigma_i^{(1)} = \sigma_f^{(1)}$$
In region 2, we have $P_2 = 0$, so that
$$D_2 = \varepsilon_0 E_2 = \sigma_f^{(2)}$$
At the same time, we have
$$D_1 = \varepsilon E_1$$
Let us look at what the total charge on the metal plates is:
\begin{eqnarray}
Q &=& \frac{A}{2} \left(\sigma_f^{(1)} + \sigma_f^{(2)} \right) \nonumber \\
&=& \frac{A}{2}\left( D_1 + D_2 \right) \nonumber \\
&=& \frac{A}{2} \left( \varepsilon E_1 + \varepsilon_0 E_2 \right) \nonumber \\
&=& \frac{A}{2} E \left( \varepsilon + \varepsilon_0 \right) \nonumber \\
&=& \frac{A}{2} \frac{\Delta V}{d} \varepsilon_0 (\varepsilon_r + 1) \nonumber
\end{eqnarray}
Pr definition, the capacitance of the capacitor is
$$C = \frac{Q}{\Delta V}$$
so here we have
$$C = \frac{A}{2d} \, \varepsilon_0 (\varepsilon_r + 1) = \frac{\varepsilon_r + 1}{2} C_0$$
with $C_0 = \varepsilon_0 A/d$.
                                                                                                                                            
\noindent
It should perhaps not come as a big surprise that this capacitor can be viewed as
two capacitors coupled in parallel, both with plate area $A/2$,
plate distance $d$, and one of them filled with vacuum and the other filled with
a dielectric with permittivity
$\varepsilon = \varepsilon_r \varepsilon_0$.
Thus, we could actually have written down the result more or less directly, using the result
in question 12.


\vspace{0.4cm}
\noindent
$19)$ This question is similar to the previous one, except that we have here effectively
two capacitors coupled in series instead of parallel. Let us try the simple solution first,
equipped with the wisdom learnt in question 18. The two capacitances, coupled in series both
have plate area $A$ and distance between the plates $d/2$. One of them is filled with air/vacuum
and the other is filled with a dielectric with permittivity
$\varepsilon = \varepsilon_r \varepsilon_0$. Then we may use the result found in question 13.
The capacitance of the half filled with dielectric is
$$C_1 = \varepsilon \frac{A}{d/2} = 2\varepsilon_r \varepsilon_0 \frac{A}{d}$$
whereas the capacitance of the half filled with air is
$$C_2 = \varepsilon_0 \frac{A}{d/2} = 2\varepsilon_0 \frac{A}{d}$$
Total capacitance is then, according to question 13
\begin{eqnarray}
C &=& \left(\frac{1}{C_1} + \frac{1}{C_2} \right)^{-1} \nonumber \\
&=& \varepsilon_0 \frac{A}{d} \left( \frac{1}{2\varepsilon_r} + \frac{1}{2} \right)^{-1} \nonumber \\
&=& \varepsilon_0 \frac{A}{d} \frac{2\varepsilon_r}{\varepsilon_r + 1} \nonumber \\
&=& \frac{2\varepsilon_r}{\varepsilon_r + 1} C_0 \nonumber
\end{eqnarray}
Also in this question, we may go the long way via electric displacement, induced and free
charge etc., and determine the relation between total free charge $Q$ and the potential
difference $\Delta V$.
We have:
$$D = \sigma_f = Q/A$$
In this case, $D$ is constant throughout the region between the plates since the free
charge is uniformly distributed over the metal plates (no difference between right and left here).
Next, we have
$$D = \varepsilon E_1$$
for the relation between electric displacement and electric field in
region 1 (i.e., the lower half with dielectric). Furthermore,
$$D = \varepsilon_0 E_2$$
for the relation between electric displacement and electric field in
region 2 (i.e., the upper half, with vacuum). The potential difference between
the plates is
$$\Delta V = E_1 \cdot \frac{d}{2} + E_2 \cdot \frac{d}{2}$$
which is found by taking the line integral of $\bold{E}$ from one of the plates to the other.
But now we have more or less what we need:
\begin{eqnarray}
\Delta V &=& E_1 \cdot \frac{d}{2} + E_2 \cdot \frac{d}{2} \nonumber \\
&=& \frac{d}{2} \left( \frac{D}{\varepsilon} + \frac{D}{\varepsilon_0} \right) \nonumber \\
&=& \frac{d}{2\varepsilon_0} \left( \frac{Q}{A\varepsilon_r} + \frac{Q}{A} \right) \nonumber \\
&=& \frac{Qd}{2\varepsilon_0 A}\cdot \frac{1 + \varepsilon_r}{\varepsilon_r} \nonumber
\end{eqnarray}
so that
$$C = \frac{Q}{\Delta V} = \frac{2\varepsilon_0 A}{d}\cdot \frac{\varepsilon_r}{1 + \varepsilon_r}
= \frac{2\varepsilon_r}{1 + \varepsilon_r} C_0$$
In other words, the same as found above, by simply using the formula for two
capacitances coupled in series!


\vspace{0.4cm}
\noindent
$20)$ Here, the electric field in the region between the inner and outer metal cylinder
is given in the text, so it's just a question of calculating the potential difference:
\begin{eqnarray}
\Delta V &=& V_a - V_b = -\int_b^a E(r)\;dr \nonumber \\
&=& \frac{\lambda}{2\pi\varepsilon} \int_a^b \frac{dr}{r} \nonumber \\
&=& \frac{\lambda}{2\pi\varepsilon} (\ln b - \ln a ) \nonumber \\
&=& \frac{Q}{2\pi\varepsilon L} \ln \frac{b}{a} \nonumber
\end{eqnarray}
Hence, the capacitance of the cylindrical capacitor becomes
$$C = \frac{Q}{\Delta V} = \frac{2\pi\varepsilon L}{\ln b/a}$$
The biggest problem here is perhaps keeping track of the overall sign, and
for example end up with alternative C. However, then you must remember:
A capacitance is, pr definition, a {\em positive} quantity. Since
$a<b$, alternative D simply must be the correct one, because the logarithm
of a number smaller than 1 is negative.
                                                                                                                                            
\noindent
Another thing to note, perhaps, is that we will have the high value of
the electric potential at the positive conductor. This follows from the
definition of electric potential, in terms of potential energy pr unit charge.
Imagine a small positive charge. It must obviously have the largest potential
energy if we choose to put it in a position near the positively charged conductor.
Hence, here is also the highest value of the electric potential. For a small
negative charge, it is opposite: It will have the highest potential energy if we
choose to put it in a position near the negatively charged conductor. So, here we
must have the lowest value of the electric potential, so that when we multiply
the value of the potential with the negative value of the charge, we end
up with (the largest) positive value of the potential energy.
                                                                                                                                            
\vspace{1.5cm}
\noindent
One final comment: Do you think this exercise required a lot of work? I agree.
The midterm exam will contain 40 questions. Obviously, this cannot be 40 questions that require
as much work as some of the questions above. This applies in particular to some
of the last questions, which really required hard work, much because it's
new and unfamiliar topics. On the other hand: Questions like nr 5 should not take
too long time. Either you know that $E=0$ inside a metal sphere in equilibrium, or you don't!

\noindent
With 40 questions and 3 hours, you will have 4.5 minutes for each question. Since all
correctly answered questions count equally, it may be a good idea to do the "quick ones"
first.

\end{document}
