\documentclass[12pt]{article} \usepackage{epsfig} \topmargin=-0.5cm \oddsidemargin=-0.5cm \evensidemargin=-0.5cm \textheight=23cm \textwidth=17cm \begin{document} \newcommand{\bold}[1]{\mbox{\boldmath $#1$}} \begin{bf} \noindent Institutt for fysikk, NTNU \noindent TFY4155/FY1303 Elektrisitet og magnetisme \noindent V{\aa}r 2004 \end{bf} \begin{center} \begin{large} Solution to {\o}ving 8 \end{large} \end{center} \noindent Guidance week 9 and 10. \vspace{1cm} \noindent \begin{tabular}{|r|c|c|c|c|c|} \hline Exercise &\hspace{0.3cm} A \hspace{0.3cm} &\hspace{0.3cm} B \hspace{0.3cm} &\hspace{0.3cm} C \hspace{0.3cm} & \hspace{0.3cm} D \hspace{0.3cm} &\hspace{0.3cm} E \hspace{0.3cm} \\ \hline 1 \hspace{0.5cm} & & &x& & \\ \hline 2 \hspace{0.5cm} & &x& & & \\ \hline 3 \hspace{0.5cm} & & &x& & \\ \hline 4 \hspace{0.5cm} &x& & & & \\ \hline 5 \hspace{0.5cm} & & & &x&\#\#\#\#\# \\ \hline 6 \hspace{0.5cm} &x& & & &\#\#\#\#\# \\ \hline 7 \hspace{0.5cm} & & & &x&\#\#\#\#\# \\ \hline 8 \hspace{0.5cm} & & & &x&\#\#\#\#\# \\ \hline 9 \hspace{0.5cm} & & &x& & \\ \hline 10 \hspace{0.5cm} &x& & & & \\ \hline 11 \hspace{0.5cm} & & & &x& \\ \hline 12 \hspace{0.5cm} & &x& & & \\ \hline 13 \hspace{0.5cm} & & & & &x\\ \hline 14 \hspace{0.5cm} &x& & &\#\#\#\#\# &\#\#\#\#\# \\ \hline 15 \hspace{0.5cm} & &x& & &\#\#\#\#\# \\ \hline 16 \hspace{0.5cm} & &x& &\#\#\#\#\# &\#\#\#\#\# \\ \hline 17 \hspace{0.5cm} &x& & &\#\#\#\#\# &\#\#\#\#\# \\ \hline 18 \hspace{0.5cm} & & & & &x\\ \hline 19 \hspace{0.5cm} &x& & & & \\ \hline 20 \hspace{0.5cm} & & & & &x\\ \hline \end{tabular} \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 is 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 potentia $V(r)$ 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 previous question) $$V(R) = \frac{Q}{4\pi\varepsilon_0 R}$$ Hence, largest potential for smallest radius. \noindent This should also be intuitively clear: Imagine starting with a neutral metal sphere and putting on electric charge until the net charge is $Q$. Then you will experience a larger "resistance" the smaller the sphere, i.e., you must perform a larger work in order to put the charge onto a smaller sphere than a larger one. In other words, the small sphere will end up with the largest potential energy, and thereby also the largest potential. \noindent [If the charge were negative, the smallest sphere would have had the {\em smallest} electric potential, i.e., {\em the most negative} potential compared to $V(\infty)=0$. However, the smallest sphere would still have the largest potential energy, since $\Delta U = Q \Delta V = -|Q| \Delta V$ when $Q<0$.] \vspace{0.4cm} \noindent $4)$ The potential in a certain distance from the point charge $Q$ depends upon $Q$, {\em not} the value of $q$, i.e., the charge that we put in the position where we ask for the value of the potential. Remember, that was the point by introducing the quantities electric field and electric potential in the first place, as force and potential energy, respectively, pr unit charge. Then we had physical quantities that "are there" whether the "test charge" $q$ is present or not. The potential from $Q$ decays as $1/r$, so $V(2r) = V(r)/2$. All the other statements are 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 $6)$ $E=0$ inside the metal sphere, thus C and D are out of the question. The plastic is a dielectric with electric dipoles that align along the electric field from the charged metal sphere. We obtain an induced negative surface charge on the inner surface of the plastic and a positive surface charge on the outer surface. (All together the same amount of negative and positive charge, since the plastic layer is overall neutral.) These induced charges generate a contribution to the total electric field directed inwards, i.e., in the opposite direction of the field from the charge on the metal sphere. Thus, the electric field inside the plastic must be reduced when compared to the situation with vacuum everywhere. Outside the plastic layer, the field will be the same as without the plastic present, and therefore stronger than inside the plastic. You may prove this by using Gauss' law, by taking a spherical gaussian surface with radius larger than the distance from the center to the outer surface of the plastic. Then the net charge inside the gaussian surface is $Q$, so the field must be $Q/4\pi\varepsilon_0 r^2$, i.e., as for a point charge $Q$ in the center. Alternatively, if you know that a spherically symmetric charge distribution yields the same electric field (on the outside) as if all the charge were placed in the center, i.e., as for a point charge: Here, you have three spherically symmetric charge distributions: $Q$ on the surface of the metal sphere, $-q_i$ on the inner surface of the plastic layer, and $q_i$ on the outer surface of the plastic layer. All in all as a point charge $Q - q_i + q_i = Q$ in the center! (This "wisdom" is a direct result of Gauss' law, and therefore Coulomb's law. You may remember that the same thing is true for gravitational forces, i.e., the earth attracts you with the same force as if all the mass of the earth were positioned at the center.) \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 on 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 pair 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 elctric 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 electris 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 $$\Delta V = -\int_{(-)}^{(+)} \bold{E} \cdot d\bold{l} = \frac{\sigma (d-h)}{\varepsilon_0}$$ 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 $d-h$. Inside the metal sheet, $E=0$. \noindent In conclusion: The potential difference becomes 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. \noindent Is there a quick, intuitive argument for getting a reduced potential difference in this problem? Well, let us try: Because of the induced negative and positive charge on the upper and lower surface of the metal sheet, there will be {\em attractive} forces between the surfaces of the sheet and the capacitor plates. (But the {\em net force} on the metal sheet is zero, of course!) This means that if we want to remove the metal sheet, we must necessarily perform a positive work on the system: We must overcome the attractive forces between sheet and capacitor plates if we for example pull the sheet out towards the right. This must again mean that the system has a lower potential energy with the sheet "in place" than without. Hence, the potential difference between the plates must also be smaller with the metal sheet "in place". \vspace{0.4cm} \noindent $15)$ We know that the electric field from an infinitely large charged plane is independent of the distance to the plane. Hence, the field between two such planes is also independent of the distance to the two planes. Increasing the distance $d$ between the planes does not change the electric field strength. \noindent What about the alternative solutions? An increase in $d$ results in reduced capacitance ($C = \varepsilon_0 A/d$), increased potential energy ($U = \int (\varepsilon_0 E^2/2)dV = A d \varepsilon_0 E^2/2$) and increased potential difference between the planes ($V = Ed$). \vspace{0.4cm} \noindent $16)$ See question 7: The field from the metal plates will polarize the dielectric and induce surface charge on the dielectric, negative on top and positive at the bottom. This will result in more free electrons moving from the right to the left side of the lower plate, and opposite for the upper plate. Thus, $Q_1 > Q_2$. \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 the previous question, we concluded that we had the largest amount of free charge ($Q_1$) on the left half of the metal plates. \vspace{0.4cm} \noindent $18)$ Starting from questions 7, 16, 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 susceptibilitet} 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 displacment 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 $$C = \frac{A}{2} d \varepsilon_0 (\varepsilon_r + 1) = \frac{\varepsilon_r + 1}{2} C_0$$ 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 displacment, 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 elctric 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 D. However, then you must remember: A capacitance is, pr definition, a {\em positive} quantity. Since $a