Solved Problems

Solution:

Note: the resistance of a conducting object with cross-sectional area $A$ an length $L$ is given by $R=\rho \ L/A$. So first calculate the $A$ of this wire as \begin{align*}A&=\pi r^2=\pi {\left(\frac{d}{2}\right)}^2\\&=\frac{\pi }{4}{\left(1.88\times {10}^{-4} \rm{m}\right)}^2\\&=2.78\times {10}^{-8} \quad {\rm m^2}\end{align*}

 
(a) Now we can calculate the resistance of the wire at $T=20{}^\circ {\rm C}$
\begin{align*}R_0&=\rho \frac{L}{A}\\ \\&=\left(2.44\times {10}^{-8}\right)\frac{\ 0.137}{2.78\times {10}^{-8}}\\ \\&=0.120\quad \rm{\Omega }\end{align*}
(b) By definition, the variation of the resistance with temperature is given by 
\begin{align*}R&=R_0\left[1+\alpha \left(T-{T_0}\right)\right]\\ \\ &=R_0+\alpha R_0\left(T-T_0\right)\\ \\ \Rightarrow T&=T_0+\frac{R-R_0}{\alpha R}\\ \\&=20{}^\circ {\rm C}+\frac{\left(0.253-0.120\right)}{\left(3.40\times {10}^{-3}\right)\left(0.120\right)}\\ \\&=20{}^\circ {\rm C}+326{}^\circ {\rm C}\\ \\&=346{}^\circ {\rm C}\end{align*}

Solution: The current density is defined as $J=I/A$, but this quantity is related to the electric field via $\vec{E}=\rho \vec{J}$ (this is an alternative version of Ohm's law for conducting objects!) thus we have $I=JA=\frac{1}{\rho }EA\Rightarrow E=\rho I/A$
\begin{align*} \Rightarrow E&=\rho \frac{I}{A_{disk}}\\&=\rho \frac{I}{\pi r^2}\\&=\left(2.44\times {10}^{-8}\right)\frac{940\times {10}^{-3}}{\pi {\left(\frac{0.9}{2}\times {10}^{-3}\right)}^2}\\&=0.0361\, {\rm \frac Vm} \end{align*} 
 

Solution: A far as possible, first simplify the diagram as below and then calculate the equivalence resistance. 

The $6\, {\rm \Omega }$ and $12\, {\rm \Omega }$ are in parallel 
\[\frac{1}{R_1}=\frac{1}{6}+\frac{1}{12}\Rightarrow R_1=4\,{\rm \Omega }\] 

Three $4\, {\rm \Omega}$ resistors are in series so \[R_2=4+4+4=12\,{\rm \Omega }\] 

The $12\, {\rm \Omega }$ and $6\,{\rm \Omega }$ are in parallel so \[\frac{1}{R_3}=\frac{1}{12}+\frac{1}{6}\Rightarrow R_3=4\,{\rm \Omega }\] 

Therefore the equivalent resistance is $R_{eq}=2+4+2=8\,{\rm \Omega }$

By definition of the electric current  $I=Q/\Delta t\ $and surface charge density $\sigma =Q/A$, we must first find the area of the plastic film during $1\, \mathrm{s}$ moving and then multiply it by the $\sigma $ to calculate the total charge during this interval time. Thus, 
\[\omega =90\ \mathrm{rpm} \xrightarrow{convert}\ 90\ \left(\frac{\mathrm{rev}}{\mathrm{min}}\right)\left(\frac{2\pi \mathrm{rad}}{\mathrm{1\ rev}}\right)\left(\frac{\mathrm{1\ min}}{\mathrm{60s}}\right)=9.425\frac{\mathrm{rad}}{\mathrm{s}}\] 
The linear speed of the plastic film is determined by $v=\omega r=9.425\times 1=9.425\frac{\mathrm{cm}}{\mathrm{s}}$
The length has the plastic traveled during $1\,\mathrm{s}$ is $x=vt=9.425\ \mathrm{cm}$ so the area of it is $A=\left(9.425\times 4\right)=37.7\ {\rm cm^2}$. Now 
\[Q=\sigma A=\left(5\times {10}^{-9}\frac{\mathrm{C}}{{\mathrm{c}\mathrm{m}}^{\mathrm{2}}}\right)\left(\mathrm{37.7\ }{\mathrm{cm}}^{\mathrm{2}}\right)\mathrm{=}188.5\mathrm{\ nC}\] 
\[I=\frac{Q}{\Delta t}=\frac{188.5}{1}=188.5\ \mathrm{nA}\] 
 

Note: the energy stored in a capacitor is determined by the following equations:
\[U=\frac{1}{2}CV^2=\frac{Q^2}{2C}=\frac{1}{2}QV\] The given information is:
$C=1\,{\rm \mu F}$ and $V_1=6\, {\rm V}\ ,\ V_2=8\,{\rm V}$, therefore, \begin{gather*} U_1=\frac{1}{2}C_1V^2_1=\frac{1}{2}\left(1\right){\left(6\right)}^2=18\,{\rm \mu J} \\  \\ U_2=\frac{1}{2}C_2V^2_2=\frac{1}{2}\left(1\right){\left(8\right)}^2=32\,{\rm \mu J}  \\ \\ \Rightarrow \Delta U=U_2-U_1=14\,{\rm \mu J}\end{gather*} Use each of them, where appropriate.

(a) The capacitance of two parallel plates with area $A$ that is separated by $d$ is determine by $C=k{\epsilon }_0\frac{A}{d}$ where $k$ is the dielectric constant. Since there is air between the plates (without dielectric) so $k=1$ \begin{align*} C&={\epsilon }_0\frac{A}{d}\\\\&=(8.85\times {10}^{-12}) \frac{0.1\times 0.1}{0.005}\\ \\&=1.77\times {10}^{-11}\,\rm F \\ \\ &=17.7\,{\rm pF}\end{align*} 
(b) Recall from section of the electric potential problems, the voltage or potential difference between two plates is given by $\Delta V=Ed$. Therefore \[E=\frac{\Delta V}{d}=\frac{200}{0.005}=4\times {10}^4\,{\rm N/C}\] 
(c) By definition, the capacitance of a capacitor is $C=Q/V$, so \begin{align*}Q&=C\Delta V\\ \\&=(17.7\times {10}^{-12})(200)\\ \\&=35.4\times {10}^{-10}\quad {\rm C}\\ \\&=3.54\quad {\rm nC}\end{align*}

(d) By definition, the potential energy stored in a capacitor is $U=\frac{1}{2}CV^{2}=\frac{Q^2}{2C}=\frac{1}{2}QV$
\begin{align*}U&=\frac{1}{2}CV^2\\ \\&=\frac{1}{2}\left(17.7\times {10}^{-12}\right)(200)^{2}\\ \\&=35.4\times {10}^{-8}\,{\rm J}\\ \\&=354\quad{\rm nJ} \end{align*}
(e) Apply the kinetic-energy theorem $\Delta K=W_{net}$
Recall that because the charge of the electron is negative, the electric force $F_e=-eE$ acting on it is in the opposite direction of the electric field. So the angle between the force and displacement is $180^\circ$ and the physical work done on the electron is  $W_e=F_ex\,{\cos 180^\circ}=-F_ex$. Therefore, we have \begin{gather*} K_2-K_1=W_{net}=W_e\\ \\ \frac{1}{2}mv^2_f-\frac{1}{2}mv^2_i=-\left(-eE\right)x=e\underbrace{Ex}_{\Delta V}\\ \\ \Rightarrow v_f=\sqrt{v^2_i+\frac{2e}{m}\mathrm{\Delta }V} \\ \\ =\sqrt{{\left(1.5\times {10}^7\right)}^2+\frac{2\left(1.6\times {10}^{-19}\right)}{9.11\times {10}^{-31}}(200)}\\ \\ =1.72\times {10}^7\quad {\rm m/s}\end{gather*}
(f) As above, we can write \begin{align*}\Delta K&=W_e=-(-eE)x\\ \\&=e\,\Delta V\\ \\&=e(200)\\ &=200\quad{\rm eV}\end{align*}

(a) Use Ohm's law to find the resistance of a conductor: $R=\frac{V}{I}=\frac{4}{2}=2\,\mathrm{\Omega }$
(b) The electrical resistance of a wire or conductor with resistivity $\rho $, length $L$ and area cross section $A$ is defined as $R=\rho \ L/A$. Thus,\begin{gather*} A=\pi {\left(\frac{d}{2}\right)}^2=\rho \frac{L}{R}\\ \\ \Rightarrow d=\sqrt{4\rho \frac{L}{R\pi }}\\ \\ =\sqrt{4\times \left(1.5\times {10}^{-6}\right)\frac{0.50}{2\pi }}\\ \\ =0.69\times {10}^{-3}\quad {\rm m}\\ \\ =0.69\quad {\rm mm}\end{gather*}
(c) Recall that the power dissipated in a resistor is defined by the following relations
\begin{gather*}P=VI=\frac{V^2}{R}=RI^2\\ \\ \Rightarrow P=VI=4\times 2=8\quad {\rm W}\end{gather*}
(d) Use the definitions of electric current, $I=\frac{\Delta }{q}{\Delta t}v$ ,and electric charge, $q=ne$, we have \begin{align*} I&=\frac{ne}{\Delta t}\\ \\ \Rightarrow n&=\frac{I\Delta t}{e}\\ \\&=\frac{2\times 1}{1.6\times {10}^{-19}}\\ \\&=1.25\times {10}^{19}\end{align*}

The power delivered to a conductor (resistor is a conductor) is given by: $P=\frac{V^2}{R}=VI=RI^2$ and the resistance of a cylindrical conductor is related to its resistivity $\rho $, length $l$ and cross-sectional area $A$ by$R=\rho \frac{l}{A}$. Therefore, 
\[\frac{P_A}{P_B}=\frac{R_B}{R_A}=\frac{l_B}{l_A}\frac{A_A}{A_B}=\frac{1}{2}{\left(2\right)}^2=2\] 
Where $A=\pi r^2$

(a) Given the power, we can calculate the energy delivered per pulse by the definition of the power as \begin{align*} P&=\frac{energy}{time}=\frac{U}{t} \\ \Rightarrow U&=Pt\\&=255\times {10}^3\times \left(12\times {10}^{-9}\right)\\ &=0.0031\quad \rm{J=3.1\,mJ\ per\ pulse}\end{align*} 
(b) Frequency and wavelength are related to each other by
\[f=\frac{c}{\lambda }=\frac{3\times {10}^8}{632\times {10}^{-9}}=4.75\times {10}^{14}\mathrm{Hz}\] 
 

(a) Recall from the capacitance problems section that the capacitance of a parallel-plate capacitor is $C=\epsilon_0\kappa A/d$. So before the slab is inserted $C={\epsilon }_0A/d$ and $V=\frac{Q}{C}$ so, \[Q=V{\epsilon }_0A/d\]
(b) After insertion of the slab, one can show that there is two capacitor in series.
\begin{gather*} \frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}=\frac{1}{\frac{{\epsilon }_0A}{\frac{d}{2}}}+\frac{1}{\frac{\kappa {\epsilon }_0A}{\frac{d}{2}}} \\\\ \Rightarrow \frac{1}{C_{eq}}=\frac{d}{2{\epsilon }_0A}\left(1+\frac{1}{\kappa }\right) \\\\  \Rightarrow C_{eq}=\frac{2{\epsilon }_0A}{d}\left(\frac{\kappa }{\kappa +1}\right)\end{gather*} Now, use the definition of capacitance $C=\frac{Q}{V}$, and find the charge stored on its plates. \[\therefore Q'=C_{eq}V=\frac{2{\epsilon }_0A}{d}\left(\frac{\kappa }{\kappa +1}\right)V\] 
(c) By definition, the work done on a system is equal to the change in the potential energy of that system. Therefore, we must calculate the difference in potential energy between two situations above:
\begin{align*} W&=U_{after}-U_{befor}\\\\ &=\frac{1}{2}C_{eq}V^2-\frac{1}{2}CV^2\\\\&=\frac{1}{2}V^2\left(\frac{{\epsilon }_0A}{d}\right)\left(\frac{2\kappa }{\kappa +1}-1\right) \end{align*} Thus, the work done on the system is obtained as \[W=\frac{V^2}{2}\left(\frac{{\epsilon }_0A}{d}\right)\frac{\kappa -1}{\kappa +1}\] 
 

(a) Let $E_1$ be the electric field in the air and $E_2$ be in the dielectric.The electric field in presence of a dielectric is $E_2=E_1/K\ $. If the electric field between two points to be uniform, then the potential difference between them is equal to the field strength $E$ multiplied by the points separation $L$ i.e.$\Delta V=EL$
Therefore, the potential difference between the plates are 
\[\Delta V=E_1\left(L-d\right)+E_2d\] 
Substituting $E_1/k$ for $E_2$, we obtain
\[\Delta V=\frac{E_1}{K}d+E_1\left(L-d\right)=E_1\left(\frac{d}{K}+\left(L-d\right)\right)\] 
\[\therefore E_1=\frac{\mathrm{V}}{\left(\frac{d}{K}+\left(L-d\right)\right)}\] 
\[E_2=\frac{E_1}{K}=\frac{V}{d+\left(L-d\right)}\] 
(b) The electric field between two parallel conducting plates is $E=\sigma /{\epsilon }_0$, where $\sigma $ is the surface charge density of the plates. Therefore, 
\[E_1=\frac{\sigma }{{\epsilon }_0}\to \sigma ={\epsilon }_0E_1={\epsilon }_0\frac{\mathrm{V}}{\left(\frac{d}{K}+\left(L-d\right)\right)}\] 
(c) Given the charge density, we can find the amount of charge placed on the plates by $Q=\sigma A={\epsilon }_0\frac{\mathrm{V}}{\left(\frac{d}{K}+\left(L-d\right)\right)}A$. Now use the definition of the capacitance as $C=\frac{Q}{V}$ to find the capacitance of this capacitor. 
\[C=\frac{Q}{V}={\epsilon }_0\frac{\mathrm{A}}{\left(\frac{d}{K}+\left(L-d\right)\right)}\] 
(d) The above configuration can be illustrated as two parallel plate capacitors $C_1\ ,\ C_2$ that are in series. We know that the capacitance of such a capacitor is $C=\frac{k{\epsilon }_0A}{d}$ so 
\[C_1=\frac{{\epsilon }_0A}{L-d}\ \ ,\ \ C_2=\frac{K{\epsilon }_0A}{d}\] 
\[\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}=\frac{1}{{\epsilon }_0A}\left(\frac{d}{K}+\left(L-d\right)\right)\] 
\[C_{eq}=\frac{{\epsilon }_0A}{\frac{d}{K}+\left(L-d\right)}\] 

In figure a, the configuration of the dielectric slabs are like the series connected capacitors so recall that the equivalent capacitance of two capacitors that are connected in series is given by \begin{gather*} \frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}\\\\ \Rightarrow C_{eq}=\frac{C_1C_2}{C_1+C_2}\end{gather*}
Where $C_1={\kappa }_1{\epsilon }_0A/(\frac{d}{2})$ is the capacitance of the parallel plate capacitor. Therefore, \begin{align*}C_{eq}&=\frac{\left(\frac{{\kappa }_1{\epsilon }_0A}{\frac{d}{2}}\right)\left(\frac{{\kappa }_2{\epsilon }_0A}{\frac{d}{2}}\right)}{\left(\frac{{\kappa }_1{\epsilon }_0A}{\frac{d}{2}}\right)+\left(\frac{{\kappa }_2{\epsilon }_0A}{\frac{d}{2}}\right)}\\\\&=\frac{{\left(\frac{{\epsilon }_0A}{\frac{d}{2}}\right)}^2\left({\kappa }_1{\kappa }_2\right)}{\frac{2{\epsilon }_0A}{d}\left({\kappa }_1+{\kappa }_2\right)}\\\\&=\left(\frac{{\kappa }_1{\kappa }_2}{{\kappa }_1+{\kappa }_2}\right)\frac{2{\epsilon }_0A}{d}\end{align*}
In figure b, the slabs are connected in parallel so the equivalent capacitance of them is \begin{align*} C_{eq}&=C_1+C_2\\\\ &=\frac{{\kappa }_1{\epsilon }_0A}{\frac{d}{2}}+\frac{{\kappa }_2{\epsilon }_0A}{\frac{d}{2}}\\\\&=2\frac{{\epsilon }_0A}{d}({\kappa}_1+{\kappa}_2)\end{align*}

The two left capacitors in the upper branch are in series and have equivalent capacitance of $C^{'}=\frac{C_1C_1}{C_1+C_1}=\frac{C_1}{2}=5\,\mu\mathrm{F}$.

This is in parallel with the $10\,{\rm \mu F}$ in the facing branch so the equivalent capacitance of these two branch is $C^{''}=C^{'}+C_1=5+10=15\,{\rm \mu F}$. The equivalent capacitor of the top loop is in series with the lower capacitor $C_1$ and thus the total equivalent capacitance of the circuit is 
\begin{align*} C_{eq}&=\frac{C^{''}C_1}{C^{''}+C_1}\\&=\frac{15\times 10}{15+10}\\&=6\quad {\rm \mu F}\end{align*} Recall that the capacitors in series have equal charges so the total charge of the circuit is \begin{align*} q_{tot}&=C_{eq}V=6\times 10=60\quad {\rm \mu C}\\ q_{tot}&=q_{C_1}=q_{C^{''}}=60\quad {\rm \mu C}\end{align*} The potential drop across the lower capacitor $C_1$ is $V_{C_1}=\frac{q_{C_1}}{C_1}=\frac{60}{10}=6\ \mathrm{V}$. we know that the sum of the potential drops across the capacitors in parallel is the total potential drop of the circuit, by using this fact, we find the potential drop across the upper network as follows \begin{align*} V&=V_{C_1}+V_{C^{''}}\\ \Rightarrow 10&=6+V_{C^{''}}\\\Rightarrow \ V_{C^{''}}&=4\ \mathrm{V}\end{align*} The upper network has two branches which are in parallel with each other so the potential drop across each of them is $4\, {\rm V}$. By knowing the voltage, we can find the charge stored in the left branch of the upper network as $q_{C^{'}}=C^{'}V_{C^{''}}=5\times 4=20\,\mu\mathrm{C}$. Therefore, the charge of the red capacitor is $20\,{\rm \mu C}$.

Using the definition of the current density $J$, we can compute the total current through the cross-sectional area of the wire
\[J=\frac{I}{A}\Rightarrow I=\int{\vec{J}.\hat{n}dA}\] 
In this case, the current density is in the radial direction.

A cylindrical wire has three surfaces; top, bottom, and curved. The unit vector for the top and bottom is $\hat{n}=+\hat{z}\ ,\ -\hat{z}$, respectively and for the curved surface $\hat{n}=\hat{r}$. Therefore, the dot product of the top and bottom surfaces with $\vec{J}$ becomes zero.
\begin{align*}I_{tot}&=\int^R_0{\underbrace{ar\hat{r}}_{J(r)}\cdot\underbrace{\hat{r}}_{\hat{n}}\underbrace{(2\pi rdr)}_{dA}}\\ \\&=2\pi a\int^R_0{r^2dr}\\ \\&=\frac{2}{3}\pi aR^3\end{align*} 
 

(a) At time $t=0$, since the capacitor is uncharged, it acts as a short circuit like a wire with no resistance so we can drop it out of the circuit (figure a) . In such a case, the two resistors in the left and middle branches are in parallel and their combinations with the third is in series. Thus, 
\[R_{eq}=R+{\left(\frac{1}{R}+\frac{1}{R}\right)}^{-1}=\frac{3}{2}R\] 
The main current in the circuit $I_1$ is found as 
\[I_1=\frac{\mathcal{E}}{R_{eq}}=\frac{2\mathcal{E}}{3R}\] 
The potential drop across the parallel resistors are the same $V_2=V_3$and the sum of the currents through them is equal to the total current in the circuit i.e. $I_1=I_2+I_3$. Recall that the sum of the potential drop across the series resistors is equal to the emf of the circuit i.e. $\mathcal{E}=V_1+V_{23}$, where $V_{23}$ is the voltage of the parallel branches. Using Ohm's law, we obtain \begin{align*} \mathcal{E}&=V_1+V_{23}\\ \\\to V_{23}&=\mathcal{E}-I_1R\\ \\&=\mathcal{E}-\frac{2\mathcal{E}}{3R}R\\ \\ \Rightarrow V_{23}&=\frac{1}{3}\mathcal{E}\end{align*}
We know $V_{23}=V_2=V_3$, therefore, \begin{gather*} I_2=\frac{V_2}{R}=\frac{\mathcal{E}}{3R} \\ \\ I_3=\frac{V_3}{R}=\frac{\mathcal{E}}{3R}\end{gather*} 
(b) At $t=\infty $, the capacitor is fully charged, so it acts like a wire with infinity resistance, using the Ohm's law, $I=V/R$, the current through it is zero i.e. $I_3=0$. In this case, indeed there is only two series resistors (see figure b). 
\begin{gather*}R_{eq}=R+R=2R\\\ \\to I_{tot}=\frac{\mathcal{E}}{R_{eq}}=\frac{\mathcal{E}}{2R}\end{gather*} 
Recall that in the series resistors, $I_{tot}=I_1=I_2$
(c) At $t=\infty $, there is no current in the branch containing the capacitor, so there is no potential drop across the resistor in this branch but, by definition of the capacitance $C=Q/V_C$, there must be potential difference across the capacitor. This branch and the middle branch are in parallel, so $V_2=V_C$. From part (a) we know that 
\[V_2=V_{23}=\frac{1}{3}\mathcal{E}\] 
Thus, $V_C=\mathcal{E}/3$. 

The resistors $R_1$ and $R_2$ are in series which we call their combinations as $R_{12}=R_1+R_2=2R$. The $R_{12}$ and $R_3$ are in parallel, so their combination is \begin{align*} \frac{1}{R_{123}}&=\frac{1}{R_{12}}+\frac{1}{R_3}\\ \to R_{123}&=\frac{R_{12}R_3}{R_{12}+R_3}\\&=\frac{R^2}{3R}=\frac{R}{3}\end{align*} $R_{123}$ and $R_4$ are in series \begin{align*}R_{1234}&=R_{123}+R_4\\&=\frac{2}{3}R+R\\&=\frac{5}{3}R\end{align*} $R_{1234}$ and $R_5$ are in parallel \begin{align*} R_{12345}&=\frac{R_{1234}R_5}{R_{1234}+R_5}\\&=\frac{\frac{5}{3}R\times R}{\frac{5}{3}R+R}\\&=\frac{5}{8}R\end{align*} $R_{12345}$ and $R_6$ are in series, thus the equivalent resistor of the circuit is \[R_{eq}=\frac{5}{8}R+R=\frac{13}{8}R\] 

(a) The energy stored in a capacitor is found by $U=\frac{1}{2}CV^2=\frac{1}{2}QV=Q^2/2C$. Since $C$ and $V$ has given, we can choose \begin{align*}U_0&=\frac{1}{2}C_0V^2_0\\ \\ &=\frac{1}{2}(20 {\rm \mu {F}}){\left(\mathrm{60V}\right)}^{2}\\ \\&=36000\quad {\rm \mu J}\\ \\&=36\quad mJ\end{align*}
(b) With dielectric $\kappa $, the capacitance of capacitors becomes $\kappa $ times the case without the dielectric that is 
\[\kappa =\frac{C_{with}}{C_{without}}=\frac{20}{10}=2\] 
(c) After disconnecting from battery, since there is no source to build up the charges so the total amount of charges on the plates of capacitors must be constant. By definition of capacitance, we have \begin{align*} Q_0&=C_0V_0\\&=(20\,\rm{ \mu F})(60)\\&=1200\quad {\rm \mu C}=Q_1 \\ \\ V_1&=\frac{Q_1}{C_1}\\ &=\frac{1200}{10}\\&=120\quad {\rm V}\\ \\ U_1&=\frac{1}{2}C_1V^2_1\\&=\frac{1}{2}(10)(120)^{2}\\&=72\quad {\rm mJ}\end{align*}
(d) Without explicit calculation we could expect the energy stored in the capacitor must be increase since removing dielectric requires some work does on the capacitor. This work deposited into the energy of the capacitor. 

The capacitance of a cylindrical capacitor, comprising of a long cylindrical conductor with
radius $r_a$ and linear charge density $+\lambda $, and a coaxial cylindrical conducting shell with radius $r_b$ , $r_b>r_a$, and linear charge density $-\lambda $ is calculated as
\[C=\frac{2\pi {\epsilon }_0L}{{\mathrm{ln} r_b\ }-{\mathrm{ln} r_a\ }}=\frac{2\pi {\epsilon }_0L}{{\mathrm{ln} \left(\frac{r_b}{r_a}\right)\ }}\] 
Where $L$ is the length of the cylinder. As you can see, the capacitance is inversely proportional to the difference of logarithm of the inner and outer cylinder. Therefore by increasing the inner radius $r_a$, its logarithms is also increasing. As a result, difference between the two logarithms is decreasing and consequently the capacitance of the system is increased. 

 The correct answer is B.
 

The stored energy in a capacitor with any shape is found with one of the following formulas
\[U=\frac{1}{2}CV^2=\frac{Q^2}{2C}=\frac{1}{2}QV\] 
Where $C$ is the capacitance of the capacitor and $Q$ is the total charge loaded on its planes. So \begin{align*} U&=\frac{1}{2}CV^2\\ &=\frac{1}{2}\left(20\times {10}^{-6}\right){(200)}^2\\&=0.4\quad {\rm J}\end{align*}

The correct answer is C. 
 

In the section of Ohm's law questions, we were familiarized with the ohmic materials. The resistance in such materials is defined as the ratio of the potential difference across a conductor to the current in the conductor \[R=\frac{\Delta V}{I}\] That is resistance is the slope of the $V-I$ graph. Based upon experimental evidence, ohmic materials have a linear current-potential difference relationship over a broad range of applied potential differences. Therefore, 

The correct answer is B.