Electrostatic Potential And Capacitance Practice Questions

$C_{e q}=\;\frac{\epsilon _0 A}{d-\;\frac{d}{2}+\;\frac{d}{2 k}}=\;\frac{\epsilon _0 A}{\frac{d}{2}}=\;\frac{2 \epsilon _0 A}{d}$If $C=\;\frac{\epsilon _0 A}{d}$ $\Rightarrow C_{e q}=2 C \;\text{ or }\; \;{C_{\;\text{new }\;}}/{C_{\;\text{old }\;}}=\;\frac{2}{1}$

${\text{q}}= {\text{C}} {\text{V}}_{100 Ω}$ $=(1.1 \times 10^{-6})(\;{10}/{ {\text{R}}+ {\text{r}}} {\text{R}})$ $=1.1 \times 10^{-6}(\;\frac{10}{110} \times 100)$ $=10 \mu {\text{C}}$

Equivalent $C=∑ C_i$ $=10 \mu {\text{F}}$ $\Rightarrow \;\text{ Charge }\; Q=C V$ $=200 \mu C$

Field inside the dielectric $=\;\frac{\sigma }{k \epsilon _0}$According to the given information,$\;\frac{\sigma }{k \epsilon _0}=3.6 \times 10^7$ $\Rightarrow \;\frac{\;\frac{Q}{A}}{k \epsilon _0}=3.6 \times 10^7$ $\Rightarrow k=2.33$

$U=\;\frac{1}{2} (k C_0) V^2$ $\Rightarrow \;\frac{U^{'}}{U}=1.5$ $\Rightarrow$ Energy increases by $50 \%$

The given situation can be shown as belowLet, $v_{1}$ and $v_{2}$ be the incoming and outgoing velocities of electron into the capacitor and out of the capacitor, respectively. Since, electric field is along $X$-axis, hence electric force on electron along $Y$-axis, $(F_{y})=0$ $\therefore$ Change in momentum along $Y$-axis, $\Delta p_{y}=0$ i.e. $\;\; p_{1}=p_{2}$ (along $Y$-axis) $\Rightarrow \;\; m_{1} v_{1} \cos \alpha =m_{2} v_{2} \cos \beta$ $\Rightarrow \;\; v_{1} / v_{2}=\cos \beta / \cos \alpha$ $\because$ Kinetic energy $(K)=1 / 2 {mv}^{2}$ If mass is same, $K \propto v^{2}$ $\therefore \;\; \;\frac{K_{1}}{K_{2}}=(\;\frac{v_{1}}{v_{2}})^{2}=(\;\frac{\cos \beta }{\cos \alpha })^{2}=\;\frac{\cos ^{2} \beta }{\cos ^{2} \alpha }$

${V}={V}_{0}(1-{e}^{-\frac{{t}}{{RC}}})$ $50=100(1-{e}^{-\frac{{t}}{{RC}}})$ ${t}=0.69 \\times 10^{-4} {sec}$

Given, The area of the parallel plate capacitor, $A=100 {m}^{2}$ The distance between the parallel plate of the capacitor, $d=10 {m}$ The dielectric constant of the material, $K=10$ The thickness of the space between the plates, $t=5 {m}$ As we know, the expression of the capacitor with dielectric constant $C=\;\frac{A \epsilon _{0}}{(d-t+\;\frac{t}{K})}$ $\Rightarrow \;\; C=\;\frac{100 \times 8.85 \times 10^{-12}}{(10-5+\;\frac{5}{10})}$ $\Rightarrow \;\; C=161 \times 10^{-12} {F}=161 {pF}$ Hence, the resultant capacitance of the system is $161 {pF}$. So, the value of $x$ to the nearest integer is 161 .

When a dielectric is inserted in a capacitor Due to free charge ${{E}}^{\to }={\vec{E}}_{0}$ only After dielectric ${E}'=\frac{{E}_{0}}{{k}}$ ${q}_{{B}}={q}_{{f}}(1-\frac{1}{{k}})$

The target answer suggests that $Q_2 = Q_3$. This implies that capacitors $C_2$ and $C_3$ are connected in series.

1. First, calculate the equivalent capacitance of $C_2$ and $C_3$ in series:
   $C_{23}' = \frac{C_2 C_3}{C_2 + C_3} = \frac{(6 \mu F)(12 \mu F)}{6 \mu F + 12 \mu F} = \frac{72}{18} \mu F = 4 \mu F$
2. In a series connection, the charge across each capacitor is the same, so $Q_2 = Q_3$. The charge on this equivalent series combination is $Q_{23}' = Q_2 = Q_3$.
3. The configuration must be that $C_1$ is in parallel with the series combination of $C_2$ and $C_3$. Let the voltage across this parallel combination be $V$.
4. For parallel components, the voltage across them is the same. So, the voltage across $C_1$ is $V_1 = V$, and the voltage across the series combination $C_{23}'$ is also $V_{23}' = V$.
5. Now, calculate the charges:
   $Q_1 = C_1 V_1 = (2 \mu F) V$
   $Q_2 = Q_3 = Q_{23}' = C_{23}' V_{23}' = (4 \mu F) V$
6. Form the ratio of the charges $Q_1 : Q_2 : Q_3$:
   $Q_1 : Q_2 : Q_3 = (2V) : (4V) : (4V) = 2 : 4 : 4 = 1 : 2 : 2$

The given figure can be shown with capacitors as Let $C_{0}$ be the capacitance of each capacitor. In the above figure capacitance $C_{1}$ is in series combination with equivalent of parallel combination of capacitance $C_{2}$ and $C_{3}$. $C_{e q}^{'}=C_{2}+C_{3} \Rightarrow C_{e q}^{'}=C_{0}+C_{0}=2 C_{0}$ $\Rightarrow \;\; \;{1}/{C_{ {eq}}}=\;\frac{1}{C_{1}}+\;{1}/{C_{ {eq}}^{'}}=\;\frac{1}{C_{0}}+\;\frac{1}{2 C_{0}}=\;\frac{3}{2 C_{0}}$ $\Rightarrow \;\; \;\; C_{\;\text{eq }\;}=\;\frac{2 C_{0}}{3}$....(i) As, $C_{0}=\;\frac{\epsilon _{0} A}{d}$...(ii) From Eqs. (i) and (ii), we get $C_{\;\text{eq }\;}=\;\frac{2}{3}(\;\frac{\epsilon _{0} A}{d}) \Rightarrow C_{\;\text{eq }\;}=\;\frac{2 \epsilon _{0}}{3 d}(I \times b)=\;\frac{2 \epsilon _{0}}{3 d}(2 \times \;\frac{3}{2})$ $\Rightarrow \;\; C_{\;\text{eq }\;}=\;\frac{2 \epsilon _{0}}{d}$...(iii) According to question, the equivalent capacitance between $A$ and $C$ is $\;\frac{x \epsilon _{0}}{d}$. So, comparing it with Eq. (iii), we get $x=2$

Let $V$ is the voltage across each capacitor, Now, after removing the battery the capacitor is connected. So, using the law of conservation of charge, $2 V+8 V=2 \times 10=20$ $\Rightarrow \;\; 10 {V}=20 \Rightarrow V=2 {V}$ As we know, $Q=C V=8 \times 2=16 \mu C$ Hence, the magnitude of the charge in $C_{2}$ on equilibrium condition is $16 \mu {C}$.

Given, area of dielectrics with dielectric constant $\text{K}_{1}$ and $\text{K}_{2}$, $\text{A}_{1}=\text{A}_{2}=\text{A} / 2$ Thickness of dielectrics with dielectric constant $\text{K}_{1}$ and $\text{K}_{2}$, $d_{1}=d_{2}=d / 2$ Other half of the capacitor is still filled with air, so distance betweenplates in other half of capacitor will be d but the area of plates for airsection will be $\text{A}/2$. Calculating the capacitance of section of capacitor filled with air. $\text{C}_{a}={\epsilon _{0}(\frac{\text{A}}{2})}{d}=\frac{\epsilon _{0} \text{A}}{2 d}$ Now, the capacitance of section of capacitor filled with dielectricconstant $\text{K}_1$ can be calculated as $\text{C}_{1}=\frac{\text{K}_{1} \epsilon _{0} \text{A}_{1}}{d_{1}}={\text{K}_{1} \epsilon _{0}(\frac{\text{A}}{2})}/{\frac{d}{2}}=\frac{\text{K}_{1} \epsilon _{0} \text{A}}{d}$ Similarly, the capacitance of section of capacitor filled withdielectric constant $\text{K}_2$ can be calculated as $\text{C}_{2}=\frac{\text{K}_{2} \epsilon _{0} \text{A}_{2}}{d_{2}}={\text{K}_{2} \epsilon _{0}(\frac{\text{A}}{2})}/{\frac{d}{2}}=\frac{\text{K}_{2} \epsilon _{0} \text{A}}{d}$ Now, the dielectric sections are in series with each other, so equivalent capacitance of combination can be calculated as ${1}/{\text{C}_{{\text{eq}}}}=\frac{1}{\text{C}_{1}}+\frac{1}{\text{C}_{2}}=\frac{d}{\text{K}_{1} \epsilon _{0} \text{A}}+\frac{d}{\text{K}_{2} \epsilon _{0} \text{A}}$ $\Rightarrow \text{C}_{{\text{eq} }}=\frac{\epsilon _{0} \text{A}}{d}[\frac{\text{K}_{1}\text{K}_{2}}{\text{K}_{1}+\text{K}_{2}}]$ Now, $\text{C}_{{\text{eq} }}$ is in parallel with the section of capacitor filled with air. Total capacitance of combination can be calculated as $\text{C}_{\text{T}}=\text{C}_{a}+\text{C}_{e q}=\frac{\epsilon _{0} \text{A}}{2 d}+\frac{\epsilon _{0} \text{A}}{d}[\frac{\text{K}_{1} \text{K}_{2}}{\text{K}_{1}+\text{K}_{2}}]$ $\Rightarrow \text{C}_{\text{T}}=\frac{\epsilon _{0} \text{A}}{d}[\frac{1}{2}+\frac{\text{K}_{1} \text{K}_{2}}{\text{K}_{1}+\text{K}_{2}}]$

$\frac{1}{C_{e f f}}=\frac{d}{K \epsilon _{0} A}+\frac{2 d}{3 K \epsilon _{0} A}+\frac{3 d}{5 K \epsilon _{0} A}$ $C_{\text{eff }}=\frac{15 K \epsilon _{0} A}{34 d}$

Given, capacitance of capacitor, $C=14 {pF}$ Potential difference, $V=12 {V}$ Energy of capacitor, $E_{i}=\;\frac{1}{2} C V^{2}$ $=\;\frac{1}{2} \times 14 \times 12 \times 12=1008 {pJ}$ Now, the charging battery is disconnected and a porcelain plate with $K=7$ is inserted between the plates $\therefore \;\; E_{f}=\;\frac{E_{i}}{7} \Rightarrow f=\;\frac{1008}{7} {pj} \Rightarrow f=144 {pj}$ Mechanical energy with which the plate would oscillate back and forth between the plates will be $=(1008-144) {pJ}=864 {pJ}$

The given situation is shown below Resistivity, ρ = 200 Ωm $\text{C}=2 {\text{pF}}=2 \times 10^{-12} {\text{F}}$ V = 40 V K or $\epsilon _r$ = 50 $\text{i}_{{\text{leakage} }}=?$ $\text{C}=\frac{\text{K} \epsilon _{0} \text{A}}{d}$(K is the dielectric constant or relative permittivity) and $\text{R}=\frac{\rho d}{A}$ Now, charge will be discharged through the resistance between theplates. Now, time constant (τ) of discharging, $\tau =\text{RC}=\frac{\rho d}{\text{A}} \times \frac{\text{K} \epsilon _{0} \text{A}}{d}$ $\tau = \rho \text{K}\epsilon _0$ For a given R-C circuit, the discharged current is given by $\text{i}=\frac{\text{Q}}{\text{R} \text{C}} {\text{e}}^{-\frac{t}{\text{RC}}}$ $\text{i}=\frac{\text{Q}}{\text{p} \text{K} \epsilon _{0}} e^{-\frac{t}{\text{p} \text{K} \epsilon _{0}}}$ The above discharge current is the leakage current, $\text{i}_{{\text{leakage} }}=\frac{\text{Q}}{\rho \text{K} \epsilon _{0}} e^{-\frac{t}{\rho \text{K} \epsilon _{0}}}$ Maximum leakage current, $(\text{i}_{0})_{{\text{leakage} }}=\frac{\text{Q}}{\rho \text{K} \epsilon _{0}}=\frac{\text{C} \text{V}}{\rho \text{K} \epsilon _{0}}$ $=\frac{2 \times 10^{-12} 40}{200 \times 50 \times 8.85 \times 10^{-12}}$ = 903 µA = 0.9 mA ∴ $(\text{i}_{0})_{{\text{leakage} }}=0.9 {\text{mA}}$

Taking an element of width ${dx}$ at a distance ${x}({x} < {d} / 2)$ from left plate ${dc}=\frac{(\epsilon _{0}+{kx}) {A}}{{dx}}$ Capacitance of half of the capacitor $\frac{1}{{C}}=\int _{0} ^{{d} / 2} \frac{1}{{dc}}=\frac{1}{{A}} ∫_{0} ^{{d} / 2} \frac{{dx}}{\epsilon _{0}+{kx}}$ $\frac{1}{{C}}=\frac{1}{{kA}} \ln (\frac{\epsilon _{0}+{kd} / 2}{\epsilon _{0}})$ Capacitance of second half will be same $C_{e q}=\frac{C}{2}={k A}/{2 \ln (\frac{2 \epsilon _{0}+k d}{2 \epsilon _{0}})}$

Given, the peak voltage of the battery, $\text{V}_0$ = 20V The voltage of the battery, V = 2V Time, $\text{t}=1 \\mu {\text{s}}=1 \times 10^{-6} {\text{s}}$ Resistance of the capacitor, R = 10 Ω As we know that, $\text{V}=\text{V}_{0}(1-{\text{e}}^{-\text{t} /\text{R} \text{C}})$ Substituting the values in the above equation, we get $2=20(1-{e}^{-\text{t} /\text{R} \text{C}})$ \Rightarrow $\frac{\text{t}}{\text{RC}}=\ln (\frac{10}{9})$ \Rightarrow $\text{C}=\frac{\text{t}}{\text{R} \ln (10 / 9)}=\frac{10^{-6}}{10 \times \ln (10 / 9)}$ $=0.95 \\mu {\text{F}}$ ∴ The capacitance of the capacitor is 0.95 μ F.