Electrostatic Potential And Capacitance Practice Questions

Given, $C_{1}=C_{2}=C$ When both capacitors are connected in series, their equivalent capacitance will be $\;\frac{1}{C_{s}}=\;\frac{1}{C}+\;\frac{1}{C}=\;\frac{2}{C}$ $\;C_{s}=\;\frac{C}{2}$ $\Rightarrow \;\; C_{s}=\;\frac{C}{2}$ When both capacitors are connected in parallel, their equivalent capacitance will be $C_{p}=C+C=2 C$ $\therefore$ The ratio of equivalent capacitance in series and parallel combination is $\;\frac{C_{s}}{C_{p}} \;\; =\;\frac{C / 2}{2 C}=\;\frac{1}{4}$ $\Rightarrow \;\; C_{s}: C_{p} \;\; =1: 4$

${V}_{{C}}=\frac{2 {CV}+{CV}}{{KC}+2 {C}}$ $=\frac{3 {V}}{{K}+2}$

${Q}=[\frac{{C}_{1} {C}_{2}}{{C}_{1}+{C}_{2}}][{V}_{1}+{V}_{2}]$ ${E}=\frac{{Q}}{{A} \in _{0}}=[\frac{{C}_{1} {C}_{2}}{{C}_{1}+{C}_{2}}] {[{V}_{1}+{V}_{2}]}/{{A} \in _{{o}}}$ ${C}_{1}={\in _{{o}} {A}}/{{d}-{t}} \Rightarrow{E}=\frac{{C}_{2}[{V}_{1}+{V}_{2}]}{({C}_{1}+{C}_{2})({d}-{t})}$ Now $\theta =\tan ^{-1}[\frac{{q} \.{E}}{{mg}}]$ $\theta =\tan ^{-1}[\frac{{q}}{{mg}} \\times \frac{{C}_{2}({V}_{2}+{V}_{1})}{({C}_{1}+{C}_{2})({d}-{t})}]$

The equivalent capacitance of the capacitor when dielectric material is partially filled, is given as $C=\;\frac{\epsilon _{0} A}{(d-t)+\;\frac{t}{k}}=\;{\epsilon _{0} A}/{(d-\;\frac{d}{2})+\;\frac{d {/} 2}{k}}=\;\frac{\epsilon _{0} A}{\frac{d}{2 k}+\;\frac{d}{2}}$ where, $\epsilon _{0}=$ absolute electrical permittivity of free space, $A=$ area of the plates of capacitor $=2 {m}^{2}$ $K=$ dielectric constant $=3.2$ $t=$ thickness of dielectric material $=\;\frac{d}{2}$ and $d=$ distance between the plates $=2 {m}$. $C=\;\frac{2 \epsilon _{0} A}{\frac{d}{k}+d}=\;\frac{2 \times 2 \epsilon _{0}}{\frac{1}{3.2}+1}=\;\frac{4 \times 3.2}{4.2} \epsilon _{0}$ $\Rightarrow \;\; C=3.04 \epsilon _{0}$ The required value after rounding off to the nearest integer is 3 .

Given, capacity of capacitor, C = 200µF EMF of battery = Potential difference across capacitor, V = 200 V Dielectric constant of slab, K = 2 $\text{U}_{1}=\frac{1}{2} \text{CV}^{2}=\frac{1}{2} \times 200 \times (200)^{2} \mu\text{J}=4 \text{J}$ New capacity with dielectric slab, C′ = KC = 200 × 2 = 400µF Since, the battery remains connected. Then, V′ = V = 200 V Now, electrostatic energy in capacitor with dielectric $\text{U}_{2}=\frac{1}{2} \text{C} '\text{V}'=\frac{1}{2} \times 400 \times (200)^{2} \mu\text{J}$ = 8 J Hence, change in the electrostatic energy, $\Delta\text{U}=\text{U}_{2}-\text{U}_{1}=8-4=4 {\text{J}}$ Thus, the correct answer is 4.

According to given circuit diagram, At $t=0$, the key is in closed position. Current through the resistor will be maximum. Using Ohm's law, $I_{\max }=\;\frac{V}{R} \Rightarrow I_{\max }=(\;\frac{Q}{C}) \times \;\frac{1}{R}$ $\Rightarrow \;\; I_{\max }=(\;\frac{30 \times 10^{-6}}{3 \times 10^{-6}}) \times \;\frac{1}{5 \times 10^{6}}$ $I_{\max }=2 \times 10^{-6} {A}$ $I_{\max }=2 \mu {A}$ The value of the current flowing through the $5 Ω$ resistor is $2 \mu {A}$. Hence, the value of the $x$ to the nearest integer is 2 .

The given circuit diagram is shown below The battery connected in the circuit is a DC source and capacitor does not allow direct current to pass through it. Therefore, there will no current pass through upper branch of circuit. The equivalent resistance of remaining circuit will be $\text{R}_{{\text{eq}}}=\text{R}_{{\in}}+4 \Omega$ Here, $\text{R}_{{\in}}$ is internal resistance of battery. $\text{R}_{{\text{eq}}}=1+5=5 \Omega$ From Ohm's law, we know that $\text{I}={\text{V}}/{\text{R}_{{\text{eq}}}}=\frac{5}{5}=1 {\text{A}}$ Now, voltage drop across branch AB is calculated as $\text{V}_{\text{AB}}=1 \times 4=4 {\text{V}}$ Voltage across upper branch will also be 4V as the connection is parallel. Now, charge flowing from upper branch is given as $\text{Q}=\text{C}_{{\text{eq}}} \text{V}_{\text{AB}}$ ...(i) Calculating the equivalent capacitance, ${1}/{\text{C}_{{\text{eq}}}}=\frac{1}{4}+\frac{1}{2+2}=\frac{1}{4}+\frac{1}{4}$ ${\text{C}_{{\text{eq}}}}=2$ Substituting the value of equivalent capacitance and voltage drop in Eq. (i), we get $\text{Q}=2 \times 4 {\text{V}}=8 \\mu {\text{C}}$ Therefore, the charge stored in 4µF will be 8µC.

${q}={CV}$ $[{C}]=[\frac{{q}}{{V}}]=\frac{({A} \\times {T})^{2}}{{ML}^{2} {T}^{-2}}$ $={M}^{-1} {L}^{-2} {T}^{4} {A}^{2}$ $[{E}]=[\frac{{F}}{{q}}]=\frac{{MLT}^{-2}}{{AT}}$ $={MLT}^{-3} {A}^{-1}$ ${F}={{q}_{1} {q}_{2}}/{4 \pi \in _{{o}} {r}^{2}}$ $[\in _{0}]={M}^{-1} {L}^{-3} {T}^{4} {A}^{2}$ Speed of light ${c}={1}/{\\sqrt{\mu_{0} \in _{0}}}$ $\mu_{0}=\frac{1}{\in _{0} {c}^{2}}$ [\mu_{0}]=\frac{1}{[{M}^{-1} {L}^{-3} {T}^{4} {A}^{2}]\$[{LT}^{-1}]^{2}}$$=[{M}^{1} {L}^{1} {T}^{-2} {A}^{-2}]$

$({*})$ Let, equivalent capacitance of two capacitors $C_{1}$ and $C_{2}$ connected in parallel be $C_{a}$ and equivalent capacitance of same, when connected in series be $C_{b}$. According to given data, $C_{ {a}}=\;\frac{15}{4} C_{b}$ . . . (i) Since, equivalent capacitance in parallel combination, $C_{ {eq}}=C_{1}+C_{2}$ $C_{ {a}}=C_{1}+C_{2}$ . . . (ii) and equivalent capacitance in series combination, $\;{1}/{C_{ {eq}}^{'}} \;\; =\;\frac{1}{C_{1}}+\;\frac{1}{C_{2}}$ $\;\frac{1}{C_{b}} \;\; =\;\frac{1}{C_{1}}+\;\frac{1}{C_{2}}=\;\frac{C_{2}+C_{1}}{C_{1} C_{2}}$ $C_{b} \;\; =\;\frac{C_{1} C_{2}}{C_{1}+C_{2}}$. . . (iii) Substituting Eqs. (ii) and (iii) in Eq. (i), we get $C_{1}+C_{2}=\;\frac{15}{4} \;\frac{C_{1} C_{2}}{C_{1}}+C_{2}$ $4(C_{1}+C_{2})^{2}=15 C_{1} C_{2}$ $\Rightarrow \;\; 4 C_{1}^{2}+4 C_{2}^{2}+8 C_{1} C_{2}=15 C_{1} C_{2}$ $\Rightarrow \;\; 4 C_{1}^{2}+4 C_{2}^{2}-7 C_{1} C_{2}=0$ On dividing both sides by $C_{1}^{2}$, we get $4+4(\;\frac{C_{2}}{C_{1}})^{2}-7 \;\frac{C_{2}}{C_{1}}=0$ $\;\text{ or }\; \;\; 4(\;\frac{C_{2}}{C_{1}})^{2}-7(\;\frac{C_{2}}{C_{1}})+4=0$ $\;\text{ If }\; \;\frac{C_{2}}{C_{1}}=x, \;\text{ then }\; 4 x^{2}-7 x+4=0$ By using the concept of quadratic equation, $\;x=\;{-b ± \sqrt{b^{2}-4 a c}}/{2 a} \Rightarrow x=\;\frac{7 ± \sqrt{49-64}}{8}$ $\Rightarrow \;\; x=\;\frac{C_{2}}{C_{1}}=\;\frac{7 ± \sqrt{-15}}{8}=\;\frac{7 ± \sqrt{15} i}{8}$

According to given circuit diagram, Capacitance $=50 \\mu {\text{F}}=50 \times 10^{-6} {\text{F}}$ Supply voltage, V = 6 VIn steady state, capacitor will act as open circuit, ∴ Equivalent resistance $\text{R}_{{\text{eq}}}=(2+2+2) {\text{k}} \Omega=6 {\text{k}} \Omega$ Circuit current, $\text{I}={\text{V}}/{\text{R}_{{\text{eq} }}}=\frac{6}{6 \times 1000}=10^{-3} {\text{A}}$ ∴ Voltage across $2 {k} \Omega=1 \times 2=10^{-3} \times 2 \times 10^{3}=2 {\text{V}}$ Now, charge on capacitor, $q=\text{CV}=50 \times 10^{-6} \times 2$ $=100 \times 10^{-6} {\text{C}}$ $=100 \\mu {\text{C}}$

Let potential of point ${O}\;\;\;\; {v}_{0}=0$ Now, using junction analysis We can say, $q_{1}+q_{2}+q_{3}=0$ $2(x-6)+4(x-6)+5(x)=0$ ${x}= \frac{36}{11} \;\;\;{q}_{3}= \frac{36(5)}{11}= \frac{180}{11}$ ${q}_{3}=16.36 \mu{C}$

◆ $\ {R}_{1}$ to $\ {R}_{5} \to$ each $2 \Omega$ ◆ Cap. is fully charged ◆ So no current is there in branch ADB ◆ Effective circuit of current flow : ${R}_{ {eq}}=(\frac{4 \\times 2}{4+2} )+2$ ${R}_{ {eq}}=\frac{4}{3}+2= \frac{10}{3} \Omega$ ${i}= \frac{10}{10 / 3}=3 \ {A}$ So potential different across AEB $\Rightarrow 2 \\times 1+2 \\times 3=8 \ {V}$ Hence potential difference across Capacitor $=\Delta {V}= {V}_{{AEB}}=8 {V}$

As K is variable we take a plate element of Area A and thickness dx at distance x Capacitance of element $dC=\frac{(A)K(1+\alpha x)\epsilon _0}{dx}$Now all such elements are is series so equivalent capacitance $1/C=∫1/{dC}=∫ ^{d}_{0}\frac{dx}{AK\epsilon _0(1+\alpha x)}$ $=1/{\alpha AK\epsilon _0}\ln({1+\alpha d}/1)$ $=1/{\alpha AK\epsilon _0}(\alpha d-(\alpha d)^2/2+(\alpha d)^3/3+...)$ $= \frac{\alpha d}{\alpha AK\epsilon _0}(1-{\alpha d}/2+(\alpha d)^2/3+..)$ $1/C= \frac{ d}{ AK\epsilon _0}(1-{\alpha d}/2 )$ $C={AK\epsilon _0}/d (1+{\alpha d}/2)$

${\text{V}}_{0}({t})={\text{V}}_{{\in}}(1-{e}^{-{{t}}/{{\text{RC}}}})$ at ${t}=5 \mu{s}$ ${\text{V}}_{0}({t})=5(1-{e}^{-\frac{5 \times 10^{-6}}{10^{3} \times 10 \times 10^{-9}}})$ $=5(1-{e}^{-0.5})=2 {\text{V}}$ Now ${\text{V}}_{{\in }}=0$ means discharging ${\text{V}}_{0}({t})=2 {e}^{-{t}/{{\text{RC}}}}=2 {e}^{-0.5}$ $=1.21 {\text{V}}$ ${\text{V}}_{0}({t})=5-3.79 {e}^{-{1}/{{\text{RC}}}}$ ${\text{V}}_{0}({t})=2.79 {\text{Volt}} \approx 3 {\text{V}}$ Most approperiate Ans. (1)

${u}_{ {i}}= \frac{1}{2} \\times 5 \\times 10^{-6}(220)^{2}$ Final common potential ${v}= \frac{220 \\times 5+0 \\times 2.5}{5+2.5}$ $=220 \\times \frac{2}{3}$ ${u}_{ {f}}= \frac{1}{2}(5+2.5) \\times 10^{-6} (220 \\times \frac{2}{3} )^{2}$ $\Delta{u}= {u}_{ {f}}- {u}_ { {i}}$ $\Deltau=-403.33 \\times 10^{-4}$ $\Rightarrow -403.33 \\times 10^{-4}=\frac{{X}}{100}$ $X=-4.03$ or magnitude or value of $\ {X}$ is approximate 4

Before inserting slab ${C}_{ {i}}= \frac{\epsilon _{0} {A}}{ {d}}$ ${C}_{ {i}}= \frac{\epsilon _{0} l{w}}{ {d}}$ After inserting dielectric slab ${C}_{ {f}}= {C}_{1}+ {C}_{2}$ ${C}_{ {f}}=\frac{{K} \epsilon _{0} { A}_{1}}{ {d}}+ \frac{\epsilon _{0} { A}_{2}}{ {d}}$ ${C}_{ {f}}=\frac{{K} \epsilon _{0} { wx}}{ {d}}+ \frac{\epsilon _{0} {w}(l- {x})}{ {d}}$ ${C}_{ {f}}=2 {C}_{ {i}} \Rightarrow \frac{ {K} \epsilon _{0} {wx}}{{x}}+\ \frac{\epsilon _{0} {w}(l - {x})}{ {d}}$ $= \frac{2 \epsilon _{0}l{w}}{ {d}}$ $4 x+l-x=2l$ $[x= \frac{l}{3}]$

$\Rightarrow$ By conservation of charge ${q}_{ {i}}= {q}_{ {f}}$ ${Q}_{1}+ {Q}_{2}={q}_{1}+ {q}_{2}$ $4 {CV}-{CV}=( {C}+2 {C}) {V}_{ {C}}$ ${V}_{ {C}}= \frac{3 {CV}}{3 {C}} \Rightarrow {V}$ $\Rightarrow \frac{1}{2} \times (3 {C}) \times {V}_{{C}}^{2}$ $= \frac{1}{2}\times 3 {C} \times {V}^{2}= \frac{3}{2} {CV}^{2}$

${q}_{3}=20 \\times 8=160 \mu{C}$ $\therefore {q}_{2}=750-160$ $=590 \mu{C}$

$C_1 + C _2 = 10$ $1/2C_2V^2=4\times 1/2C_1V^2$ $\therefore C_2=4C_1$ $\therefore C_1=2$ & $C_2=8$For series combination$E_\text{eq}=\frac{C_1C_2}{C_1+C_2}=1.6$