Electrostatic Potential And Capacitance Practice Questions

$\;\text{ Energy density }\;\;=\;\frac{1}{2} \epsilon _0 E^2$ $\;=\;\frac{1}{2} \epsilon _0 (\;\frac{V}{d}) ^2$ $\;=\;\frac{1}{2} (8.85 \times 10^{-12}) (\;\frac{20}{10^{-6}}) ^2 J / m ^3$ $\;≃ 1.8 \times 10^3 J / m ^3$

$\;\;\frac{1}{C}=\;\frac{1}{C_1}+\;\frac{1}{C_2}=\;\frac{d / 2}{A k_1 \epsilon _0}+\;\frac{d / 2}{A k_2 \epsilon _0}$ $\;= (\;\frac{1}{k_1}+\;\frac{1}{k_2}) \;\frac{d}{2 A \epsilon _0}$ $\;= (\;\frac{1}{3}+\;\frac{1}{5}) \;\frac{d}{2 A \epsilon _0}$ $\;\;\frac{1}{C}=\;\frac{4}{15} \;\frac{d}{A \epsilon _0}$ $\;C=\;\frac{15}{4} \times \;\frac{A \epsilon _0}{d}=\;\frac{15}{4} \times \;\frac{4 \times 10^{-4} \times 8.85 \times 10^{-12}}{1.77 \times 10^{-3}}$ $\;=7.5 {\text{pF}}$ $\;C+C_3=(7.5+7.5) {\text{pF}}=15 {\text{pF}}$

So to increase the capacitance by 10 factor $(\;\frac{A}{d})$ has to increase by 10 fator.For option (A) $\;\; C^{'}=\;\frac{(30 ℓ) b \epsilon _0}{ (\;\frac{d}{3}) }=30 C$For option (B) $\;\; C^{'}=\;\frac{ℓ b \epsilon _0}{10 d}=\;\frac{C}{10}$For option (C) $C^{'}=\;\frac{(2 ℓ) 5 b \epsilon _0}{d}=10 C$For option (D) $C^{'}=\;\frac{ (\frac{ℓ}{3}) b \epsilon _0}{ (\;\frac{10 d}{3}) }=\;\frac{c}{10}$For option (E) $\;\; C^{'}=\;\frac{ (\frac{ℓ}{3}) 5(2 b) \epsilon _0}{ (\;\frac{d}{3}) }=10 c$Clearly (C) and (E) are the situation for 10 C

$\;V_{10 Ω}=V_c=\;\frac{5}{25} \times 10=2 V$ $\;Q=8 \times 2=16 \mu C$

Horizontal component of velocity will remain same.$t\;=\;\frac{l}{v_x}=\;\frac{0.1}{10^6}$ $v_y\;=a_y t=\;\frac{e E}{m} t$ $\;=\;\frac{1.6 \times 10^{-19} \times 9.1 \times 10^2}{9.1 \times 10^{-31}} \times \;\frac{0.1}{10^6}$ $\;=1.6 \times 10^7$

For a capacitor at steady state$q=C V \;\text{ and }\; U=\;\frac{1}{2} C V^2$Since $C_1$ and $C_2$ are connected in parallel, $V_1=V_2$.Also from graph $q_1< q_2$ $\Rightarrow C_1 V_1< C_2 V_2$i.e. $C_1< C_2$ or $C_2>C_1$ $\;\frac{U_1}{U_2}=\;\frac{C_1 V_1^2}{C_2 V_2^2}=\;\frac{C_1}{C_2}< 1$or $U_1< U_2$ or $U_2>U_1$

$\;I=\;{V}/{R_{\;\text{eq }\;}} I=\;{V}/{R_{\;\text{eq }\;}}$ $=\;\frac{9}{1+\;\frac{12 \times 4}{12+4}}=\;\frac{9}{4}$ $\;I_1=\;\frac{9}{4} \times \;\frac{4}{16}=\;\frac{9}{16}$ $\;V_A-V_B=I_1 \times 8$ $=\;\frac{9}{16} \times 8=\;\frac{9}{2} V$ $\; \therefore U=\;\frac{1}{2} \times 4 \times \;\frac{81}{4} \mu J$ $\; \therefore U=\;\frac{81}{2} \mu J$ $\; \therefore x=81$

In steady state there will be no current in branch of capacitor, so no voltage drop across ${\text{R}}_2=5 Ω$ $\; {\text{I}}_2=0$ $\; {\text{I}}_1= {\text{I}}_3=\;\frac{10}{4+6}=1 {\text{A}}$ $\; {\text{V}}_{ {\text{R}}_3}= {\text{V}}_{ {\text{c}}}+ {\text{V}}_{ {\text{R}}_2} \;\; {\text{V}}_{ {\text{R}}_2}=0$ $\; {\text{I}}_3 {\text{R}}_3= {\text{V}}_{ {\text{c}}}$ $\; {\text{V}}_{ {\text{c}}}=1 \times 6=6 {\text{volt}}$ $\; {\text{q}}_{ {\text{c}}}= {\text{CV}}_{ {\text{c}}}=10 \times 6=60 \mu {\text{C}}$

At steady state, capacitor behaves as an open circuit and current flows in circuit as shown in the diagram. $\; {\text{R}}_{ {\text{cq}}}=9 Ω$ $\; {\text{i}}=\;{9 {\text{V}}}/{9 Ω}=1 {\text{A}}$ $\;\Delta {\text{V}}_{6 Ω}=1 \times 6=6 {\text{V}}$ $\; {\text{V}}_{ {\text{A}}}=3 {\text{V}}$So, potential difference across $6 \mu {\text{F}}$ is $6 {\text{V}}$.Hence$Q\;=C \Delta V$ $\;=6 \times 6 \times 10^{-6} {\text{C}}$ $\;=36 \mu {\text{C}}$

$\; {\text{C}}_{ {\text{eq}}}= {\text{C}}_1+ {\text{C}}_2$ $\; {\text{C}}_1=\;{2 E_0 {\text{A}}}/{2 \times {\text{d}}}=10 \mu {\text{F}}$ $\; {\text{C}}_2=\;{3 E_0 {\text{A}}}/{2 {\text{d}}}=15 \mu {\text{F}}$ $\; {\text{C}}_{ {\text{eq}}}=25 \mu {\text{F}}$Now the charge on$\; {\text{C}}_1=10 {\text{V}} \mu {\text{c}}$ $\; {\text{C}}_2=1.5 {\text{V}} \mu {\text{C}} .$Now force between the plates $[ {\text{F}}=\;{ {\text{Q}}^2}/{2 {\text{A}} \in _0}]$ $\;\;{100 {\text{V}}^2 \times 10^{-12}}/{2 \times 2 \times 10^{-4} E_0}+\;{225 {\text{V}}^2 \times 10^{-12}}/{2 \times 2 \times 10^{-4} \times E_0}=8$ $\;325 {\text{V}}^2=8 \times 4 \times 10^{-4} \times 8.85$ $\; {\text{V}}^2=\;\frac{32 \times 8.85 \times 10^{-4}}{325}$ $\; \therefore {\text{V}}=\sqrt{\;\frac{283.2 \times 10^{-4}}{325}}$ $\; {\text{V}}=0.93 \times 10^{-2}$

$\;V_A+\;\frac{10}{3}(1)-6(1)=V_B$ $\;V_A-V_B=6-\;\frac{10}{3}=\;\frac{8}{3} {\text{volt}}$ $\;Q=C (V_A-V_B)$ $\;=150 \times \;\frac{8}{3}=400 \mu {\text{C}}$

$\;\;{ {\text{A}} \epsilon _{ {\text{o}}}}/{ {\text{d}}}=\;{ {\text{A}} \epsilon _{ {\text{o}}}}/{ (0.2+\;{ {\text{d}}}/{ {\text{k}}}) }$ $\;0.6=0.2+\;{0.6}/{ {\text{k}}}$ $\; {\text{k}}=\;\frac{3}{2}$

Without dielectric${\text{Q}}=\;{ {\text{A}} \in _0}/{ {\text{d}}} {\text{V}}$with dielectric${\text{Q}}=\;{ {\text{A}} \in _0 {\text{V}}}/{ {\text{d}}- {\text{t}}+\;{ {\text{t}}}/{ {\text{K}}}}$given$\;\;{ {\text{A}} \in _0 {\text{V}}}/{ {\text{d}}- {\text{t}}+\;{ {\text{t}}}/{ {\text{K}}}}=(1.25) \;{ {\text{A}} E_0 {\text{V}}}/{ {\text{d}}}$ $\;\Rightarrow 1.25 (3+\;{2}/{ {\text{K}}}) =5$ $\;\Rightarrow {\text{K}}=2$

Before inserting dielectric capacitance is given ${\text{C}}_0=12.5 {\text{pF}}$ and charge on the capacitor ${\text{Q}}= {\text{C}}_0 {\text{V}}$ After inserting dielectric capacitance will become $E_{ {\text{r}}} {\text{C}}_0$.Change in potential energy of the capacitor$\;=E_{ {\text{i}}}- {\text{E}}_{ {\text{f}}}$ $\;=\;{ {\text{Q}}^2}/{2 {\text{C}}_{ {\text{i}}}}-\;{ {\text{Q}}^2}/{2 {\text{C}}_{ {\text{f}}}}=\;{ {\text{Q}}^2}/{2 {\text{C}}_0} [1-\;{1}/{E_{ {\text{r}}}}]$ $\;=\;{ ( {\text{C}}_0 {\text{V}}) ^2}/{2 {\text{C}}_0} [1-\;{1}/{E_{ {\text{r}}}}] =\;\frac{1}{2} {\text{C}}_0 {\text{V}}^2 [1-\;{1}/{E_{ {\text{r}}}}]$Using ${\text{C}}_0=12.5 {\text{pF}}, {\text{V}}=12 {\text{V}}, E_{ {\text{r}}}=6$ $\;=\;\frac{1}{2}(12.5) \times 12^2 [1-\;\frac{1}{6}] =\;\frac{1}{2}(12.5) \times 12^2 \times \;\frac{5}{6}$ $\;=750 {\text{pJ}}=750 \times 10^{-12} {\text{J}}$

In steady state $\;\;\text{ Req }\;=12 Ω$ $\; {\text{I}}=\;\frac{6}{12}=0.5 {\text{A}}$ $\;\;\text{ P.D across }\; {\text{C}}_1=3 {\text{V}}$ $\;\;\text{ P.D acoross }\; {\text{C}}_2=4 {\text{V}}$ $\; {\text{q}}_1= {\text{C}}_1 {\text{V}}_1=12 \mu {\text{C}}$ $\; {\text{q}}_2= {\text{C}}_2 {\text{V}}_2=24 \mu {\text{C}}$ $\;\;{ {\text{q}}_1}/{ {\text{q}}_2}=\;\frac{1}{2}$

In parallel combination : Potential difference is same across all$\;\;\text{ Energy }\;=\;\frac{1}{2} ( {\text{C}}_1+ {\text{C}}_2+ {\text{C}}_3) {\text{V}}^2$ $\;=\;\frac{1}{2}(25+30+45) \times (100)^2 \times 10^{-6}=0.5= {\text{E}}$In series combination: Charge is same on all.$\;\;{1}/{ {\text{C}}_{\;\text{equ }\;}}=\;{1}/{ {\text{C}}_1}+\;{1}/{ {\text{C}}_2}+\;{1}/{ {\text{C}}_3}=\;\frac{1}{25}+\;\frac{1}{30}+\;\frac{1}{45}$ $\;\;{1}/{ {\text{C}}_{\;\text{equ }\;}}=\;\frac{(18+15+10)}{450}=\;\frac{43}{450} \Rightarrow {\text{C}}_{\;\text{equ }\;}=\;\frac{450}{43}$ $\;\;\text{ Energy }\;=\;{ {\text{Q}}^2}/{2 {\text{C}}_1}+\;{ {\text{Q}}^2}/{2 {\text{C}}_2}+\;{ {\text{Q}}^2}/{2 {\text{C}}_3}$ $\;=\;{ {\text{Q}}^2}/{2} [\;{1}/{ {\text{C}}_1}+\;{1}/{ {\text{C}}_2}+\;{1}/{ {\text{C}}_3}]$ $\;\;{ ( {\text{V}} \times {\text{C}}_{\;\text{equ }\;}) ^2}/{2} \times \;{1}/{ {\text{C}}_{\;\text{equ }\;}}=\;{ {\text{V}}^2 {\text{C}}_{\;\text{equ }\;}}/{2}$ $\;\;\frac{(100)^2}{2} \times \;\frac{450}{43} \times 10^{-6}$ $\Rightarrow \;\frac{4.5}{86}=\;\frac{9}{x} E=\;\frac{9}{x} \times 0.5 \Rightarrow x=86$

$\; {\text{E}}=\;{ {\text{E}}_0}/{3} \Rightarrow {\text{V}}=\;{ {\text{V}}_0}/{3}$ $\;\;{ {\text{V}}_0}/{3}= {\text{V}}_0 {\text{e}}^{-\;{ {\text{t}}}/{\tau }}$ $\; {\text{t}}=\tau ℓ {\text{n}} 3$ $\;6.6 \times 10^{-6}= {\text{R}} (1.5 \times 10^{-6}) (1.1)$ $\; {\text{R}}=\;\frac{6}{1.5}=4 Ω$