Sharing of Charge in Capacitor Circuit Questions in English

(B) The energy stored in a capacitor is given by $U = \frac{1}{2} \frac{Q^2}{C}$.
Initially,the capacitor is charged by a $100 \ V$ battery. The charge $Q$ on the capacitor is $Q = CV = 900 \times 10^{-12} \ F \times 100 \ V = 9 \times 10^{-8} \ C$.
The initial energy is $U_i = \frac{1}{2} \frac{Q^2}{C} = 4.5 \times 10^{-6} \ J$.
When the charged capacitor is connected to an identical uncharged capacitor of $900 \ pF$ in parallel,the total charge $Q$ is redistributed between them.
The equivalent capacitance of the system becomes $C' = C + C = 2C = 1800 \ pF$.
Since the charge $Q$ is conserved,the final energy of the system is $U_f = \frac{1}{2} \frac{Q^2}{C'} = \frac{1}{2} \frac{Q^2}{2C} = \frac{1}{2} U_i$.
Therefore,$U_f = \frac{4.5 \times 10^{-6} \ J}{2} = 2.25 \times 10^{-6} \ J$.

(C) Initial capacitance $C_1 = 0.2 \, \mu F$ and initial potential $V_1 = 600 \, V$.
Initial charge on the capacitor $Q = C_1 V_1 = 0.2 \, \mu F \times 600 \, V = 120 \, \mu C$.
When this capacitor is connected in parallel to an uncharged capacitor $C_2 = 1.0 \, \mu F$,the total charge $Q$ remains conserved.
The equivalent capacitance of the parallel combination is $C_{eq} = C_1 + C_2 = 0.2 \, \mu F + 1.0 \, \mu F = 1.2 \, \mu F$.
The final potential $V_f$ is given by $V_f = \frac{Q}{C_{eq}}$.
$V_f = \frac{120 \, \mu C}{1.2 \, \mu F} = 100 \, V$.

(D) Total charge $Q = Q_1 + Q_2 = 80 \ \mu C + 40 \ \mu C = 120 \ \mu C$.
When connected by a conducting wire,the spheres reach a common potential $V$.
The new charges $Q_1'$ and $Q_2'$ on the spheres of radii $r_1 = 4 \ cm$ and $r_2 = 6 \ cm$ are given by:
$Q_1' = Q \left( \frac{r_1}{r_1 + r_2} \right) = 120 \left( \frac{4}{4 + 6} \right) = 120 \times 0.4 = 48 \ \mu C$.
The initial charge on the $4 \ cm$ sphere was $80 \ \mu C$.
The charge that flows from the $4 \ cm$ sphere to the $6 \ cm$ sphere is $\Delta Q = Q_1 - Q_1' = 80 \ \mu C - 48 \ \mu C = 32 \ \mu C$.

(C) Given: $C_1 = 2\ \mu F = 2 \times 10^{-6}\ F$, $V_1 = 300\ V$, $C_2 = 3\ \mu F = 3 \times 10^{-6}\ F$, $V_2 = 500\ V$.
The formula for energy lost when two capacitors are connected in parallel is given by:
$\Delta U = \frac{C_1 C_2}{2(C_1 + C_2)} (V_1 - V_2)^2$
Substituting the values:
$\Delta U = \frac{(2 \times 10^{-6}) \times (3 \times 10^{-6})}{2(2 \times 10^{-6} + 3 \times 10^{-6})} (300 - 500)^2$
$\Delta U = \frac{6 \times 10^{-12}}{2(5 \times 10^{-6})} (40000)$
$\Delta U = \frac{6 \times 10^{-12}}{10 \times 10^{-6}} (40000)$
$\Delta U = 0.6 \times 10^{-6} \times 40000 = 0.6 \times 0.04 = 0.024\ J$
Note: Re-calculating, $\Delta U = \frac{6 \times 10^{-12} \times 4 \times 10^4}{10 \times 10^{-6}} = \frac{24 \times 10^{-8}}{10^{-5}} = 2.4 \times 10^{-2} = 0.024\ J$.
Given the options provided, the closest value is $0.0375\ J$ (likely due to a typo in the question's source values), but mathematically the result is $0.024\ J$.

(B) When two capacitors are connected in parallel,the charge is conserved,and the common potential $V$ is given by the formula:
$V = \frac{C_1V_1 + C_2V_2}{C_1 + C_2}$
Given:
$C_1 = 20\,\mu F, V_1 = 500\,V$
$C_2 = 10\,\mu F, V_2 = 200\,V$
Substituting the values:
$V = \frac{(20 \times 500) + (10 \times 200)}{20 + 10}$
$V = \frac{10000 + 2000}{30}$
$V = \frac{12000}{30} = 400\,V$
Thus,the common potential is $400\,V$.

(B) Initial charges on the capacitors are:
$Q_1 = C_1 V = 1\,\mu F \times 1200\,V = 1200\,\mu C$
$Q_2 = C_2 V = 2\,\mu F \times 1200\,V = 2400\,\mu C$
When connected with opposite polarities,the net charge available to redistribute is $Q_{net} = |Q_2 - Q_1| = |2400 - 1200| = 1200\,\mu C$.
The common potential $V'$ after redistribution is given by:
$V' = \frac{Q_{net}}{C_1 + C_2} = \frac{1200\,\mu C}{1\,\mu F + 2\,\mu F} = \frac{1200}{3} = 400\,V$.
The new charges on the capacitors are:
$Q_1' = C_1 V' = 1\,\mu F \times 400\,V = 400\,\mu C$
$Q_2' = C_2 V' = 2\,\mu F \times 400\,V = 800\,\mu C$.

(C) When two metallic spheres are connected by a conducting wire,charge flows until both spheres reach a common potential $V$.
The total charge $Q_{total} = q_1 + q_2 = -1 \times 10^{-2} + 5 \times 10^{-2} = 4 \times 10^{-2} \, C$.
The total capacitance $C_{total} = C_1 + C_2 = 4 \pi \varepsilon_0 R_1 + 4 \pi \varepsilon_0 R_2 = 4 \pi \varepsilon_0 (R_1 + R_2)$.
The common potential $V = \frac{Q_{total}}{C_{total}} = \frac{4 \times 10^{-2}}{4 \pi \varepsilon_0 (1 \times 10^{-2} + 3 \times 10^{-2})} = \frac{4 \times 10^{-2}}{4 \pi \varepsilon_0 (4 \times 10^{-2})} = \frac{1}{4 \pi \varepsilon_0}$.
The final charge on the bigger sphere $(R_2 = 3 \, cm)$ is $q_2' = C_2 V = (4 \pi \varepsilon_0 R_2) \times V$.
$q_2' = (4 \pi \varepsilon_0 \times 3 \times 10^{-2}) \times \frac{1}{4 \pi \varepsilon_0} = 3 \times 10^{-2} \, C$.

(C) Initially,the energy stored in the $2\,\mu F$ capacitor is:
$U_{i} = \frac{1}{2} C V^{2} = \frac{1}{2} (2 \times 10^{-6}) V^{2} = V^{2} \times 10^{-6} \, J$
Initially,the charge stored in the $2\,\mu F$ capacitor is:
$Q_{i} = C V = (2 \times 10^{-6}) V = 2V \times 10^{-6} \, C$
When the switch $S$ is turned to position $2,$ the charge flows and both capacitors share charges until a common potential $V_{c}$ is reached.
$V_{c} = \frac{\text{Total Charge}}{\text{Total Capacitance}} = \frac{2V \times 10^{-6}}{(2 + 8) \times 10^{-6}} = \frac{V}{5} \, V$
Finally,the energy stored in both capacitors is:
$U_{f} = \frac{1}{2} (C_{1} + C_{2}) V_{c}^{2} = \frac{1}{2} (10 \times 10^{-6}) \left(\frac{V}{5}\right)^{2} = 5 \times 10^{-6} \times \frac{V^{2}}{25} = \frac{V^{2}}{5} \times 10^{-6} \, J$
Percentage loss of energy is given by:
$\Delta U \% = \frac{U_{i} - U_{f}}{U_{i}} \times 100 \%$
$\Delta U \% = \frac{V^{2} \times 10^{-6} - \frac{V^{2}}{5} \times 10^{-6}}{V^{2} \times 10^{-6}} \times 100 \% = \left(1 - \frac{1}{5}\right) \times 100 \% = \frac{4}{5} \times 100 \% = 80 \%$

(B) Let the initial capacitance be $C$ and the potential difference provided by the battery be $V$. The initial charge on the capacitor is $q = CV$.
The initial electrostatic energy stored in the capacitor is $U_i = \frac{1}{2} CV^2$.
When the battery is removed and an identical uncharged capacitor of capacitance $C$ is connected in parallel,the total charge $q$ is shared between the two capacitors. Since the capacitors are identical,the charge on each capacitor becomes $q' = q/2 = CV/2$.
The common potential difference across the parallel combination is $V_c = \frac{q}{C_{eq}} = \frac{CV}{C + C} = \frac{V}{2}$.
The final electrostatic energy of the system is the sum of the energy stored in both capacitors:
$U_f = \frac{1}{2} C V_c^2 + \frac{1}{2} C V_c^2 = C V_c^2$.
Substituting $V_c = V/2$:
$U_f = C \left(\frac{V}{2}\right)^2 = C \left(\frac{V^2}{4}\right) = \frac{1}{4} CV^2$.
Comparing the final energy with the initial energy:
$U_f = \frac{1}{2} \left(\frac{1}{2} CV^2\right) = \frac{1}{2} U_i$.
Thus,the total electrostatic energy of the resulting system decreases by a factor of $2$.

(A) When body $1$ (charge $Q$,capacitance $C$) is touched with body $2$ (charge $0$,capacitance $2C$),the total charge $Q$ is shared in the ratio of their capacitances. The new charge on body $1$ is $Q_1' = Q \times \frac{C}{C + 2C} = \frac{Q}{3}$.
After touching body $2$ with body $3$ (capacitance $C_3 \rightarrow \infty$),body $2$ becomes uncharged because all its charge flows to the infinite reservoir of body $3$.
Thus,after one complete operation,the charge on body $1$ becomes $Q_1 = \frac{Q}{3}$.
After the second operation,the charge on body $1$ becomes $Q_2 = \frac{Q_1}{3} = \frac{Q}{3^2}$.
Repeating this process $N$ times,the charge on body $1$ after $N$ operations will be $Q_N = \frac{Q}{3^N}$.

(B) Initial charges on the capacitors are:
$q_1 = C \times V = CV$
$q_2 = 2C \times 2V = 4CV$
When connected with opposite polarities,the net charge on the system is:
$Q_{net} = |q_2 - q_1| = |4CV - CV| = 3CV$
The equivalent capacitance of the capacitors in parallel is:
$C_{eq} = C + 2C = 3C$
The common potential $V'$ across the combination is:
$V' = \frac{Q_{net}}{C_{eq}} = \frac{3CV}{3C} = V$
The final energy stored in the configuration is:
$U = \frac{1}{2} C_{eq} (V')^2 = \frac{1}{2} (3C) (V)^2 = \frac{3}{2} CV^2$

(B) Initial charges on the capacitors are:
$Q_1 = C_1 V_1 = 2 \mu F \times 10 \ V = 20 \mu C$
$Q_2 = C_2 V_2 = 4 \mu F \times 20 \ V = 80 \mu C$
Initial energy stored in the capacitors is:
$U_i = \frac{1}{2} C_1 V_1^2 + \frac{1}{2} C_2 V_2^2 = \frac{1}{2} (2 \times 10^{-6}) (10)^2 + \frac{1}{2} (4 \times 10^{-6}) (20)^2 = 100 \mu J + 800 \mu J = 900 \mu J$
Since the plates are connected with opposite polarity,the net charge is:
$Q_{net} = |Q_2 - Q_1| = |80 \mu C - 20 \mu C| = 60 \mu C$
The common potential $V$ after connection is:
$V = \frac{Q_{net}}{C_1 + C_2} = \frac{60 \mu C}{2 \mu F + 4 \mu F} = \frac{60}{6} V = 10 \ V$
Final energy stored in the capacitors is:
$U_f = \frac{1}{2} (C_1 + C_2) V^2 = \frac{1}{2} (6 \mu F) (10 \ V)^2 = 300 \mu J$
Heat evolved in the circuit is the loss in energy:
$H = U_i - U_f = 900 \mu J - 300 \mu J = 600 \mu J$

(C) When two capacitors are connected such that their plates with opposite polarities are joined together,the total charge on the connected plates must be zero for the final potential to be zero.
The initial charges on the capacitors are $q_1 = C_1 V_1 = 120 C_1$ and $q_2 = C_2 V_2 = 200 C_2$.
For the final potential to be zero,the net charge must be zero,which implies that the magnitudes of the charges must be equal:
$120 C_1 = 200 C_2$
Dividing both sides by $40$,we get:
$3 C_1 = 5 C_2$

(A) Initially,the capacitor $C$ connected across $AB$ is charged to $Q$. The initial energy stored is $U_i = \frac{Q^2}{2C}$.
When the switch is shifted to position $2$,the capacitor $C$ is connected in parallel with two other capacitors,each of capacitance $C$. The total equivalent capacitance of the circuit becomes $C_{eq} = C + C + C = 3C$.
The charge $Q$ is now redistributed among these three capacitors. Since they are in parallel,the final potential difference across each is the same. The final energy stored in the system is $U_f = \frac{Q^2}{2C_{eq}} = \frac{Q^2}{2(3C)} = \frac{Q^2}{6C}$.
The heat released is the difference between the initial and final energy: $\text{Heat} = U_i - U_f = \frac{Q^2}{2C} - \frac{Q^2}{6C} = \frac{3Q^2 - Q^2}{6C} = \frac{2Q^2}{6C} = \frac{Q^2}{3C}$.

(B) The electric potential $V$ of a conductor is given by the formula $V = \frac{Q}{C}$,where $Q$ is the charge and $C$ is the capacitance.
For conductor $A$:
$V_A = \frac{Q_A}{C_A} = \frac{30 \, \mu\text{C}}{10 \, \mu\text{F}} = 3 \, \text{V}$
For conductor $B$:
$V_B = \frac{Q_B}{C_B} = \frac{20 \, \mu\text{C}}{5 \, \mu\text{F}} = 4 \, \text{V}$
Since positive charge always flows from a body at a higher potential to a body at a lower potential,and here $V_B > V_A$ $(4 \, \text{V} > 3 \, \text{V})$,the positive charge will flow from $B$ to $A$.

(C) When two capacitors are connected in parallel such that their similarly charged plates are joined together,the total charge is conserved and the capacitors attain a common potential $V$.
The total charge $Q_{total} = Q_1 + Q_2 = C_1 V_1 + C_2 V_2$.
The total capacitance $C_{total} = C_1 + C_2$.
The common potential $V$ is given by:
$V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$
Given $C_1 = 20 \, \mu F$,$V_1 = 30 \, V$,$C_2 = 30 \, \mu F$,$V_2 = 20 \, V$.
$V = \frac{(20 \times 30) + (30 \times 20)}{20 + 30} \, V$
$V = \frac{600 + 600}{50} \, V$
$V = \frac{1200}{50} \, V = 24 \, V$.

(B) Initial charges on the capacitors are:
$Q_1 = C_1 V_1 = 2\,\mu F \times 10\,V = 20\,\mu C$
$Q_2 = C_2 V_2 = 4\,\mu F \times 20\,V = 80\,\mu C$
Since the positive plate of one is connected to the negative plate of the other,the net charge is $Q_{net} = |Q_2 - Q_1| = |80 - 20| = 60\,\mu C$.
The common potential $V$ after connection is $V = \frac{Q_{net}}{C_1 + C_2} = \frac{60\,\mu C}{2\,\mu F + 4\,\mu F} = \frac{60}{6} = 10\,V$.
Initial energy $U_i = \frac{1}{2} C_1 V_1^2 + \frac{1}{2} C_2 V_2^2 = \frac{1}{2}(2)(10)^2 + \frac{1}{2}(4)(20)^2 = 100 + 800 = 900\,\mu J$.
Final energy $U_f = \frac{1}{2} (C_1 + C_2) V^2 = \frac{1}{2}(2 + 4)(10)^2 = \frac{1}{2}(6)(100) = 300\,\mu J$.
Heat evolved $H = U_i - U_f = 900\,\mu J - 300\,\mu J = 600\,\mu J$.

(B) The charge on the first capacitor is $Q_1 = C \times V = CV$.
The charge on the second capacitor is $Q_2 = 2C \times 2V = 4CV$.
Since they are connected with opposite polarity,the net charge is $Q_{net} = |Q_2 - Q_1| = |4CV - CV| = 3CV$.
The total capacitance is $C_{eq} = C + 2C = 3C$.
After the parallel connection,the common potential difference is $V' = \frac{Q_{net}}{C_{eq}} = \frac{3CV}{3C} = V$.
The final energy of the configuration is $U_f = \frac{1}{2} C_{eq} (V')^2 = \frac{1}{2} (3C) V^2 = \frac{3}{2} CV^2$.

(C) The capacitor $C$ is initially charged to a potential difference $V_0$. The initial charge on the capacitor is $Q = CV_0$.
When this capacitor is connected to an uncharged capacitor of capacitance $C_x$,the total charge $Q$ is conserved and distributed between the two capacitors.
Since they are connected in parallel,they will share a common potential difference $V$.
The total charge remains $Q = (C + C_x)V$.
Equating the initial and final charges: $CV_0 = (C + C_x)V$.
Rearranging for $C_x$: $C + C_x = \frac{CV_0}{V}$.
Therefore,$C_x = C\left( \frac{V_0}{V} - 1 \right)$.

(B) When the switch $S$ is open,the two capacitors $C$ (labeled $A$ and $B$) are in series with the battery $E$. The equivalent capacitance is $C_{eq} = \frac{C \times C}{C + C} = \frac{C}{2}$.
The initial charge supplied by the battery is $Q_1 = C_{eq}E = \frac{CE}{2}$.
When the switch $S$ is closed,the capacitor $B$ is short-circuited. Only capacitor $A$ remains connected across the battery $E$.
The final charge supplied by the battery is $Q_2 = CE$.
The charge that flows through the battery is the change in charge: $\Delta Q = Q_2 - Q_1 = CE - \frac{CE}{2} = \frac{CE}{2}$.

(D) The potential of a conducting sphere is given by $V = \frac{Q}{4 \pi \varepsilon_0 R}$.
When two spheres are brought into contact,charge flows until their potentials become equal.
The initial energy of the system is $U_i = \frac{1}{2} \frac{Q_1^2}{C_1} + \frac{1}{2} \frac{Q_2^2}{C_2}$,where $C_1 = 4 \pi \varepsilon_0 R_1$ and $C_2 = 4 \pi \varepsilon_0 R_2$.
The final energy of the system is $U_f = \frac{1}{2} (C_1 + C_2) V^2$,where $V = \frac{Q_1 + Q_2}{C_1 + C_2}$.
The loss in energy is given by $\Delta U = U_i - U_f = \frac{C_1 C_2}{2(C_1 + C_2)} (V_1 - V_2)^2$.
Since $(V_1 - V_2)^2$ is always positive,$\Delta U > 0$ if $V_1 \neq V_2$.
$V_1 = V_2$ implies $\frac{Q_1}{R_1} = \frac{Q_2}{R_2}$,which means $Q_1 R_2 = Q_2 R_1$.
Therefore,if $Q_1 R_2 \neq Q_2 R_1$,there is a decrease in the energy of the system.

(C) The energy loss $\Delta U$ when two capacitors with capacitances $C_1$ and $C_2$ at potentials $V_1$ and $V_2$ are connected is given by the formula:
$\Delta U = \frac{1}{2} \frac{C_1 C_2 (V_1 - V_2)^2}{C_1 + C_2}$
Given: $C_1 = 3\,\mu F = 3 \times 10^{-6}\,F$,$V_1 = 300\,V$,$C_2 = 5\,\mu F = 5 \times 10^{-6}\,F$,$V_2 = 500\,V$.
Substituting the values:
$\Delta U = \frac{1}{2} \times \frac{(3 \times 10^{-6}) \times (5 \times 10^{-6}) \times (500 - 300)^2}{(3 \times 10^{-6} + 5 \times 10^{-6})}$
$\Delta U = \frac{1}{2} \times \frac{15 \times 10^{-12} \times (200)^2}{8 \times 10^{-6}}$
$\Delta U = \frac{15 \times 10^{-12} \times 40000}{16 \times 10^{-6}}$
$\Delta U = \frac{600000 \times 10^{-12}}{16 \times 10^{-6}}$
$\Delta U = \frac{6 \times 10^5 \times 10^{-12}}{16 \times 10^{-6}} = \frac{6}{16} \times 10^{-1} = 0.375 \times 10^{-1} = 0.0375\,J$.

(C) Charge flows from higher potential to lower potential. The potential of a capacitor is given by $V = \frac{Q}{C}$.
For circuit $(a)$:
Potential of left capacitor $V_L = \frac{2Q}{3C} = \frac{2}{3} \frac{Q}{C}$.
Potential of right capacitor $V_R = \frac{Q}{C}$.
Since $V_R > V_L$, charge flows from $R$ to $L$.
For circuit $(b)$:
Potential of left capacitor $V_L = \frac{2Q}{2C} = \frac{Q}{C}$.
Potential of right capacitor $V_R = \frac{Q}{2C} = 0.5 \frac{Q}{C}$.
Since $V_L > V_R$, charge flows from $L$ to $R$.

(C) Initial charge on $C_1$ when $S_1$ is closed is given by:
$q = C_1 V = 6\,\mu F \times 20\,V = 120\,\mu C$.
When $S_1$ is opened,the charge $q = 120\,\mu C$ remains stored on $C_1$.
When $S_2$ is closed,the charge is shared between $C_1$ and $C_2$ connected in parallel. Let the common potential be $V'$.
The total charge is conserved: $q = (C_1 + C_2) V'$.
$120\,\mu C = (6\,\mu F + 3\,\mu F) V' = 9\,\mu F \times V'$.
$V' = \frac{120}{9} = \frac{40}{3}\,V$.
The final charge on $C_2$ is:
$q_2 = C_2 V' = 3\,\mu F \times \frac{40}{3}\,V = 40\,\mu C$.

(B) Initially,the capacitor of capacitance $C$ is charged to a potential $V$ by a battery. The energy stored is $U = \frac{1}{2} CV^2 = \frac{Q^2}{2C}$,where $Q$ is the initial charge.
When the battery is disconnected and an identical capacitor $C$ is connected in parallel,the total charge $Q$ is redistributed between the two capacitors.
Since the capacitors are in parallel,they will share the charge equally,so each capacitor now has a charge $Q' = \frac{Q}{2}$.
The new potential difference across each capacitor is $V' = \frac{Q'}{C} = \frac{Q}{2C} = \frac{V}{2}$.
The energy stored in each capacitor is $U' = \frac{1}{2} C(V')^2 = \frac{1}{2} C \left(\frac{V}{2}\right)^2 = \frac{1}{2} C \frac{V^2}{4} = \frac{1}{4} \left(\frac{1}{2} CV^2\right) = \frac{U}{4}$.

(A) Initial charge on the capacitor $Q = C_1 V_1 = 0.2 \, F \times 600 \, V = 120 \, C$.
When the battery is removed and the capacitor is connected to another capacitor of $C_2 = 1 \, F$,the total charge $Q$ is conserved and distributed across both capacitors.
The new potential $V'$ is given by $V' = \frac{Q}{C_1 + C_2}$.
Substituting the values: $V' = \frac{120 \, C}{0.2 \, F + 1 \, F} = \frac{120}{1.2} \, V$.
$V' = 100 \, V$.

(C) The initial charge $q$ on the $10\,\mu F$ capacitor is given by $q = C_1 V_1 = 10\,\mu F \times 1000\, V = 10000\,\mu C$.
When the two capacitors are connected in parallel,the total charge $q$ is redistributed between them.
The total capacitance of the system is $C_{eq} = C_1 + C_2 = 10\,\mu F + 6\,\mu F = 16\,\mu F$.
The final potential difference $V'$ across each capacitor is given by $V' = q / C_{eq}$.
$V' = 10000\,\mu C / 16\,\mu F = 625\, V$.

(D) Let the capacitance of the air-cored capacitor be $C$. Then the capacitance of the capacitor with dielectric $K$ is $C_K = KC$.
When connected,the total charge $Q = C \times V + KC \times 0 = CV$.
The total capacitance of the system is $C_{total} = C + KC = C(1 + K)$.
The common potential $V'$ is given by $V' = \frac{Q}{C_{total}} = \frac{CV}{C(1 + K)} = \frac{V}{1 + K}$.
Rearranging for $K$: $1 + K = \frac{V}{V'}$.
Therefore,$K = \frac{V}{V'} - 1 = \frac{V - V'}{V'}$.

(D) Initially,capacitor $C_1$ has charge $Q$ and $C_2$ is uncharged. When the switch is closed,charge flows until both capacitors reach the same potential $V$.
In the steady state,the potential difference across both capacitors is equal:
$V = \frac{Q_1}{C_1} = \frac{Q_2}{C_2}$
Given $C_2 = 2C_1$,we have:
$\frac{Q_1}{C_1} = \frac{Q_2}{2C_1} \implies Q_2 = 2Q_1$
By the law of conservation of charge,the total charge remains constant:
$Q_1 + Q_2 = Q$
Substituting $Q_2 = 2Q_1$ into the equation:
$Q_1 + 2Q_1 = Q \implies 3Q_1 = Q \implies Q_1 = \frac{Q}{3}$
Then,$Q_2 = 2Q_1 = \frac{2Q}{3}$
Thus,the charges on the capacitors are $\frac{Q}{3}$ and $\frac{2Q}{3}$.

(A) The initial energy stored in the capacitor $C_1$ is given by:
$U_1 = \frac{1}{2} C_1 V_0^2$
When the switch $S$ is closed,the charge $Q = C_1 V_0$ is redistributed between the two capacitors until they reach a common potential $V$. By the principle of conservation of charge:
$C_1 V_0 = (C_1 + C_2) V$
$V = \frac{C_1 V_0}{C_1 + C_2}$
The final energy stored in the system is:
$U_2 = \frac{1}{2} (C_1 + C_2) V^2 = \frac{1}{2} (C_1 + C_2) \left( \frac{C_1 V_0}{C_1 + C_2} \right)^2 = \frac{1}{2} \frac{C_1^2 V_0^2}{C_1 + C_2}$
The ratio of the energy before and after the connection is:
$\frac{U_1}{U_2} = \frac{\frac{1}{2} C_1 V_0^2}{\frac{1}{2} \frac{C_1^2 V_0^2}{C_1 + C_2}} = \frac{C_1 (C_1 + C_2)}{C_1^2} = \frac{C_1 + C_2}{C_1}$

(A) Initial energy stored in the capacitor: $U_i = \frac{1}{2} C V^2 = \frac{1}{2} \times (400 \times 10^{-12} \, F) \times (100 \, V)^2 = 2 \times 10^{-6} \, J$.
When the charged capacitor is connected to an identical uncharged capacitor,the charge redistributes until the potential difference across both is the same.
The common potential is $V' = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} = \frac{400 \times 100 + 400 \times 0}{400 + 400} = 50 \, V$.
The final energy stored in the system is $U_f = \frac{1}{2} (C_1 + C_2) (V')^2 = \frac{1}{2} \times (800 \times 10^{-12} \, F) \times (50 \, V)^2 = 1 \times 10^{-6} \, J$.
The energy loss is $\Delta U = U_i - U_f = 2 \times 10^{-6} \, J - 1 \times 10^{-6} \, J = 1 \times 10^{-6} \, J$.

(B) The initial energy stored in the capacitor is $U = \frac{1}{2}CV^2$,where $C$ is the capacitance and $V$ is the potential difference.
When an uncharged capacitor of the same capacity $C$ is connected in parallel,the total charge $Q = CV$ is redistributed between the two capacitors.
Since the capacitors are in parallel,the new potential difference $V'$ across each capacitor is given by $V' = \frac{Q}{C_{eq}} = \frac{CV}{C + C} = \frac{V}{2}$.
The new energy $U'$ stored in each capacitor is $U' = \frac{1}{2}C(V')^2$.
Substituting $V' = \frac{V}{2}$,we get $U' = \frac{1}{2}C(\frac{V}{2})^2 = \frac{1}{2}C(\frac{V^2}{4}) = \frac{1}{4}(\frac{1}{2}CV^2) = \frac{U}{4}$.

(C) Initially,capacitor $C_{1}$ is charged to a potential $V$ by the battery. The initial energy stored in $C_{1}$ is $U_{i} = \frac{1}{2} C V^{2}$.
When the battery is disconnected and $C_{1}$ is connected in parallel with an uncharged capacitor $C_{2}$ (where $C_{1} = C_{2} = C$),charge is redistributed.
The common potential $V'$ after connection is given by $V' = \frac{C_{1}V + C_{2}(0)}{C_{1} + C_{2}} = \frac{CV}{2C} = \frac{V}{2}$.
The final energy stored in the system is $U_{f} = \frac{1}{2} (C_{1} + C_{2}) (V')^{2} = \frac{1}{2} (2C) (\frac{V}{2})^{2} = C \cdot \frac{V^{2}}{4} = \frac{1}{4} C V^{2}$.
The loss in energy is $\Delta U = U_{i} - U_{f} = \frac{1}{2} C V^{2} - \frac{1}{4} C V^{2} = \frac{1}{4} C V^{2}$.
The percentage loss of energy is $\frac{\Delta U}{U_{i}} \times 100 = \frac{\frac{1}{4} C V^{2}}{\frac{1}{2} C V^{2}} \times 100 = 50 \%$.

(B) Initial charge on the capacitor: $Q = CV = 60 \times 10^{-12} \; F \times 20 \; V = 1200 \times 10^{-12} \; C = 1.2 \times 10^{-9} \; C$.
Initial electrostatic energy stored: $U_i = \frac{1}{2} CV^2 = \frac{1}{2} \times 60 \times 10^{-12} \times (20)^2 = 30 \times 10^{-12} \times 400 = 12000 \times 10^{-12} \; J = 12 \; nJ$.
When connected in parallel,the total capacitance becomes $C_{eq} = C + C = 2C = 120 \; pF$. The total charge $Q$ is conserved and redistributes equally as $Q/2$ on each capacitor.
Final electrostatic energy stored: $U_f = \frac{(Q/2)^2}{2C} + \frac{(Q/2)^2}{2C} = 2 \times \frac{Q^2/4}{2C} = \frac{Q^2}{4C} = \frac{1}{2} \times \frac{Q^2}{2C} = \frac{1}{2} U_i = \frac{1}{2} \times 12 \; nJ = 6 \; nJ$.
Energy lost: $\Delta U = U_i - U_f = 12 \; nJ - 6 \; nJ = 6 \; nJ$.

(A) Initial capacitance $C_1 = 600\; pF = 600 \times 10^{-12}\; F$ and potential $V_1 = 200\; V$.
Initial energy $E_i = \frac{1}{2} C_1 V_1^2 = \frac{1}{2} \times 600 \times 10^{-12} \times (200)^2 = 1.2 \times 10^{-5}\; J$.
When connected to an identical uncharged capacitor $C_2 = 600\; pF$, the charge $Q = C_1 V_1 = 600 \times 10^{-12} \times 200 = 1.2 \times 10^{-7}\; C$ is shared.
The common potential $V'$ is given by $V' = \frac{Q}{C_1 + C_2} = \frac{1.2 \times 10^{-7}}{1200 \times 10^{-12}} = 100\; V$.
Final energy $E_f = \frac{1}{2} (C_1 + C_2) (V')^2 = \frac{1}{2} \times 1200 \times 10^{-12} \times (100)^2 = 0.6 \times 10^{-5}\; J$.
Energy loss $\Delta E = E_i - E_f = 1.2 \times 10^{-5} - 0.6 \times 10^{-5} = 0.6 \times 10^{-5}\; J = 6 \times 10^{-6}\; J$.

(A) Initial capacitance $C_1 = 4 \;\mu F = 4 \times 10^{-6} \;F$ and supply voltage $V_1 = 200 \;V$.
Initial electrostatic energy $E_1 = \frac{1}{2} C_1 V_1^2 = \frac{1}{2} \times 4 \times 10^{-6} \times (200)^2 = 8 \times 10^{-2} \;J$.
When connected to an uncharged capacitor $C_2 = 2 \;\mu F$,the common potential $V$ is given by $V = \frac{C_1 V_1}{C_1 + C_2} = \frac{4 \times 200}{4 + 2} = \frac{800}{6} = \frac{400}{3} \;V$.
Final energy $E_2 = \frac{1}{2} (C_1 + C_2) V^2 = \frac{1}{2} \times (6 \times 10^{-6}) \times (\frac{400}{3})^2 = 3 \times 10^{-6} \times \frac{160000}{9} = \frac{16}{3} \times 10^{-2} \approx 5.33 \times 10^{-2} \;J$.
Energy lost $\Delta E = E_1 - E_2 = 8 \times 10^{-2} - 5.33 \times 10^{-2} = 2.67 \times 10^{-2} \;J$.

(N/A) Case $1$: $K_1$ is closed and $K_2$ is open.
The capacitors $C_1$ and $C_2$ are in series with the battery $E = 9 \, V$.
The equivalent capacitance is $C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{(6C)(3C)}{6C + 3C} = \frac{18C^2}{9C} = 2C = 2 \,\mu F$.
The charge on each capacitor is $Q = C_{eq} E = 2 \,\mu F \times 9 \, V = 18 \,\mu C$.
Thus,$Q_1 = 18 \,\mu C$,$Q_2 = 18 \,\mu C$,and $Q_3 = 0 \,\mu C$.
Case $2$: $K_1$ is opened and $K_2$ is closed.
Now,$C_1$ is disconnected from the battery. The capacitor $C_2$ (charged to $18 \,\mu C$) is connected in parallel with $C_3$ (initially uncharged).
The total charge $Q_{total} = 18 \,\mu C$ is shared between $C_2$ and $C_3$.
Since $C_2 = 3C$ and $C_3 = 3C$,the charge is shared equally.
$Q_2' = Q_3' = \frac{Q_{total}}{2} = \frac{18 \,\mu C}{2} = 9 \,\mu C$.
Since $K_1$ is open,$C_1$ remains charged at $18 \,\mu C$.
Final charges: $Q_1 = 18 \,\mu C$,$Q_2 = 9 \,\mu C$,$Q_3 = 9 \,\mu C$.

(B) Initially, the charge on the $10\,\mu F$ capacitor is:
$Q = C_1 V_1 = (10\,\mu F)(50\, V) = 500\,\mu C$
When this capacitor is connected in parallel to an uncharged capacitor of capacitance $C_2$, the total charge remains conserved.
Let the final potential difference be $V = 20\, V$.
The total charge $Q$ is distributed between the two capacitors:
$Q = (C_1 + C_2)V$
$500\,\mu C = (10\,\mu F + C_2)(20\, V)$
Dividing both sides by $20\, V$:
$25\,\mu F = 10\,\mu F + C_2$
$C_2 = 25\,\mu F - 10\,\mu F = 15\,\mu F$

(A) Initial energy stored in the capacitor: $U_i = \frac{1}{2} C_1 V^2 = \frac{1}{2} \times (5 \times 10^{-6}) \times (220)^2 = 0.121 \, J$.
When connected to an uncharged capacitor $C_2 = 2.5 \, \mu F$,the charge $Q = C_1 V = 5 \times 10^{-6} \times 220 = 1.1 \times 10^{-3} \, C$ redistributes.
The common potential $V'$ is given by $V' = \frac{Q}{C_1 + C_2} = \frac{1.1 \times 10^{-3}}{(5 + 2.5) \times 10^{-6}} = \frac{1100}{7.5} = \frac{440}{3} \, V$.
Final energy stored: $U_f = \frac{1}{2} (C_1 + C_2) (V')^2 = \frac{1}{2} \times (7.5 \times 10^{-6}) \times (\frac{440}{3})^2 = 0.08066 \, J$.
Energy change $\Delta U = U_f - U_i = 0.08066 - 0.121 = -0.04033 \, J$.
Given $\Delta U = -\frac{X}{100} \, J$ (considering magnitude of loss),we have $\frac{X}{100} = 0.04033$,so $X \approx 4$.

(A) When two conducting spheres are connected by a wire,charge flows until their potentials become equal.
Let the final charges be $Q_{1}'$ and $Q_{2}'$.
According to the law of conservation of charge,the total charge remains constant:
$Q_{1}' + Q_{2}' = Q_{1} + Q_{2} = 12\, \mu C - 3\, \mu C = 9\, \mu C$.
Since the potentials are equal $(V_{1} = V_{2})$,we have:
$\frac{K Q_{1}'}{R_{1}} = \frac{K Q_{2}'}{R_{2}} \Rightarrow \frac{Q_{1}'}{2R/3} = \frac{Q_{2}'}{R/3}$.
This simplifies to $Q_{1}' = 2 Q_{2}'$.
Substituting this into the conservation equation:
$2 Q_{2}' + Q_{2}' = 9\, \mu C \Rightarrow 3 Q_{2}' = 9\, \mu C \Rightarrow Q_{2}' = 3\, \mu C$.
Then,$Q_{1}' = 2 \times 3\, \mu C = 6\, \mu C$.
Thus,the final charges are $6\, \mu C$ and $3\, \mu C$.

(A) Initial charge on capacitor $C$ is $Q = CV_{0}$.
Initial energy stored in the capacitor is $U_{i} = \frac{1}{2}CV_{0}^{2}$.
When connected in parallel,the common potential $V$ is given by $V = \frac{Q_{total}}{C_{total}} = \frac{CV_{0}}{C + C/2} = \frac{CV_{0}}{3C/2} = \frac{2}{3}V_{0}$.
The final energy stored in the system is $U_{f} = \frac{1}{2}(C + C/2)V^{2} = \frac{1}{2}(\frac{3C}{2})(\frac{2}{3}V_{0})^{2} = \frac{1}{2}(\frac{3C}{2})(\frac{4}{9}V_{0}^{2}) = \frac{1}{3}CV_{0}^{2}$.
The energy loss is $\Delta U = U_{i} - U_{f} = \frac{1}{2}CV_{0}^{2} - \frac{1}{3}CV_{0}^{2} = \frac{1}{6}CV_{0}^{2}$.
Since $\frac{1}{6} \approx 0.166$,the correct option is $A$.

(D) Initial charges on the capacitors are:
$Q_1 = C \times V = CV$
$Q_2 = 2C \times 2V = 4CV$
Since they are connected with opposite polarities (positive to negative),the net charge available is:
$Q_{net} = Q_2 - Q_1 = 4CV - CV = 3CV$
When connected in parallel,the equivalent capacitance is:
$C_{eq} = C + 2C = 3C$
The common potential $V_c$ is given by:
$V_c = \frac{Q_{net}}{C_{eq}} = \frac{3CV}{3C} = V$
The final energy $U_f$ stored in the configuration is:
$U_f = \frac{1}{2} C_{eq} V_c^2 = \frac{1}{2} \times (3C) \times V^2 = 1.5 CV^2$

(D) $1$. Initial charge on capacitor $C_{1}$ is $Q = C_{1}V = 2\, \mu F \times 10\, V = 20\, \mu C$.
$2$. When the battery is removed and $C_{1}$ is connected to the uncharged capacitor $C_{2}$,the total charge $Q$ is conserved and shared between the two capacitors.
$3$. The common potential $V'$ after connection is given by $V' = \frac{Q}{C_{1} + C_{2}} = \frac{20\, \mu C}{2\, \mu F + 8\, \mu F} = \frac{20}{10} = 2\, V$.
$4$. The charge on capacitor $C_{2}$ at equilibrium is $Q_{2} = C_{2}V' = 8\, \mu F \times 2\, V = 16\, \mu C$.

(A) Initially,the total charge in the circuit is stored in capacitor $C_{1}$.
$Q_{\text{total}} = C_{1} V$
When the switch $S$ is closed,the charge redistributes between $C_{1}$ and $C_{2}$ until they reach a common potential $V'$.
Since the capacitors are connected in parallel,the common potential is given by:
$V' = \frac{Q_{\text{total}}}{C_{\text{eq}}} = \frac{C_{1} V}{C_{1} + C_{2}}$
The charge on capacitor $C_{2}$ at equilibrium is:
$Q_{2} = C_{2} V' = C_{2} \left( \frac{C_{1} V}{C_{1} + C_{2}} \right) = \frac{C_{1} C_{2} V}{C_{1} + C_{2}}$

(D) The initial energy stored in the first capacitor is $U_i = \frac{1}{2} C V^2 = \frac{1}{2} \times 50 \times 10^{-12} \times (100)^2 = 250 \times 10^{-9} \; J = 250 \; nJ$.
When connected to an identical uncharged capacitor,the charge redistributes until the potential across both is $V' = \frac{V}{2} = 50 \; V$.
The final energy stored in the system is $U_f = 2 \times (\frac{1}{2} C V'^2) = C \times (\frac{V}{2})^2 = 50 \times 10^{-12} \times 2500 = 125 \times 10^{-9} \; J = 125 \; nJ$.
The energy loss is $\Delta U = U_i - U_f = 250 \; nJ - 125 \; nJ = 125 \; nJ$.
Alternatively,using the formula for energy loss: $\Delta U = \frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} (V_1 - V_2)^2 = \frac{1}{2} \frac{50 \times 50}{50 + 50} \times 10^{-12} \times (100 - 0)^2 = 125 \times 10^{-9} \; J = 125 \; nJ$.

(A) Before closing the switch,the total charge $Q$ on capacitor $C_{1}$ is given by:
$Q = C_{1} V_{0} = 5\,\mu F \times 30\,V = 150\,\mu C$
After closing the switch,the charge redistributes until both capacitors reach a common potential $V$. Since the total charge is conserved:
$V = \frac{Q}{C_{1} + C_{2}} = \frac{150\,\mu C}{5\,\mu F + 10\,\mu F} = \frac{150}{15}\,V = 10\,V$
Now,the charge on capacitor $C_{2}$ at equilibrium is:
$Q_{2} = C_{2} V = 10\,\mu F \times 10\,V = 100\,\mu C$

(D) Let the capacitance of the air-cored capacitor be $C$. The capacitance of the capacitor with dielectric $K$ is $K C$.
Initially,the charge on the air-cored capacitor is $Q = C V$ and the charge on the uncharged capacitor is $0$.
When connected in parallel,the total charge is conserved and the capacitors reach a common potential $V^{\prime}$.
The total charge $Q_{total} = C V + 0 = C V$.
The total capacitance $C_{total} = C + K C = C(1 + K)$.
Using the formula for common potential $V^{\prime} = \frac{Q_{total}}{C_{total}}$,we get:
$V^{\prime} = \frac{C V}{C(1 + K)} = \frac{V}{1 + K}$.
Rearranging for $K$:
$1 + K = \frac{V}{V^{\prime}}$
$K = \frac{V}{V^{\prime}} - 1 = \frac{V - V^{\prime}}{V^{\prime}}$.

(B) Initial energy stored in the two capacitors is:
$U_i = \frac{1}{2} C V^2 + \frac{1}{2} C V^2 = C V^2$
When the capacitors are connected with opposite polarities,the net charge on the system becomes:
$Q_{net} = (+CV) + (-CV) = 0$
Since the total charge is zero,the final potential difference across the capacitors is zero,and thus the final energy stored is:
$U_f = 0$
The loss of energy is given by:
$\Delta U = U_i - U_f = C V^2 - 0 = C V^2$

(D) Let the capacitance of the air-filled capacitor be $C$. Its charge is $Q = CV$.
Since the capacitors are identical,the capacitance of the dielectric-filled capacitor is $C' = KC$.
When the two capacitors are connected in parallel,the total charge $Q_{total}$ is conserved.
$Q_{total} = Q_1 + Q_2 = 0 + CV = CV$.
The equivalent capacitance of the parallel combination is $C_{eq} = C + KC = C(1+K)$.
The common potential $V'$ is given by the formula $V' = \frac{Q_{total}}{C_{eq}}$.
Substituting the values,we get $V' = \frac{CV}{C(1+K)}$.
Therefore,$V' = \frac{V}{1+K}$.

(B) In the steady state,the current in the circuit becomes zero,and both capacitors are connected in parallel across the same potential difference. Let the new charges on $C_1$ and $C_2$ be $Q_1$ and $Q_2$ respectively.
Since they are in parallel,their potentials must be equal:
$V_1 = V_2 \implies \frac{Q_1}{C_1} = \frac{Q_2}{C_2}$
Given $C_2 = 2C_1$,we have:
$\frac{Q_1}{C_1} = \frac{Q_2}{2C_1} \implies Q_2 = 2Q_1$
According to the law of conservation of charge,the total charge remains constant:
$Q_1 + Q_2 = Q$
Substituting $Q_2 = 2Q_1$ into the conservation equation:
$Q_1 + 2Q_1 = Q \implies 3Q_1 = Q \implies Q_1 = \frac{Q}{3}$
Then,$Q_2 = 2Q_1 = \frac{2Q}{3}$
Thus,the charges on the capacitors are $\frac{Q}{3}$ and $\frac{2Q}{3}$.