Sharing of Charge in Capacitor Circuit Questions in English

(D) The total charge on the system is $Q = 80\,\mu C + 40\,\mu C = 120\,\mu C$.
When two spheres are connected by a wire,charge flows until they reach the same potential $V$.
The potential of a sphere is given by $V = \frac{kQ}{r}$.
After connection,the new charges $Q'_A$ and $Q'_B$ are distributed in proportion to their radii: $\frac{Q'_A}{Q'_B} = \frac{r_A}{r_B}$.
Using the formula $Q'_A = Q \left( \frac{r_A}{r_A + r_B} \right)$:
$Q'_A = 120\,\mu C \left( \frac{4}{4 + 6} \right) = 120 \times 0.4 = 48\,\mu C$.
The charge that flowed from $A$ is $\Delta Q = Q_{initial} - Q_{final} = 80\,\mu C - 48\,\mu C = 32\,\mu C$.
Since the charge on $A$ decreased,$32\,\mu C$ of charge flowed from $A$ to $B$.

(D) When two charged conductors are connected,charge flows from the conductor at a higher potential to the conductor at a lower potential until they reach a common potential.
Potential of sphere $1$ is $V_1 = \frac{k Q_1}{R_1}$ and potential of sphere $2$ is $V_2 = \frac{k Q_2}{R_2}$.
If $V_1 \neq V_2$ (i.e.,$\frac{Q_1}{R_1} \neq \frac{Q_2}{R_2}$),charge redistribution occurs.
During this process,some energy is always dissipated as heat in the connecting wire due to resistance.
Therefore,the total energy of the system decreases unless the potentials are already equal,which corresponds to the condition $\frac{Q_1}{R_1} = \frac{Q_2}{R_2}$ or $Q_1 R_2 = Q_2 R_1$.

(A) When two charged conductors are connected by a conducting wire,charge flows until they reach a common potential $V$.
The common potential is given by the formula: $V = \frac{Q_1 + Q_2}{C_1 + C_2}$.
Given: $Q_1 = Q_2 = 150 \times 10^{-6}\,C$,$r_1 = 0.2\,m$,$r_2 = 0.1\,m$.
The capacitance of an isolated spherical conductor is $C = 4\pi \varepsilon_0 r$.
Thus,$C_1 = 4\pi \varepsilon_0 (0.2)$ and $C_2 = 4\pi \varepsilon_0 (0.1)$.
Total charge $Q_{total} = 150 \times 10^{-6} + 150 \times 10^{-6} = 300 \times 10^{-6}\,C$.
Total capacitance $C_{total} = 4\pi \varepsilon_0 (0.2 + 0.1) = 4\pi \varepsilon_0 (0.3)$.
Using $k = \frac{1}{4\pi \varepsilon_0} = 9 \times 10^9\,N\cdot m^2/C^2$,we have $C_{total} = \frac{0.3}{9 \times 10^9}$.
$V = \frac{300 \times 10^{-6}}{0.3 / (9 \times 10^9)} = \frac{300 \times 10^{-6} \times 9 \times 10^9}{0.3} = 1000 \times 9 \times 10^3 = 9 \times 10^6\,V$.

(C) The initial energy of the system is $U_i = \frac{1}{2}CV^2 + \frac{1}{2}C(-V)^2 = CV^2$.
When connected by a fine wire,the total charge $Q_{total} = CV + C(-V) = 0$.
Since the total charge is $0$,the final potential of the system is $V_f = \frac{Q_{total}}{C_1 + C_2} = 0$.
The final energy of the system is $U_f = \frac{1}{2}(C + C)V_f^2 = 0$.
The loss of energy is $\Delta U = U_i - U_f = CV^2 - 0 = CV^2$.

(B) When two conductors are brought into contact,charge flows from the conductor at a higher potential to the conductor at a lower potential until their potentials become equal.
Let the final common potential be $V$.
The charge on the first sphere is $Q_1 = C_1 V$.
The charge on the second sphere is $Q_2 = C_2 V$.
Taking the ratio of the two charges,we get $\frac{Q_1}{Q_2} = \frac{C_1 V}{C_2 V} = \frac{C_1}{C_2}$.
Therefore,the final charges on the spheres satisfy the relation $\frac{Q_1}{Q_2} = \frac{C_1}{C_2}$.

(B) When two conducting spheres are connected by a conducting wire, charge flows until both spheres reach the same electric potential.
Let the radii be $r_1 = 5\, cm$ and $r_2 = 10\, cm$.
The total charge $Q_{total} = 15\,\mu C + 15\,\mu C = 30\,\mu C$.
After connection, the potential $V$ is the same for both: $V = \frac{k q_1}{r_1} = \frac{k q_2}{r_2}$.
This implies $\frac{q_1}{q_2} = \frac{r_1}{r_2} = \frac{5}{10} = \frac{1}{2}$.
The charge on the smaller sphere $q_1$ is given by $q_1 = Q_{total} \left( \frac{r_1}{r_1 + r_2} \right)$.
Substituting the values: $q_1 = 30\,\mu C \left( \frac{5}{5 + 10} \right) = 30 \left( \frac{5}{15} \right) = 30 \left( \frac{1}{3} \right) = 10\,\mu C$.

(A) Initial energy of the capacitor with capacitance $C_1 = 4\,\mu F$ charged to $V_1 = 80\,V$ is given by $U_i = \frac{1}{2} C_1 V_1^2 = \frac{1}{2} \times (4 \times 10^{-6}) \times (80)^2 = 0.0128\,J = 12.8\,mJ$.
When the two capacitors are connected,the common potential $V$ is given by $V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} = \frac{4 \times 80 + 6 \times 30}{4 + 6} = \frac{320 + 180}{10} = 50\,V$.
The final energy of the $4\,\mu F$ capacitor at potential $V = 50\,V$ is $U_f = \frac{1}{2} C_1 V^2 = \frac{1}{2} \times (4 \times 10^{-6}) \times (50)^2 = 0.005\,J = 5.0\,mJ$.
The energy lost by the $4\,\mu F$ capacitor is $\Delta U = U_i - U_f = 12.8\,mJ - 5.0\,mJ = 7.8\,mJ$.

(C) Let the radius of each small drop be $r$ and the charge on each be $q$. The capacitance of a small drop is $C = 4\pi\epsilon_0 r$.
The energy stored in a small drop is $u = \frac{q^2}{2C} = \frac{q^2}{8\pi\epsilon_0 r}$.
When $n$ drops coalesce, the total volume remains constant: $\frac{4}{3}\pi R^3 = n \times \frac{4}{3}\pi r^3$, so $R = n^{1/3}r$.
The total charge on the big drop is $Q = nq$.
The capacitance of the big drop is $C' = 4\pi\epsilon_0 R = 4\pi\epsilon_0 (n^{1/3}r) = n^{1/3}C$.
The energy stored in the big drop is $U = \frac{Q^2}{2C'} = \frac{(nq)^2}{2(n^{1/3}C)} = \frac{n^2 q^2}{n^{1/3} 2C} = n^{5/3} \left( \frac{q^2}{2C} \right) = n^{5/3} u$.
Therefore, the ratio $U : u = n^{5/3} : 1$.

(D) Given: $r_1 = 1\,cm$,$r_2 = 2\,cm$,$Q_1 = 10^{-2}\,C$,$Q_2 = 5 \times 10^{-2}\,C$.
When two spheres are connected by a conducting wire,charge flows until both reach the same potential.
The total charge $Q_{total} = Q_1 + Q_2 = 10^{-2} + 5 \times 10^{-2} = 6 \times 10^{-2}\,C$.
The final charge on the smaller sphere $(Q'_1)$ is given by the formula $Q'_1 = Q_{total} \times \frac{r_1}{r_1 + r_2}$.
Substituting the values: $Q'_1 = (6 \times 10^{-2}) \times \frac{1}{1 + 2} = (6 \times 10^{-2}) \times \frac{1}{3} = 2 \times 10^{-2}\,C$.

(D) Let the capacitance of each capacitor be $C$. Initially,they are connected in parallel and charged to potential $V$. The charge on each capacitor is $Q = CV$.
When they are separated and connected in series such that the positive plate of one is connected to the negative plate of the other,the total charge on the connected plates is $+Q + (-Q) = 0$.
Since the plates are isolated,the charge on each capacitor remains $Q = CV$.
In the series combination,the potential difference across each capacitor is $V$. Since they are connected in series with opposite polarities,the total potential difference across the free plates is $V + V = 2\,V$.

(A) Initial energy stored in capacitor $C_1$ is $U_i = \frac{1}{2} C_1 V_0^2 = \frac{q^2}{2C_1}$,where $q = C_1 V_0$ is the charge on the capacitor.
When the switch $S$ is closed,the charge $q$ is shared between the two capacitors $C_1$ and $C_2$ connected in parallel. The total capacitance becomes $C_{eq} = C_1 + C_2$.
The final energy stored in the system is $U_f = \frac{q^2}{2 C_{eq}} = \frac{q^2}{2(C_1 + C_2)}$.
The ratio of the energies is $\frac{U_i}{U_f} = \frac{q^2 / (2C_1)}{q^2 / (2(C_1 + C_2))} = \frac{C_1 + C_2}{C_1}$.

(C) Initial energy $U_i = \frac{1}{2} C_1 V_1^2 + \frac{1}{2} C_2 V_2^2$.
When connected,the common potential $V$ is given by $V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$.
Final energy $U_f = \frac{1}{2} (C_1 + C_2) V^2 = \frac{1}{2} (C_1 + C_2) \left( \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} \right)^2 = \frac{(C_1 V_1 + C_2 V_2)^2}{2(C_1 + C_2)}$.
Energy loss $\Delta U = U_i - U_f = \left( \frac{1}{2} C_1 V_1^2 + \frac{1}{2} C_2 V_2^2 \right) - \frac{(C_1 V_1 + C_2 V_2)^2}{2(C_1 + C_2)}$.
Simplifying this expression: $\Delta U = \frac{C_1 C_2 (V_1 - V_2)^2}{2(C_1 + C_2)}$.

(B) Initially,the charge on the first capacitor is $Q = CV$.
When this charged capacitor is connected in parallel to an identical uncharged capacitor,the total charge $Q$ is shared between the two capacitors.
Since the capacitors are identical and connected in parallel,the potential difference across both capacitors becomes equal,say $V'$.
By the principle of conservation of charge,the total charge remains constant: $Q_{total} = Q_1 + Q_2$.
Since they are identical,$C_1 = C_2 = C$,the charge distributes equally: $Q_1 = Q_2 = Q / 2$.
Therefore,the new charge on each capacitor is $CV / 2$.

(C) The initial charge on the first capacitor is $Q_1 = C_1 V_1 = 2\,\mu F \times 200\;V = 400\,\mu C$.
When the second capacitor $C_2$ (initially uncharged,$V_2 = 0$) is connected in parallel,the total charge remains conserved.
The common potential $V$ is given by the formula: $V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$.
Substituting the given values: $20 = \frac{2 \times 200 + C_2 \times 0}{2 + C_2}$.
$20 = \frac{400}{2 + C_2}$.
$2 + C_2 = \frac{400}{20} = 20$.
$C_2 = 20 - 2 = 18\,\mu F$.

(C) Initial charge on the $6\,\mu F$ capacitor is $Q = C_1 V_0 = 6\,\mu F \times 100\,V = 600\,\mu C$.
When connected in parallel,the capacitors share charge until they reach a common potential $V$.
The common potential $V$ is given by $V = \frac{Q_{total}}{C_1 + C_2} = \frac{600\,\mu C}{6\,\mu F + 14\,\mu F} = \frac{600}{20} = 30\,V$.
The charges on the capacitors are $q_1 = C_1 V = 6\,\mu F \times 30\,V = 180\,\mu C$ and $q_2 = C_2 V = 14\,\mu F \times 30\,V = 420\,\mu C$.
The ratio of the charges is $\frac{q_1}{q_2} = \frac{C_1}{C_2} = \frac{6}{14}$.
Thus,the ratio is $6/14$ and the potential is $30\,V$.

(A) The initial charge $Q$ on the $0.2\,F$ capacitor is given by $Q = C_1 V_1 = 0.2\,F \times 600\,V = 120\,C$.
When the battery is removed and the capacitor is connected to another capacitor of $C_2 = 1\,F$,the total charge is conserved.
Let the common potential be $V$. The total capacitance of the parallel combination is $C_{eq} = C_1 + C_2 = 0.2\,F + 1\,F = 1.2\,F$.
Using the principle of charge conservation: $Q = C_{eq} V$.
$120\,C = 1.2\,F \times V$.
$V = \frac{120}{1.2} = 100\,V$.

(A) Given: Capacitance of the first capacitor $C_1 = 10\,\mu F$,initial potential $V_1 = 100\,V$.
Capacitance of the second capacitor $C_2 = ?$,initial potential $V_2 = 0\,V$.
Common potential $V = 40\,V$.
When two capacitors are connected in parallel,the total charge is conserved.
The common potential is given by the formula: $V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$.
Substituting the values: $40 = \frac{10 \times 100 + C_2 \times 0}{10 + C_2}$.
$40 = \frac{1000}{10 + C_2}$.
$40(10 + C_2) = 1000$.
$400 + 40C_2 = 1000$.
$40C_2 = 600$.
$C_2 = \frac{600}{40} = 15\,\mu F$.

(A) The total energy before connection is given by $U_i = \frac{1}{2} C_1 V_1^2 + \frac{1}{2} C_2 V_2^2$.
$U_i = \frac{1}{2} \times 4 \times 10^{-6} \times (50)^2 + \frac{1}{2} \times 2 \times 10^{-6} \times (100)^2 = 50 \times 10^{-4} + 100 \times 10^{-4} = 1.5 \times 10^{-2}\,J$.
When connected in parallel with like charges,the common potential $V$ is given by $V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} = \frac{4 \times 50 + 2 \times 100}{4 + 2} = \frac{400}{6} = \frac{200}{3}\,V$.
The total energy after connection is $U_f = \frac{1}{2} (C_1 + C_2) V^2$.
$U_f = \frac{1}{2} \times (4 + 2) \times 10^{-6} \times (\frac{200}{3})^2 = 3 \times 10^{-6} \times \frac{40000}{9} = \frac{40000}{3} \times 10^{-6} = 1.33 \times 10^{-2}\,J$.

(B) Initially,both capacitors have capacitance $C$. After inserting the dielectric slab of constant $K=4$ in the second capacitor,its new capacitance becomes $C_2 = KC = 4C$. The first capacitor remains $C_1 = C$.
Since they are in series,the equivalent capacitance is $C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{C \times 4C}{C + 4C} = \frac{4C}{5}$.
The total charge stored in the series combination is $Q = C_{eq} V = \frac{4C}{5} \times 100 = 80C$.
In a series circuit,the charge $Q$ is the same for both capacitors.
The potential difference across the first capacitor is $V_1 = \frac{Q}{C_1} = \frac{80C}{C} = 80\,V$.
The potential difference across the second capacitor is $V_2 = \frac{Q}{C_2} = \frac{80C}{4C} = 20\,V$.
Thus,the potential differences are $80\,V$ and $20\,V$ respectively.

(A) The initial energy stored in the capacitor $C_1$ is $U_0 = \frac{1}{2} C_1 V_0^2$.
When connected in parallel to an uncharged capacitor $C_2$,the charge redistributes until both capacitors reach a common potential $V = \frac{C_1 V_0}{C_1 + C_2}$.
The final energy stored in the system is $U_f = \frac{1}{2} (C_1 + C_2) V^2 = \frac{1}{2} (C_1 + C_2) \left( \frac{C_1 V_0}{C_1 + C_2} \right)^2 = \frac{C_1^2 V_0^2}{2(C_1 + C_2)}$.
The energy dissipated (loss of energy) is $\Delta U = U_0 - U_f$.
$\Delta U = \frac{1}{2} C_1 V_0^2 - \frac{C_1^2 V_0^2}{2(C_1 + C_2)} = \frac{C_1 V_0^2}{2} \left( 1 - \frac{C_1}{C_1 + C_2} \right) = \frac{C_1 V_0^2}{2} \left( \frac{C_2}{C_1 + C_2} \right)$.
Since $U_0 = \frac{1}{2} C_1 V_0^2$,we substitute this into the expression:
$\Delta U = \frac{C_2}{C_1 + C_2} U_0$.

(C) $1$. When switch $S_1$ is closed,capacitor $C_1$ gets charged by the battery $B = 20\,V$. The charge on $C_1$ is $Q_1 = C_1 \times V = 6\,\mu F \times 20\,V = 120\,\mu C$.
$2$. When $S_1$ is opened,the charge $Q_1 = 120\,\mu C$ remains stored on $C_1$. Capacitor $C_2$ is initially uncharged $(Q_2 = 0)$.
$3$. When switch $S_2$ is closed,the charge $Q_1$ is shared between $C_1$ and $C_2$ until they reach a common potential $V'$.
$4$. The common potential $V'$ is given by $V' = \frac{Q_{total}}{C_{total}} = \frac{Q_1 + Q_2}{C_1 + C_2} = \frac{120\,\mu C + 0}{6\,\mu F + 3\,\mu F} = \frac{120}{9}\,V = \frac{40}{3}\,V$.
$5$. The final charge on $C_2$ is $Q_2' = C_2 \times V' = 3\,\mu F \times \frac{40}{3}\,V = 40\,\mu C$.

(D) When two capacitors of capacitance $C$ are charged to potentials $V_1$ and $V_2$ respectively,their initial charges are $q_1 = CV_1$ and $q_2 = CV_2$.
When they are connected in parallel,the common potential $V$ is given by $V = \frac{q_1 + q_2}{C + C} = \frac{C(V_1 + V_2)}{2C} = \frac{V_1 + V_2}{2}$.
Option $A$ is correct because charge is conserved.
Option $B$ is correct because energy is lost as heat during charge redistribution.
Option $C$ is correct because the final potential is the average,not the sum.
Option $D$ is incorrect because the final potential is $\frac{V_1 + V_2}{2}$,not $V_1 + V_2$.

(B) Initially,when connected in series,the charge $Q$ on each capacitor is the same.
Equivalent capacitance $C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{10 \times 20}{10 + 20} = \frac{200}{30} = \frac{20}{3}\,\mu F$.
The total charge $Q = C_{eq} V = \frac{20}{3} \times 200 = \frac{4000}{3}\,\mu C$.
The potential difference across $C_1$ is $V_1 = \frac{Q}{C_1} = \frac{4000/3}{10} = \frac{400}{3}\,V$.
The potential difference across $C_2$ is $V_2 = \frac{Q}{C_2} = \frac{4000/3}{20} = \frac{200}{3}\,V$.
When reconnected in parallel with positive plates together,the total charge $Q_{total} = Q_1 + Q_2 = \frac{4000}{3} + \frac{4000}{3} = \frac{8000}{3}\,\mu C$ (since both capacitors were charged with the same magnitude of charge $Q$ in series).
The common potential $V = \frac{Q_{total}}{C_1 + C_2} = \frac{8000/3}{10 + 20} = \frac{8000}{3 \times 30} = \frac{800}{9}\,V$.

(D) When two capacitors are connected in parallel,the total charge is conserved.
Let $C_1 = 10\,\mu F$ and $V_1 = 50\;V$.
Let the second capacitor have capacitance $C_2$ and initial potential $V_2 = 0\;V$.
The common potential $V$ is given by the formula:
$V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$
Substituting the given values:
$20 = \frac{10 \times 50 + C_2 \times 0}{10 + C_2}$
$20 = \frac{500}{10 + C_2}$
$20(10 + C_2) = 500$
$200 + 20C_2 = 500$
$20C_2 = 300$
$C_2 = 15\,\mu F$

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

(D) The initial energy stored in the $20\,F$ capacitor is given by $U_i = \frac{1}{2} C_1 V_1^2 = \frac{1}{2} \times 20 \times 5^2 = 250\,J$.
When connected in parallel,the common potential $V$ is given by $V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} = \frac{20 \times 5 + 30 \times 0}{20 + 30} = \frac{100}{50} = 2\,V$.
The final energy of the system is $U_f = \frac{1}{2} (C_1 + C_2) V^2 = \frac{1}{2} \times (20 + 30) \times 2^2 = \frac{1}{2} \times 50 \times 4 = 100\,J$.
The decrease in energy is $\Delta U = U_i - U_f = 250\,J - 100\,J = 150\,J$.
Alternatively,using the formula $\Delta U = \frac{C_1 C_2}{2(C_1 + C_2)} (V_1 - V_2)^2 = \frac{20 \times 30}{2(20 + 30)} (5 - 0)^2 = \frac{600}{100} \times 25 = 6 \times 25 = 150\,J$.

(B) Initially,capacitor $C_1$ has charge $Q$ and $C_2$ is uncharged.
When the switch $S$ is closed,the capacitors $C_1$ and $C_2$ are connected in parallel.
In a parallel combination,the potential difference across both capacitors becomes equal.
Let the final charges on $C_1$ and $C_2$ be $Q_1$ and $Q_2$ respectively.
Since the total charge is conserved,$Q_1 + Q_2 = Q$.
For parallel connection,$V = Q_1 / C_1 = Q_2 / C_2$.
Substituting $C_2 = 2C_1$,we get $Q_1 / C_1 = Q_2 / (2C_1)$,which implies $Q_2 = 2Q_1$.
Substituting this into the conservation equation: $Q_1 + 2Q_1 = Q$,so $3Q_1 = Q$.
Thus,$Q_1 = Q/3$ and $Q_2 = 2Q/3$.

(B) The initial charges on the capacitors are $Q_1 = C_1 V_1 = 3 \mu F \times 12 V = 36 \mu C$ and $Q_2 = C_2 V_2 = 6 \mu F \times 12 V = 72 \mu C$.
Since the positive plate of one is connected to the negative plate of the other,the net charge on the system is $Q_{net} = |Q_2 - Q_1| = |72 \mu C - 36 \mu C| = 36 \mu C$.
The equivalent capacitance of the parallel combination is $C_{eq} = C_1 + C_2 = 3 \mu F + 6 \mu F = 9 \mu F$.
The common potential difference across the capacitors is $V = \frac{Q_{net}}{C_{eq}} = \frac{36 \mu C}{9 \mu F} = 4 V$.

(D) When two charged capacitors are connected in parallel,the total charge is conserved.
The total charge $Q_{total} = Q_1 + Q_2 = C_1V_1 + C_2V_2$.
The total capacitance of the parallel combination is $C_{eq} = C_1 + C_2$.
The common potential $V$ is given by the formula $V = \frac{C_1V_1 + C_2V_2}{C_1 + C_2}$.
Substituting the given values: $C_1 = 10\,\mu F$,$V_1 = 250\,V$,$C_2 = 5\,\mu F$,$V_2 = 100\,V$.
$V = \frac{(10 \times 250) + (5 \times 100)}{10 + 5} = \frac{2500 + 500}{15} = \frac{3000}{15} = 200\,V$.

(C) When two capacitors are connected,charge flows from the higher potential to the lower potential until they reach a common potential $V'$.
By the principle of conservation of charge,the total charge before connection equals the total charge after connection.
Initial charge $Q_{total} = C_1 V + C_2 (0) = C_1 V$.
Total capacitance after connection $C_{total} = C_1 + C_2$.
Common potential $V' = \frac{Q_{total}}{C_{total}} = \frac{C_1 V}{C_1 + C_2}$.

(C) The initial charge on the $10\ \mu F$ capacitor is given by $Q = C_1 V_1$.
$Q = 10 \times 10^{-6}\ F \times 1000\ V = 10^{-2}\ C$.
When the two capacitors are connected in parallel,the charge is redistributed until they reach a common potential $V$.
The common potential is given by the formula $V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$.
Since the second capacitor is uncharged,$V_2 = 0$.
$V = \frac{10 \times 10^{-6} \times 1000 + 6 \times 10^{-6} \times 0}{10 \times 10^{-6} + 6 \times 10^{-6}} = \frac{10^{-2}}{16 \times 10^{-6}} = \frac{10000}{16} = 625\ V$.

(A) The charges $\pm q$ on the plates of capacitor $A$ are bound charges due to the electric field between them.
Since capacitor $B$ is initially uncharged and is isolated from any external source of charge,closing the switch $S$ connects the inner plate of $A$ to the inner plate of $B$.
However,because the charges on the plates of $A$ are bound by the electrostatic attraction between them,they cannot move to plate $B$.
Therefore,no charge will flow to capacitor $B$,and it will remain uncharged.
Thus,the charge on $B$ is zero.

(D) When two charged conductors are brought into contact,charge flows from the conductor at a higher potential to the one at a lower potential until their potentials become equal.
Let the initial potentials be $V_1 = \frac{kQ_1}{R_1}$ and $V_2 = \frac{kQ_2}{R_2}$.
If $V_1 \neq V_2$ (i.e.,$\frac{Q_1}{R_1} \neq \frac{Q_2}{R_2}$ or $Q_1R_2 \neq Q_2R_1$),charge redistribution occurs.
The total energy of the system is $U = \frac{1}{2} C_1 V_1^2 + \frac{1}{2} C_2 V_2^2$.
When connected,the final potential $V$ is given by $V = \frac{Q_1 + Q_2}{C_1 + C_2}$.
The final energy is $U' = \frac{1}{2} (C_1 + C_2) V^2 = \frac{1}{2} \frac{(Q_1 + Q_2)^2}{C_1 + C_2}$.
It can be mathematically shown that the loss in energy $\Delta U = U - U' = \frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} (V_1 - V_2)^2$.
Since $(V_1 - V_2)^2$ is always positive,$\Delta U > 0$,meaning energy is always lost (decreases) when charge flows between conductors at different potentials.

(D) Charge flows from higher potential to lower potential. The potential of a capacitor is given by $V = Q/C$.
For circuit $(a)$:
Potential of left capacitor $V_L = (2Q)/(3C) = 0.67(Q/C)$.
Potential of right capacitor $V_R = Q/C = 1.0(Q/C)$.
Since $V_R > V_L$,charge flows from $R$ to $L$.
For circuit $(b)$:
Potential of left capacitor $V_L = (2Q)/(2C) = Q/C$.
Potential of right capacitor $V_R = Q/(2C) = 0.5(Q/C)$.
Since $V_L > V_R$,charge flows from $L$ to $R$.

(C) The flow of charge between two conductors connected by a wire depends on their electric potentials,not their total charge. Charge flows from a higher potential to a lower potential.
The potential $V$ of a spherical conductor is given by $V = \frac{kQ}{R}$.
For the first sphere $(R_1 = 1 \, cm = 0.01 \, m)$:
$Q_1 = 1.5 \times 10^{-8} \, C$
$V_1 = \frac{k \times 1.5 \times 10^{-8}}{0.01} = k \times 1.5 \times 10^{-6} \, V$
For the second sphere $(R_2 = 2 \, cm = 0.02 \, m)$:
$Q_2 = 0.3 \times 10^{-7} = 3 \times 10^{-8} \, C$
$V_2 = \frac{k \times 3 \times 10^{-8}}{0.02} = k \times 1.5 \times 10^{-6} \, V$
Since $V_1 = V_2$,there is no potential difference between the two spheres. Therefore,no charge will flow between them.

(B) When two charged spheres are connected by a conducting wire,charge flows until both spheres reach the same electric potential,$V$.
The electric potential $V$ on the surface of a sphere of radius $r$ is given by $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}$.
The electric field $E$ on the surface of a sphere of radius $r$ is given by $E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}$.
By substituting $Q = V(4\pi\epsilon_0 r)$ into the electric field equation,we get $E = \frac{V}{r}$.
Since $V$ is the same for both spheres,the electric field $E$ is inversely proportional to the radius $r$ $(E \propto 1/r)$.
Therefore,the ratio of the electric fields is $\frac{E_a}{E_b} = \frac{b}{a}$.

(B) Initially,the charge on the first capacitor is $Q = CV$ and the second capacitor is uncharged $(Q_2 = 0)$.
When connected in parallel,the total charge $Q_{total} = CV + 0 = CV$ is shared between the two identical capacitors.
Since the capacitors are identical,the potential difference across both capacitors becomes equal,and the charge is distributed equally.
Therefore,the charge on each capacitor is $Q' = \frac{Q_{total}}{2} = \frac{CV}{2}$.

(B) Let the initial charge on each sphere be $q$. The total charge of the system is $q + q = 2q$.
When connected,the spheres reach a common potential $V$. The final charges on the spheres are $q_1 = C_1 V = \frac{R_1 V}{k}$ and $q_2 = C_2 V = \frac{R_2 V}{k}$,where $k = \frac{1}{4\pi\epsilon_0}$.
By the law of conservation of charge,the total initial charge equals the total final charge:
$2q = q_1 + q_2$
$2q = \frac{R_1 V}{k} + \frac{R_2 V}{k}$
$2q = \frac{V}{k}(R_1 + R_2)$
$q = \frac{V}{2k}(R_1 + R_2)$

(B) Initially,capacitor $C_1$ has charge $Q$ and capacitor $C_2$ is uncharged. When the switch $S$ is closed,the two capacitors are connected in parallel across each other (or effectively form a closed loop where charge redistributes until potentials are equal).
Since the capacitors are connected in parallel,they will have the same potential difference $V$ across them.
The total charge $Q_{total} = Q$ is conserved.
Let the final charges be $Q_1'$ and $Q_2'$.
$Q_1' = C_1 V$ and $Q_2' = C_2 V = 2C_1 V = 2Q_1'$.
Since $Q_1' + Q_2' = Q$,we have $Q_1' + 2Q_1' = Q$,which gives $3Q_1' = Q$.
Therefore,$Q_1' = Q/3$ and $Q_2' = 2Q/3$.

(B) The capacitance of a spherical conductor is given by $C = 4\pi \epsilon_0 R$.
When two spheres are connected by a wire,charge flows until they reach a common potential $V$.
The common potential is given by $V = \frac{Q_{total}}{C_1 + C_2}$.
Here,$Q_{total} = 150 \ \mu C = 150 \times 10^{-6} \ C$,$R_1 = 0.1 \ m$,and $R_2 = 0.2 \ m$.
The capacitances are $C_1 = 4\pi \epsilon_0 R_1$ and $C_2 = 4\pi \epsilon_0 R_2$.
Thus,$V = \frac{Q_{total}}{4\pi \epsilon_0 (R_1 + R_2)} = \frac{k \cdot Q_{total}}{R_1 + R_2}$,where $k = 9 \times 10^9 \ N \cdot m^2/C^2$.
Substituting the values: $V = \frac{9 \times 10^9 \times 150 \times 10^{-6}}{0.1 + 0.2} = \frac{9 \times 150 \times 10^3}{0.3} = \frac{1350 \times 10^3}{0.3} = 4500 \times 10^3 \ V = 4.5 \times 10^6 \ V$.

(A) When two charged conductors are connected by a wire, charge flows until they reach a common potential $V$.
The common potential is given by the formula: $V = \frac{\text{Total Charge}}{\text{Total Capacitance}}$.
Total charge $Q_{total} = Q_1 + Q_2 = 150\, \mu C + 150\, \mu C = 300\, \mu C = 300 \times 10^{-6}\, C$.
The capacitance of a spherical conductor is $C = 4\pi \epsilon_0 R$.
Total capacitance $C_{total} = C_1 + C_2 = 4\pi \epsilon_0 (R_1 + R_2)$.
Given $R_1 = 0.2\, m$ and $R_2 = 0.1\, m$, so $R_1 + R_2 = 0.3\, m$.
$V = \frac{300 \times 10^{-6}}{4\pi \epsilon_0 (0.3)} = \frac{300 \times 10^{-6} \times 9 \times 10^9}{0.3}$.
$V = \frac{2700 \times 10^3}{0.3} = 9000 \times 10^3 = 9 \times 10^6\, V$.

(D) Given: $C_1 = 3 \ \mu F$,$C_2 = 6 \ \mu F$,$V_1 = 12 \ V$,$V_2 = 12 \ V$.
Since the capacitors are connected in parallel with their positive plates together and negative plates together,the common potential $V_{CM}$ is given by the formula:
$V_{CM} = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$
Substituting the values:
$V_{CM} = \frac{(3 \ \mu F \times 12 \ V) + (6 \ \mu F \times 12 \ V)}{3 \ \mu F + 6 \ \mu F}$
$V_{CM} = \frac{36 + 72}{9} = \frac{108}{9} = 12 \ V$.
Thus,the potential difference across each capacitor remains $12 \ V$.

(C) Initial charges on the capacitors are:
$Q_1 = C_1 V_1 = 3\ \mu F \times 300\ V = 900\ \mu C$
$Q_2 = C_2 V_2 = 2\ \mu F \times 200\ V = 400\ \mu C$
Since they are connected with opposite polarities,the net charge is $Q_{net} = Q_1 - Q_2 = 900\ \mu C - 400\ \mu C = 500\ \mu C$.
The common potential $V$ after connection is given by:
$V = \frac{Q_{net}}{C_1 + C_2} = \frac{500\ \mu C}{3\ \mu F + 2\ \mu F} = 100\ V$.
The final charge on the $3\ \mu F$ capacitor is $Q_1' = C_1 V = 3\ \mu F \times 100\ V = 300\ \mu C$.
The charge that flows through the wire is the change in charge on one of the capacitors:
$\Delta Q = Q_1 - Q_1' = 900\ \mu C - 300\ \mu C = 600\ \mu C$.

(C) When two capacitors are connected by a wire,charge flows until they reach a common potential,$V_{com}$.
After connection,the potential across both capacitors becomes the same $(V_{com})$.
The charge on the first capacitor is $Q'_1 = C_1 V_{com}$.
The charge on the second capacitor is $Q'_2 = C_2 V_{com}$.
The ratio of their charges is given by:
$\frac{Q'_1}{Q'_2} = \frac{C_1 V_{com}}{C_2 V_{com}} = \frac{C_1}{C_2}$.
Substituting the given values:
$\frac{Q'_1}{Q'_2} = \frac{2 \ \mu F}{5 \ \mu F} = \frac{2}{5}$.

(C) When two capacitors are connected in parallel,charge flows from the capacitor at higher potential to the one at lower potential until both reach a common potential $V$.
If the initial potentials are equal,i.e.,$V_1 = V_2$,then there is no potential difference between the plates when they are connected.
Consequently,no charge flows between the capacitors,and there is no redistribution of charge or energy.
Therefore,if $V_1 = V_2$,there will be no change in the state of the system.

(D) When two capacitors are connected in parallel,the total charge is conserved.
The formula for the common potential $V_{CM}$ is given by: $V_{CM} = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$.
Given values: $C_1 = 10 \ \mu F$,$V_1 = 50 \ V$,$V_2 = 0 \ V$ (uncharged),and $V_{CM} = 20 \ V$.
Substituting these values into the formula:
$20 = \frac{(10 \times 50) + (C_2 \times 0)}{10 + C_2}$
$20 = \frac{500}{10 + C_2}$
$20(10 + C_2) = 500$
$200 + 20 C_2 = 500$
$20 C_2 = 300$
$C_2 = \frac{300}{20} = 15 \ \mu F$.
Therefore,the capacitance of the second capacitor is $15 \ \mu F$.

(C) Initial charge on the $6\ \mu F$ capacitor is $Q = C_1 V_0 = 6\ \mu F \times 100\, V = 600\ \mu C$.
When connected in parallel,both capacitors reach a common potential $V$.
The charge on the capacitors will be $q_1 = C_1 V$ and $q_2 = C_2 V$.
The ratio of charges is $\frac{q_1}{q_2} = \frac{C_1 V}{C_2 V} = \frac{C_1}{C_2} = \frac{6}{14}$.
By conservation of charge,$q_1 + q_2 = Q = 600\ \mu C$.
$C_1 V + C_2 V = 600\ \mu C \Rightarrow V(6 + 14) = 600$.
$V = \frac{600}{20} = 30\, V$.
Thus,the ratio is $6/14$ and the potential is $30\, V$.

(A) The capacitance of a spherical conductor is given by $C = 4\pi\epsilon_0 R = \frac{R}{k}$,where $k = \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \ N \cdot m^2/C^2$.
Given: $R_1 = 0.2 \ m$,$R_2 = 0.1 \ m$,$q_1 = 150 \times 10^{-6} \ C$,$q_2 = 150 \times 10^{-6} \ C$.
When connected by a wire,the total charge $Q = q_1 + q_2 = 300 \times 10^{-6} \ C$ is shared,and the total capacitance is $C_{eq} = C_1 + C_2 = \frac{R_1 + R_2}{k}$.
The common potential $V$ is given by $V = \frac{Q}{C_{eq}} = \frac{q_1 + q_2}{(R_1 + R_2)/k} = \frac{k(q_1 + q_2)}{R_1 + R_2}$.
Substituting the values: $V = \frac{9 \times 10^9 \times (300 \times 10^{-6})}{0.2 + 0.1} = \frac{9 \times 10^9 \times 300 \times 10^{-6}}{0.3}$.
$V = \frac{2700 \times 10^3}{0.3} = 9000 \times 10^3 = 9 \times 10^6 \ V$.

(A) Initial charges on the capacitors are $Q_1 = CV$ and $Q_2 = (2C)(2V) = 4CV$.
Since they are 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 system in parallel is $C_{eq} = C + 2C = 3C$.
The common potential is $V_{com} = \frac{Q_{net}}{C_{eq}} = \frac{3CV}{3C} = V$.
The final energy stored in the system is $U = \frac{1}{2} C_{eq} V_{com}^2 = \frac{1}{2} (3C) (V)^2 = \frac{3}{2} CV^2$.

(C) Initial charges on the capacitors are $Q_1 = CV$ and $Q_2 = 2C(2V) = 4CV$.
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 parallel combination is $C_{eq} = C + 2C = 3C$.
The common potential $V'$ is given by $V' = Q_{net} / C_{eq} = 3CV / 3C = V$.
The final energy $U_f$ of the configuration is $U_f = \frac{1}{2} C_{eq} (V')^2 = \frac{1}{2} (3C) V^2 = \frac{3}{2} CV^2$.