Energy Stored in a Capacitor Questions in English

(B) When a capacitor is charged by a battery,the work done by the battery is $W = qV = CV^2$.
The energy stored in the capacitor is given by $U = \frac{1}{2} CV^2$.
Comparing the two expressions,we find that $U = \frac{W}{2}$.
Given that the work done $W = 200 \, J$,the energy stored is $U = \frac{200}{2} = 100 \, J$.
The remaining $100 \, J$ of energy is dissipated as heat in the circuit during the charging process.

(A) The energy density $u$ (energy per unit volume) stored in an electric field $E$ is given by the formula: $u = \frac{1}{2} \varepsilon_0 E^2$.
The total volume $V$ of the space between the plates of a parallel plate capacitor is the product of the area $A$ and the separation $d$,so $V = Ad$.
The total potential energy $U$ stored in the capacitor is the product of the energy density and the volume: $U = u \times V$.
Substituting the expressions,we get: $U = (\frac{1}{2} \varepsilon_0 E^2) \times (Ad)$.
Therefore,the potential energy stored is $U = \frac{1}{2} \varepsilon_0 E^2 Ad$.

(D) Since the capacitors are connected in parallel to the same battery,the potential difference $V$ across both capacitors is the same.
From the charge-time graph,we observe the maximum charge $q$ stored on the capacitors. As $t \to \infty$,the charge on the capacitor reaches its maximum value $q = CV$.
From the graph,it is clear that the steady-state charge $q_2$ for capacitor $C_2$ is greater than the steady-state charge $q_1$ for capacitor $C_1$ (i.e.,$q_2 > q_1$).
Since $q = CV$ and $V$ is constant,$q \propto C$. Therefore,$C_2 > C_1$.
The energy stored in a capacitor is given by $U = \frac{1}{2} CV^2 = \frac{q^2}{2C}$. Alternatively,using $U = \frac{1}{2} qV$,since $V$ is constant and $q_2 > q_1$,it follows that $U_2 > U_1$.
Thus,$C_2 > C_1$ and $U_2 > U_1$.

(A) Given: Capacitance $C = 1 \ \mu F = 10^{-6} \ F$,Potential difference $V = 20 \ V$,Distance $d = 1 \ \mu m = 10^{-6} \ m$.
The electric field $E$ between the plates is given by $E = \frac{V}{d} = \frac{20}{10^{-6}} = 20 \times 10^6 \ V/m$.
The energy density $u$ is given by the formula $u = \frac{1}{2} \epsilon_0 E^2$.
Substituting the values: $u = \frac{1}{2} \times (8.854 \times 10^{-12}) \times (20 \times 10^6)^2$.
$u = 0.5 \times 8.854 \times 10^{-12} \times 400 \times 10^{12}$.
$u = 0.5 \times 8.854 \times 400 = 1770.8 \ J/m^3 \approx 1.8 \times 10^3 \ J/m^3$.

(C) The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$.
Since the battery is removed,the charge $q$ on the plates remains constant.
The energy stored in a capacitor is given by $U = \frac{q^2}{2C} = \frac{q^2 d}{2 \epsilon_0 A}$.
When the separation distance $d$ is doubled $(d' = 2d)$,the new capacitance becomes $C' = \frac{\epsilon_0 A}{2d} = \frac{C}{2}$.
The new energy stored is $U' = \frac{q^2}{2C'} = \frac{q^2}{2(C/2)} = 2 \times \frac{q^2}{2C} = 2U$.
Therefore,the energy stored becomes $2U$.

(B) The work done in changing the potential difference across a capacitor is given by the change in stored energy: $\Delta U = \frac{1}{2} C(V_f^2 - V_i^2).$
In the first case,the potential changes from $15 \ V$ to $30 \ V$:
$W = \frac{1}{2} C(30^2 - 15^2) = \frac{1}{2} C(900 - 225) = \frac{1}{2} C(675) = 337.5 C.$
So,$C = \frac{W}{337.5}.$
In the second case,the potential changes from $30 \ V$ to $60 \ V$:
$W' = \frac{1}{2} C(60^2 - 30^2) = \frac{1}{2} C(3600 - 900) = \frac{1}{2} C(2700) = 1350 C.$
Substituting $C = \frac{W}{337.5}$ into the equation for $W'$:
$W' = 1350 \times \left( \frac{W}{337.5} \right) = 4W.$

(A) Consider the circuit where a capacitor of capacitance $C$ is charged by a battery of voltage $V$.
When the capacitor is fully charged,the charge on it is $Q = CV$.
The energy stored in the capacitor is $E_{\text{capacitor}} = \frac{1}{2} CV^2$.
The total work done by the battery in supplying this charge is $W = QV = (CV)V = CV^2$.
The energy lost as heat in the circuit is the difference between the work done by the battery and the energy stored in the capacitor:
$E_{\text{loss}} = W - E_{\text{capacitor}} = CV^2 - \frac{1}{2} CV^2 = \frac{1}{2} CV^2$.
Comparing this to the total energy supplied by the battery $(W = CV^2)$:
$\text{Percentage loss} = \frac{E_{\text{loss}}}{W} \times 100\% = \frac{\frac{1}{2} CV^2}{CV^2} \times 100\% = 50\%$.
Thus,$50\%$ of the energy supplied by the battery is lost.

(A) Initial capacitance $C_i = \frac{\epsilon_0 A}{d}$.
Initial charge $Q = C_i V = \frac{\epsilon_0 A V}{d}$.
Since the battery is disconnected, the charge $Q$ remains constant.
Final separation $d_f = 4d$.
Final capacitance $C_f = \frac{\epsilon_0 A}{4d} = \frac{C_i}{4}$.
Initial energy $U_i = \frac{Q^2}{2C_i} = \frac{1}{2} C_i V^2 = \frac{\epsilon_0 A V^2}{2d}$.
Final energy $U_f = \frac{Q^2}{2C_f} = \frac{Q^2}{2(C_i/4)} = 4 \left( \frac{Q^2}{2C_i} \right) = 4 U_i$.
Work done $W = U_f - U_i = 4 U_i - U_i = 3 U_i$.
Substituting $U_i$, $W = 3 \left( \frac{\epsilon_0 A V^2}{2d} \right) = \frac{3 \epsilon_0 A V^2}{2d}$.

(C) The energy stored in a capacitor is given by $U = \frac{Q^2}{2C}$,where $Q$ is the charge and $C$ is the capacitance.
Let the original charge be $Q$. The original energy is $U_1 = \frac{Q^2}{2C}$.
When the charge is increased by $3 \ C$,the new charge is $Q' = Q + 3$.
The new energy is $U_2 = \frac{(Q+3)^2}{2C}$.
Given that the energy increases by $21 \%$,we have $U_2 = U_1 + 0.21 U_1 = 1.21 U_1$.
Substituting the expressions for $U_1$ and $U_2$:
$\frac{(Q+3)^2}{2C} = 1.21 \times \frac{Q^2}{2C}$.
$(Q+3)^2 = 1.21 Q^2$.
Taking the square root on both sides:
$Q+3 = 1.1 Q$.
$3 = 1.1 Q - Q$.
$3 = 0.1 Q$.
$Q = \frac{3}{0.1} = 30 \ C$.
Therefore,the original charge on the capacitor is $30 \ C$.

(A) Let the original charge on the capacitor be $Q$ and the capacitance be $C$. The energy stored in the capacitor is given by $U = \frac{Q^2}{2C}$.
When the charge is increased by $2 \text{ C}$, the new charge becomes $Q' = Q + 2$.
The new energy stored is $U' = \frac{(Q + 2)^2}{2C}$.
Given that the energy increases by $21 \%$, we have $U' = U + 0.21U = 1.21U$.
Substituting the expressions for $U$ and $U'$, we get $\frac{(Q + 2)^2}{2C} = 1.21 \times \frac{Q^2}{2C}$.
Canceling $\frac{1}{2C}$ from both sides, we get $(Q + 2)^2 = 1.21Q^2$.
Taking the square root of both sides, $Q + 2 = 1.1Q$.
Rearranging the terms, $0.1Q = 2$, which gives $Q = \frac{2}{0.1} = 20 \text{ C}$.

(D) The energy stored in a capacitor is given by $U = \frac{1}{2} CV^2$.
The work done $W$ in changing the potential difference from $V_i$ to $V_f$ is the change in potential energy: $W = \Delta U = \frac{1}{2} C(V_f^2 - V_i^2)$.
For the first case,$V_1 = 5 \ V$ to $V_2 = 10 \ V$:
$W = \frac{1}{2} C(10^2 - 5^2) = \frac{1}{2} C(100 - 25) = \frac{75}{2} C$.
For the second case,$V_2 = 10 \ V$ to $V_3 = 15 \ V$:
$W' = \frac{1}{2} C(15^2 - 10^2) = \frac{1}{2} C(225 - 100) = \frac{125}{2} C$.
Taking the ratio:
$\frac{W'}{W} = \frac{125/2 C}{75/2 C} = \frac{125}{75} = \frac{5}{3} \approx 1.67$.
Therefore,$W' = 1.67 \ W$.

(A) For a parallel plate capacitor,the energy density $u = \frac{1}{2} \varepsilon_0 E^2$.
Since the electric field $E$ between the plates is given by $E = \frac{\sigma}{\varepsilon_0}$,where $\sigma = \frac{q}{A}$ is the surface charge density.
Substituting the value of $E$ in the energy density formula:
$u = \frac{1}{2} \varepsilon_0 \left( \frac{\sigma}{\varepsilon_0} \right)^2 = \frac{\sigma^2}{2 \varepsilon_0}$.
Now,substituting $\sigma = \frac{q}{A}$:
$u = \frac{(q/A)^2}{2 \varepsilon_0} = \frac{q^2}{2 \varepsilon_0 A^2}$.

(D) Initial capacitance $C_0 = \frac{\varepsilon_0 A}{d}$.
Initial charge $Q = C_0 V_0$.
Since the battery is disconnected,the charge $Q$ remains constant.
Final separation $d' = 3d$.
Final capacitance $C' = \frac{\varepsilon_0 A}{3d} = \frac{C_0}{3}$.
Initial potential energy $U_i = \frac{Q^2}{2C_0} = \frac{1}{2} C_0 V_0^2$.
Final potential energy $U_f = \frac{Q^2}{2C'} = \frac{Q^2}{2(C_0/3)} = \frac{3Q^2}{2C_0} = \frac{3}{2} C_0 V_0^2$.
Work done $W = U_f - U_i = \frac{3}{2} C_0 V_0^2 - \frac{1}{2} C_0 V_0^2 = C_0 V_0^2$.
Substituting $C_0 = \frac{\varepsilon_0 A}{d}$,we get $W = \frac{\varepsilon_0 A V_0^2}{d}$.

(B) The energy stored in a capacitor is given by $U = \frac{Q^2}{2C}$.
Since $C$ is constant,$U \propto Q^2$.
Let the initial charge be $Q_1$ and the initial energy be $U_1$.
When the charge is increased by $3 \ C$,the new charge is $Q_2 = Q_1 + 3$.
The new energy $U_2$ is $U_1 + 44\% \text{ of } U_1 = 1.44 \ U_1$.
Using the proportionality $U \propto Q^2$,we have:
$\frac{U_2}{U_1} = \left( \frac{Q_2}{Q_1} \right)^2$
$1.44 = \left( \frac{Q_1 + 3}{Q_1} \right)^2$
Taking the square root on both sides:
$1.2 = \frac{Q_1 + 3}{Q_1}$
$1.2 \ Q_1 = Q_1 + 3$
$0.2 \ Q_1 = 3$
$Q_1 = \frac{3}{0.2} = 15 \ C$.

(D) The circuit consists of two capacitors in parallel connected in series with two other capacitors.
First,calculate the equivalent capacitance of the two parallel capacitors:
$C_{\|} = 2 \mu F + 2 \mu F = 4 \mu F$
Now,the circuit is equivalent to three capacitors in series: $2 \mu F$,$4 \mu F$,and $2 \mu F$.
The equivalent capacitance $C_{eq}$ is given by:
$\frac{1}{C_{eq}} = \frac{1}{2 \mu F} + \frac{1}{4 \mu F} + \frac{1}{2 \mu F} = \frac{2 + 1 + 2}{4 \mu F} = \frac{5}{4 \mu F}$
$C_{eq} = \frac{4}{5} \mu F = 0.8 \times 10^{-6} \,F$
The energy stored $U$ is given by:
$U = \frac{1}{2} C_{eq} V^2$
$U = \frac{1}{2} \times (0.8 \times 10^{-6} \,F) \times (10 \,V)^2$
$U = 0.4 \times 10^{-6} \times 100 \,J = 40 \times 10^{-6} \,J = 400 \times 10^{-7} \,J$

(D) Initial capacitance $C = \frac{\varepsilon_0 A}{d}$. Initial charge $q = CV = \frac{\varepsilon_0 A V}{d}$.
Initial energy $U_i = \frac{q^2}{2C} = \frac{1}{2} CV^2 = \frac{\varepsilon_0 A V^2}{2d}$.
When the battery is disconnected,the charge $q$ remains constant.
The new separation is $d' = 4d$,so the new capacitance is $C' = \frac{\varepsilon_0 A}{4d} = \frac{C}{4}$.
Final energy $U_f = \frac{q^2}{2C'} = \frac{q^2}{2(C/4)} = \frac{4q^2}{2C} = 4U_i$.
Work done $W = U_f - U_i = 4U_i - U_i = 3U_i$.
Substituting $U_i = \frac{\varepsilon_0 A V^2}{2d}$,we get $W = 3 \times \frac{\varepsilon_0 A V^2}{2d} = \frac{3 \varepsilon_0 A V^2}{2d}$.

(C) The energy density $u$ in a parallel plate capacitor is given by $u = \frac{1}{2} \varepsilon_0 E^2$.
The volume $V$ of the space between the plates is $V = A \times d$.
The total energy $U$ stored in the capacitor is the product of energy density and volume.
$U = u \times V = (\frac{1}{2} \varepsilon_0 E^2) \times (A d) = \frac{1}{2} \varepsilon_0 E^2 A d$.

(D) The energy stored in a capacitor is given by the formula $W = \frac{1}{2} CV^2$, where $C$ is the capacitance and $V$ is the potential difference.
Initial energy $W_1 = \frac{1}{2} CV_1^2$, where $V_1 = 5 \,V$.
Final energy $W_2 = \frac{1}{2} CV_2^2$, where $V_2 = 15 \,V$.
The ratio of final energy to initial energy is $\frac{W_2}{W_1} = \frac{\frac{1}{2} CV_2^2}{\frac{1}{2} CV_1^2} = \left(\frac{V_2}{V_1}\right)^2$.
Substituting the values, we get $\frac{W_2}{W_1} = \left(\frac{15}{5}\right)^2 = (3)^2 = 9$.
Therefore, the ratio is $9 : 1$.

(B) Let the capacitance of the capacitor be $C$ and the e.m.f. of the battery be $V$.
When the capacitor is fully charged,the potential difference across the plates is $V$.
The charge stored on the capacitor is $q = CV$.
The energy stored in the capacitor is $U = \frac{1}{2} CV^2 = \frac{1}{2} qV$.
The work done by the battery in supplying charge $q$ is $W = qV = CV^2$.
The ratio of the energy stored in the capacitor to the work done by the battery is $\frac{U}{W} = \frac{\frac{1}{2} qV}{qV} = \frac{1}{2}$.

(B) The energy density $u$ of an electric field $E$ is given by $u = \frac{1}{2} \epsilon_{0} E^{2}$.
The volume $V$ of the space between the plates of a parallel plate capacitor is the product of the area $A$ and the distance $d$,so $V = Ad$.
The total energy $U$ stored in the capacitor is the product of the energy density and the volume.
Therefore,$U = u \times V = (\frac{1}{2} \epsilon_{0} E^{2}) \times (Ad) = \frac{1}{2} \epsilon_{0} E^{2} Ad$.

(D) The energy stored in a capacitor is given by $U = \frac{1}{2} C V^2$.
The work done in increasing the potential from $V_i$ to $V_f$ is $\Delta U = \frac{1}{2} C (V_f^2 - V_i^2)$.
For the first case,$W = \frac{1}{2} C (10^2 - 5^2) = \frac{1}{2} C (100 - 25) = \frac{75}{2} C$.
For the second case,$W_2 = \frac{1}{2} C (15^2 - 10^2) = \frac{1}{2} C (225 - 100) = \frac{125}{2} C$.
Taking the ratio: $\frac{W_2}{W} = \frac{125/2 C}{75/2 C} = \frac{125}{75} = \frac{5}{3} \approx 1.67$.
Therefore,$W_2 = 1.67 W$.

(A) The energy density $u$ (energy per unit volume) of a capacitor is given by the formula $u = \frac{1}{2} \varepsilon_{0} E^{2}$.
Since the electric field $E$ between the plates of a capacitor is related to the potential difference $V$ and separation $d$ by $E = \frac{V}{d}$,
Substituting this into the energy density formula,we get $u = \frac{1}{2} \varepsilon_{0} (\frac{V}{d})^{2} = \frac{1}{2} \varepsilon_{0} \frac{V^{2}}{d^{2}}$.

(A) The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$.
Given: $C = 48 \mu F = 48 \times 10^{-6} \ F$,$q_1 = 0.1 \ C$,$q_2 = 0.5 \ C$.
The change in energy $\Delta U$ is given by:
$\Delta U = U_2 - U_1 = \frac{q_2^2}{2C} - \frac{q_1^2}{2C} = \frac{1}{2C} (q_2^2 - q_1^2)$
Substituting the values:
$\Delta U = \frac{1}{2 \times 48 \times 10^{-6}} ((0.5)^2 - (0.1)^2)$
$\Delta U = \frac{1}{96 \times 10^{-6}} (0.25 - 0.01)$
$\Delta U = \frac{0.24}{96 \times 10^{-6}}$
$\Delta U = \frac{0.24 \times 10^6}{96} = \frac{240000}{96} = 2500 \ J$.

(D) The energy stored in a parallel plate capacitor is given by the formula $U = \frac{1}{2} C V^{2}$.
Here,the capacitance $C = \frac{\varepsilon_{0} A}{d}$.
The potential difference $V$ between the plates is related to the electric field $E$ and separation $d$ by $V = E d$.
Substituting these values into the energy formula:
$U = \frac{1}{2} \left( \frac{\varepsilon_{0} A}{d} \right) (E d)^{2}$
$U = \frac{1}{2} \left( \frac{\varepsilon_{0} A}{d} \right) (E^{2} d^{2})$
$U = \frac{1}{2} \varepsilon_{0} E^{2} A d$.

(B) The two capacitors are connected in parallel to the battery.
In a parallel combination,the equivalent capacitance $C_{eq}$ is the sum of individual capacitances:
$C_{eq} = C + \frac{C}{2} = \frac{3C}{2}$
The work done $W$ by the battery to charge the capacitors is equal to the energy stored in the equivalent capacitor:
$W = \frac{1}{2} C_{eq} V^2$
Substituting the value of $C_{eq}$:
$W = \frac{1}{2} \left( \frac{3C}{2} \right) V^2$
$W = \frac{3}{4} CV^2$

(A) The energy density (energy per unit volume) in an electric field is given by $u = \frac{1}{2} \epsilon_{0} E^{2}$.
The volume of the space between the plates is $V = Ad$.
Therefore,the total energy stored in the capacitor is $U = u \times V = \left( \frac{1}{2} \epsilon_{0} E^{2} \right) \times (Ad) = \frac{1}{2} \epsilon_{0} E^{2} Ad$.

(C) The energy stored in a capacitor is given by the formula $U = \frac{1}{2} CV^2$.
Given for the first capacitor: $C_1 = 900 \mu F$ and $V_1 = 100 \ V$.
The energy stored in the first capacitor is $U_1 = \frac{1}{2} C_1 V_1^2$.
When this energy is transferred to a second capacitor of capacity $C_2 = 100 \mu F$, let the new potential be $V_u$.
Since the total energy is transferred, $U_1 = U_2$, where $U_2 = \frac{1}{2} C_2 V_u^2$.
Equating the energies: $\frac{1}{2} C_1 V_1^2 = \frac{1}{2} C_2 V_u^2$.
Substituting the values: $\frac{1}{2} \times 900 \times 10^{-6} \times (100)^2 = \frac{1}{2} \times 100 \times 10^{-6} \times V_u^2$.
Simplifying the equation: $900 \times 100^2 = 100 \times V_u^2$.
$V_u^2 = \frac{900 \times 10000}{100} = 90000$.
$V_u = \sqrt{90000} = 300 \ V$.

(A) The work done $W$ in charging a capacitor is equal to the potential energy $U$ stored in it.
The formula for energy stored in a capacitor is $W = U = \frac{Q^2}{2C}$.
Given:
Charge $Q = 8 \times 10^{-18} \text{ C}$
Capacitance $C = 800 \mu\text{F} = 800 \times 10^{-6} \text{ F} = 8 \times 10^{-4} \text{ F}$.
Substituting the values:
$W = \frac{(8 \times 10^{-18})^2}{2 \times 8 \times 10^{-4}}$
$W = \frac{64 \times 10^{-36}}{16 \times 10^{-4}}$
$W = 4 \times 10^{-32} \text{ J}$.

(A) Given:
Capacitance $C = 12 \text{ pF} = 12 \times 10^{-12} \text{ F}$
Potential difference $V = 50 \text{ V}$
The electrostatic energy $U$ stored in a capacitor is given by the formula:
$U = \frac{1}{2} C V^2$
Substituting the given values:
$U = \frac{1}{2} \times (12 \times 10^{-12} \text{ F}) \times (50 \text{ V})^2$
$U = 6 \times 10^{-12} \times 2500$
$U = 15000 \times 10^{-12} \text{ J}$
$U = 1.5 \times 10^4 \times 10^{-12} \text{ J}$
$U = 1.5 \times 10^{-8} \text{ J}$
Therefore,the energy stored is $1.5 \times 10^{-8} \text{ J}$.

(B) The energy $U$ stored in a capacitor is given by the formula $U = \frac{1}{2} CV^2$.
Given:
Capacitance $C = 12 \ pF = 12 \times 10^{-12} \ F$
Potential difference $V = 50 \ V$
Substituting the values into the formula:
$U = \frac{1}{2} \times (12 \times 10^{-12} \ F) \times (50 \ V)^2$
$U = 6 \times 10^{-12} \times 2500 \ J$
$U = 15000 \times 10^{-12} \ J$
$U = 1.5 \times 10^4 \times 10^{-12} \ J$
$U = 1.5 \times 10^{-8} \ J$
Therefore,the correct option is $B$.

(A) Given: Capacitance $C = 5 \mu \text{F} = 5 \times 10^{-6} \text{ F}$.
Potential difference $V = 4 \text{ V}$.
Rate of change of potential difference $\frac{dV}{dt} = 0.6 \text{ Vs}^{-1}$.
The energy stored in a capacitor is given by $U = \frac{1}{2}CV^2$.
To find the time rate at which energy is stored,we differentiate $U$ with respect to time $t$:
$\frac{dU}{dt} = \frac{d}{dt} (\frac{1}{2}CV^2) = \frac{1}{2}C \cdot 2V \cdot \frac{dV}{dt} = CV \frac{dV}{dt}$.
Substituting the given values:
$\frac{dU}{dt} = (5 \times 10^{-6} \text{ F}) \times (4 \text{ V}) \times (0.6 \text{ Vs}^{-1})$.
$\frac{dU}{dt} = 20 \times 0.6 \times 10^{-6} \text{ W} = 12 \times 10^{-6} \text{ W} = 12 \mu \text{W}$.

(B) The energy stored in a capacitor is given by the formula:
$U = \frac{1}{2} C V^{2} \dots (i)$
where $C$ is the capacitance and $V$ is the potential difference.
From the given graph,the slope is:
$\text{Slope} = \tan \theta = \frac{Q}{V} = \frac{120 \mu C}{10 \text{ V}} = 12 \mu F \dots (ii)$
Since the capacitance $C = \frac{Q}{V}$,we have:
$C = 12 \mu F = 12 \times 10^{-6} \text{ F}$
Now,substituting the values of $C = 12 \times 10^{-6} \text{ F}$ and $V = 10 \text{ V}$ into Eq. $(i)$:
$U = \frac{1}{2} \times (12 \times 10^{-6} \text{ F}) \times (10 \text{ V})^{2}$
$U = \frac{1}{2} \times 12 \times 10^{-6} \times 100$
$U = 6 \times 10^{-4} \text{ J} = 600 \times 10^{-6} \text{ J} = 600 \mu J$
Thus,the capacitance is $12 \mu F$ and the energy stored is $600 \mu J$.

(D) The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$.
Let the initial charge be $q_1 = q$ and the final charge be $q_2 = q + 2 \text{ C}$.
The initial energy is $U_1 = \frac{q^2}{2C}$ and the final energy is $U_2 = \frac{(q+2)^2}{2C}$.
Given that the energy increases by $21 \%$,we have $U_2 = U_1 + 0.21 U_1 = 1.21 U_1$.
Substituting the expressions for $U_1$ and $U_2$:
$\frac{(q+2)^2}{2C} = 1.21 \times \frac{q^2}{2C}$.
Canceling $\frac{1}{2C}$ from both sides,we get $(q+2)^2 = 1.21 q^2$.
Taking the square root of both sides: $q + 2 = 1.1 q$.
Rearranging the terms: $1.1 q - q = 2$,which gives $0.1 q = 2$.
Therefore,$q = \frac{2}{0.1} = 20 \text{ C}$.

(C) The energy $U$ stored in a capacitor is given by the formula:
$U = \frac{1}{2} CV^2$
Given:
Capacitance $C = 10 \mu F = 10 \times 10^{-6} \text{ F}$
Potential difference $V = 10 \text{ V}$
Substituting the values into the formula:
$U = \frac{1}{2} \times (10 \times 10^{-6} \text{ F}) \times (10 \text{ V})^2$
$U = \frac{1}{2} \times 10 \times 10^{-6} \times 100$
$U = 500 \times 10^{-6} \text{ J}$
$U = 500 \mu J$

(C) The $6 \mu F$ and $3 \mu F$ capacitors are connected in series.
Therefore,the equivalent capacitance $C_1$ is given by $\frac{1}{C_1} = \frac{1}{6} + \frac{1}{3} = \frac{1+2}{6} = \frac{3}{6} = \frac{1}{2}$.
Thus,$C_1 = 2 \mu F$.
This equivalent capacitor $C_1$ is in parallel with the $2 \mu F$ capacitor.
Therefore,the total equivalent capacitance of the system is $C_{eq} = C_1 + 2 \mu F = 2 \mu F + 2 \mu F = 4 \mu F$.
The total energy $U$ stored in the system is given by $U = \frac{1}{2} C_{eq} V^2$.
Given $V = 2 \text{ V}$,we have $U = \frac{1}{2} \times 4 \mu F \times (2 \text{ V})^2 = 2 \times 4 = 8 \mu J$.

(A) The energy stored in a capacitor is given by the formula $U = \frac{1}{2} C V^{2}$.
Since $\frac{1}{2}$ is a dimensionless constant,the dimensional formula of $C V^{2}$ is equivalent to the dimensional formula of energy $(U)$.
The dimensional formula for energy is $[Work] = [Force \times Displacement] = [MLT^{-2} \times L] = [ML^{2} T^{-2}]$.
Therefore,the dimensional formula of $C V^{2}$ is $[ML^{2} T^{-2} A^{0}]$.

(D) The energy stored in a capacitor with capacitance $C$ and charge $Q$ is given by $W = \frac{Q^2}{2C}$.
When the charge is doubled,the new charge becomes $Q' = 2Q$.
The new energy stored in the capacitor is $W' = \frac{(2Q)^2}{2C} = \frac{4Q^2}{2C} = 4W$.
The additional work to be done is the change in energy,which is $\Delta W = W' - W$.
Substituting the values,we get $\Delta W = 4W - W = 3W$.

(C) The energy $U$ stored in a capacitor is given by the formula $U = \frac{1}{2} C V^2$.
Given:
Capacitance $C = 10 \mu F = 10 \times 10^{-6} \text{ F} = 10^{-5} \text{ F}$.
Potential $V = 6 \text{ kV} = 6 \times 10^3 \text{ V}$.
Substituting the values into the formula:
$U = \frac{1}{2} \times (10^{-5} \text{ F}) \times (6 \times 10^3 \text{ V})^2$
$U = \frac{1}{2} \times 10^{-5} \times 36 \times 10^6$
$U = 18 \times 10^1 \text{ J} = 180 \text{ J}$.
Therefore,the energy stored is $180 \text{ J}$.

(C) The energy stored in a capacitor is given by $E = \frac{1}{2} C V^2$.
Since the battery remains connected,the potential difference $V$ across the plates remains constant.
The capacitance of a parallel plate capacitor is $C = \frac{A \epsilon_0}{d}$,which implies $C \propto \frac{1}{d}$.
If the distance $d$ is doubled $(d' = 2d)$,the new capacitance $C'$ becomes $C' = \frac{C}{2}$.
The new energy $E'$ is given by $E' = \frac{1}{2} C' V^2$.
Substituting $C' = \frac{C}{2}$,we get $E' = \frac{1}{2} (\frac{C}{2}) V^2 = \frac{1}{2} (\frac{1}{2} C V^2) = \frac{E}{2}$.

(B) Let $V$ be the potential across the capacitor when it is fully charged. The energy stored in the capacitor is given by $U = \frac{1}{2} C V^2$.
When the capacitor is fully discharged through the resistance coil,the entire stored energy is dissipated as heat $\Delta H$ in the block.
Since the system is thermally isolated,the heat gained by the block is equal to the energy lost by the capacitor: $\Delta H = \frac{1}{2} C V^2$.
The heat gained by the block is also given by $\Delta H = m s \Delta T$,where $m$ is the mass and $s$ is the specific heat.
Equating the two expressions for $\Delta H$:
$\frac{1}{2} C V^2 = m s \Delta T$
$V^2 = \frac{2 m s \Delta T}{C}$
$V = \left(\frac{2 m s \Delta T}{C}\right)^{1 / 2}$

(B) The energy stored in a charged body (like a capacitor) is given by $E = \frac{q^2}{2C}$,where $q$ is the charge and $C$ is the capacitance. Thus,$E \propto q^2$.
Let the original charge be $q_1 = q$ and the original energy be $E_1 = E$.
When the charge is increased by $2 \ C$,the new charge is $q_2 = q + 2$.
The new energy $E_2$ increases by $21 \%$,so $E_2 = E + 0.21E = 1.21E$.
Using the ratio: $\frac{E_2}{E_1} = \left(\frac{q_2}{q_1}\right)^2$.
Substituting the values: $\frac{1.21E}{E} = \left(\frac{q+2}{q}\right)^2$.
$1.21 = \left(\frac{q+2}{q}\right)^2$.
Taking the square root on both sides: $1.1 = \frac{q+2}{q}$.
$1.1q = q + 2$.
$0.1q = 2$.
$q = \frac{2}{0.1} = 20 \ C$.

(B) The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$,where $q$ is the charge and $C$ is the capacitance.
Let the initial charge be $q_i = q$ and the final charge be $q_f = q + 2$.
The initial energy is $U_i = \frac{q^2}{2C}$ and the final energy is $U_f = \frac{(q+2)^2}{2C}$.
Given that the energy increases by $21 \%$,we have $U_f = U_i + 0.21 U_i = 1.21 U_i$.
Substituting the expressions for energy: $\frac{(q+2)^2}{2C} = 1.21 \times \frac{q^2}{2C}$.
Canceling $\frac{1}{2C}$ from both sides,we get $(q+2)^2 = 1.21 q^2$.
Taking the square root of both sides: $q + 2 = 1.1 q$.
Rearranging the terms: $1.1 q - q = 2$,which gives $0.1 q = 2$.
Therefore,$q = \frac{2}{0.1} = 20 \ C$.