Energy Stored in a Capacitor Questions in English

(D) The energy stored in a capacitor is given by the formula $U = \frac{1}{2}CV^2$.
Since $\frac{1}{2}$ is a dimensionless constant,the dimensions of $CV^2$ are the same as the dimensions of energy.
Energy is defined as force multiplied by displacement,so its dimensional formula is $[M L T^{-2}] \times [L] = [M L^2 T^{-2}]$.
Therefore,the dimensions of $CV^2$ are $[M L^2 T^{-2}]$.

(B) The electrostatic energy density $u$ is defined as the energy stored per unit volume in an electric field.
The formula for electrostatic energy density is given by $u = \frac{dU}{dV} = \frac{1}{2} \epsilon_0 E^2$,where $\epsilon_0$ is the permittivity of free space and $E$ is the electric field intensity.
From this expression,it is clear that the electrostatic energy density is directly proportional to the square of the electric field intensity,i.e.,$u \propto E^2$.

(A) The energy $U$ stored in a capacitor is given by the formula $U = \frac{1}{2}CV^2$.
Since the charge $q$ on the capacitor is given by $q = CV$,we can write $V = \frac{q}{C}$.
Substituting this value of $V$ into the energy formula:
$U = \frac{1}{2}C\left(\frac{q}{C}\right)^2 = \frac{1}{2}C\left(\frac{q^2}{C^2}\right) = \frac{q^2}{2C}$.
Therefore,the correct expression is $\frac{q^2}{2C}$.

(A) The energy stored in a capacitor is given by the formula $U = \frac{1}{2} C V^2$.
Given:
Capacitance $C = 4 \times 10^{-6} \, F$
Potential $V = 100 \, V$
Substituting the values into the formula:
$U = \frac{1}{2} \times (4 \times 10^{-6}) \times (100)^2$
$U = \frac{1}{2} \times 4 \times 10^{-6} \times 10000$
$U = 2 \times 10^{-6} \times 10^4$
$U = 2 \times 10^{-2} \, J$
$U = 0.02 \, J$
Therefore,the energy released is $0.02 \, J$.

(C) The energy stored in a capacitor is given by the expression $U = \frac{1}{2} CV^2$ or $U = \frac{1}{2} \frac{Q^2}{C}$.
In terms of the electric field $E$ between the plates,the energy density $u$ is given by $u = \frac{1}{2} \epsilon_0 E^2$.
Since the electric field outside an ideal parallel plate capacitor is zero and the field exists only in the region between the plates,the total energy resides in the electric field between the plates.

(B) The energy stored in a capacitor is the work done in charging it.
If a capacitor of capacity $C$ is charged to a potential $V$,the work done $dW$ to bring an additional charge $dq$ is given by $dW = V' dq$,where $V' = q/C$.
Integrating this from $0$ to $Q$ (where $Q = CV$):
$U = \int_{0}^{Q} \frac{q}{C} dq = \frac{1}{C} \left[ \frac{q^2}{2} \right]_{0}^{Q} = \frac{Q^2}{2C}$.
Substituting $Q = CV$,we get $U = \frac{(CV)^2}{2C} = \frac{1}{2} CV^2$.

(A) The energy $U$ stored in a capacitor is given by the formula $U = \frac{1}{2}CV^2$.
Given:
Capacitance $C = 50\,\mu F = 50 \times 10^{-6}\,F$.
Potential difference $V = 10\;V$.
Substituting the values into the formula:
$U = \frac{1}{2} \times (50 \times 10^{-6}) \times (10)^2$
$U = \frac{1}{2} \times 50 \times 10^{-6} \times 100$
$U = 25 \times 10^{-4} = 2.5 \times 10^{-3}\,J$.
Therefore,the energy stored is $2.5 \times 10^{-3}\,J$.

(A) charged capacitor stores energy in the form of an electrostatic field between its plates.
Since the charges on the capacitor plates are stationary,they produce only an electric field.
There is no current flowing through the capacitor in a steady state,so there is no magnetic field.
Therefore,the energy of a charged capacitor resides in the electric field only.

(C) The energy stored in a capacitor is given by the formula $W = \frac{Q^2}{2C}$.
When the charge is increased to $2Q$,the new energy $W'$ becomes:
$W' = \frac{(2Q)^2}{2C} = \frac{4Q^2}{2C}$.
Since $W = \frac{Q^2}{2C}$,we can substitute this into the equation:
$W' = 4 \times \left( \frac{Q^2}{2C} \right) = 4W$.
Therefore,the stored energy becomes $4W$.

(B) The energy stored in a capacitor is given by $E = \frac{1}{2}CV^2$.
The increase in energy $\Delta E$ is given by $\Delta E = E_{Final} - E_{Initial} = \frac{1}{2}C(V_{Final}^2 - V_{Initial}^2)$.
Given: $C = 6\,\mu F = 6 \times 10^{-6}\,F$,$V_{Initial} = 10\,V$,$V_{Final} = 20\,V$.
Substituting the values:
$\Delta E = \frac{1}{2} \times (6 \times 10^{-6}) \times (20^2 - 10^2)$
$\Delta E = 3 \times 10^{-6} \times (400 - 100)$
$\Delta E = 3 \times 10^{-6} \times 300$
$\Delta E = 900 \times 10^{-6}\,J = 9 \times 10^{-4}\,J$.

(B) When a capacitor of capacitance $C$ is charged by a battery of $EMF$ $V$,the total work done by the battery is $W = QV = (CV)V = CV^2$.
However,the energy stored in the capacitor is given by $U = \frac{1}{2}CV^2$.
Comparing these,we see that the energy stored in the capacitor is exactly half of the total work done (energy supplied) by the battery.
The remaining half of the energy is dissipated as heat in the connecting wires or internal resistance of the circuit.

(C) The energy stored in a charged capacitor is given by the formula $U = \frac{1}{2}CV^2$.
When the plates of the capacitor are connected by a conducting wire,the entire stored energy is dissipated as heat.
Given: $C = 2\,\mu F = 2 \times 10^{-6}\,F$ and $V = 100\,V$.
Substituting the values into the formula:
$U = \frac{1}{2} \times (2 \times 10^{-6}) \times (100)^2$
$U = 10^{-6} \times 10^4$
$U = 10^{-2}\,J = 0.01\,J$.
Therefore,the heat produced is $0.01\,J$.

(B) The energy $U$ stored in a capacitor is given by the work done in charging it.
When a charge $Q$ is transferred to a capacitor of capacitance $C$ at a potential $V$,the average potential during the process is $\frac{V}{2}$.
Therefore,the energy stored $U = \text{average potential} \times \text{charge}$.
$U = \frac{1}{2}V \times Q = \frac{1}{2}QV$.

(C) The energy stored in a capacitor is given by the formula $U = \frac{1}{2} C V^2$.
Given:
Capacitance $C = 2.0 \; \mu F = 2.0 \times 10^{-6} \; F$
Potential difference $V = 200 \; V$
Substituting the values into the formula:
$U = \frac{1}{2} \times (2.0 \times 10^{-6} \; F) \times (200 \; V)^2$
$U = 1.0 \times 10^{-6} \times 40000$
$U = 4 \times 10^{-2} \; J$
When the capacitor is discharged through a resistance wire,the entire stored energy is dissipated as heat. Therefore,the heat produced is $4 \times 10^{-2} \; J$.

(A) The energy $U$ stored in a capacitor is given by the formula $U = \frac{1}{2} C V^2$.
Given:
Capacitance $C = 10 \, pF = 10 \times 10^{-12} \, F = 10^{-11} \, F$.
Potential difference $V = 50 \, V$.
Substituting the values into the formula:
$U = \frac{1}{2} \times (10 \times 10^{-12} \, F) \times (50 \, V)^2$
$U = \frac{1}{2} \times 10^{-11} \times 2500$
$U = 0.5 \times 2.5 \times 10^{-8} \, J$
$U = 1.25 \times 10^{-8} \, J$.
Therefore,the correct option is $A$.

(D) The energy $U$ stored in a capacitor is given by the formula $U = \frac{1}{2}CV^2$.
Given:
Capacitance $C = 700\,pF = 700 \times 10^{-12}\,F = 7 \times 10^{-10}\,F$.
Potential difference $V = 50\,V$.
Substituting the values:
$U = \frac{1}{2} \times (700 \times 10^{-12}) \times (50)^2$
$U = \frac{1}{2} \times 700 \times 10^{-12} \times 2500$
$U = 350 \times 2500 \times 10^{-12}$
$U = 875000 \times 10^{-12}\,J$
$U = 8.75 \times 10^{-7}\,J$.

(D) The energy stored in a capacitor is given by the formula $U = \frac{1}{2}CV^2$.
Since the capacitor is connected to a constant voltage source of $V = 100 \ V$,the change in energy $\Delta U$ is given by:
$\Delta U = U_2 - U_1 = \frac{1}{2}C_2V^2 - \frac{1}{2}C_1V^2 = \frac{1}{2}V^2(C_2 - C_1)$.
Given $C_1 = 2 \ \mu F = 2 \times 10^{-6} \ F$,$C_2 = 10 \ \mu F = 10 \times 10^{-6} \ F$,and $V = 100 \ V$.
Substituting the values:
$\Delta U = \frac{1}{2} \times (100)^2 \times (10 - 2) \times 10^{-6} \ J$
$\Delta U = \frac{1}{2} \times 10000 \times 8 \times 10^{-6} \ J$
$\Delta U = 5000 \times 8 \times 10^{-6} \ J = 40000 \times 10^{-6} \ J = 4 \times 10^{-2} \ J$.

(A) The electrostatic 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:
$U = \frac{1}{2} \times (12 \times 10^{-12}) \times (50)^2$
$U = 6 \times 10^{-12} \times 2500$
$U = 6 \times 2500 \times 10^{-12}$
$U = 15000 \times 10^{-12}\,J$
$U = 1.5 \times 10^4 \times 10^{-12}\,J$
$U = 1.5 \times 10^{-8}\,J$.

(A) The energy stored in a capacitor is given by the formula $U = \frac{1}{2}CV^2$.
Given energy $U = 24 \, \text{watt-hour}$.
Convert energy to Joules: $U = 24 \times 60 \times 60 \, \text{J} = 86400 \, \text{J}$.
Given voltage $V = 1200 \, \text{V}$.
Substituting the values into the formula: $86400 = \frac{1}{2} \times C \times (1200)^2$.
$86400 = \frac{1}{2} \times C \times 1440000$.
$86400 = C \times 720000$.
$C = \frac{86400}{720000} \, \text{F} = 0.12 \, \text{F}$.
Converting to millifarads: $C = 0.12 \times 1000 \, \text{mF} = 120 \, \text{mF}$.

(A) The electric energy density $u$ is defined as the energy per unit volume,given by $u = \frac{1}{2} \varepsilon_0 E^2$.
For a parallel plate capacitor,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 $E$ into the energy density formula:
$u = \frac{1}{2} \varepsilon_0 \left( \frac{\sigma}{\varepsilon_0} \right)^2 = \frac{\sigma^2}{2\varepsilon_0}$.
Substituting $\sigma = \frac{q}{A}$:
$u = \frac{(q/A)^2}{2\varepsilon_0} = \frac{q^2}{2\varepsilon_0 A^2}$.
Thus,the correct option is $A$.

(B) The energy stored in a capacitor is given by the formula $U = \frac{Q^2}{2C}$.
Given,$Q = 40\,\mu C = 40 \times 10^{-6}\,C$ and $C = 10\,\mu F = 10 \times 10^{-6}\,F$.
Substituting the values,$U = \frac{(40 \times 10^{-6})^2}{2 \times 10 \times 10^{-6}} = \frac{1600 \times 10^{-12}}{20 \times 10^{-6}} = 80 \times 10^{-6}\,J = 8 \times 10^{-5}\,J$.
Since $1\,J = 10^7\,erg$,the energy in ergs is $U = 8 \times 10^{-5} \times 10^7\,erg = 800\,erg$.

(C) When a capacitor is disconnected from the battery,the charge $Q$ on its plates remains constant.
The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$.
As the plates are separated,the distance $d$ increases,so the capacitance $C$ decreases.
The energy stored in a capacitor is given by $U = \frac{Q^2}{2C}$.
Since $Q$ is constant and $C$ decreases,the energy $U$ increases.
The work done by the external agent is equal to the change in the potential energy of the capacitor: $W = \Delta U = U_f - U_i$.
If the capacitance changes from $C$ to $C'$,the work done is $W = \frac{Q^2}{2C'} - \frac{Q^2}{2C}$.
Given the options provided,the standard expression for energy stored in a capacitor is $\frac{1}{2}CV^2$.

(D) Initial capacitance $C_i = \frac{\varepsilon_0 A}{d}$. Initial energy $U_i = \frac{1}{2} C_i V_0^2 = \frac{1}{2} \frac{\varepsilon_0 A}{d} V_0^2$.
Since the battery is disconnected,the charge $Q$ on the plates remains constant. $Q = C_i V_0 = \frac{\varepsilon_0 A V_0}{d}$.
New separation $d' = 3d$. New capacitance $C_f = \frac{\varepsilon_0 A}{3d} = \frac{C_i}{3}$.
Final energy $U_f = \frac{Q^2}{2 C_f} = \frac{(C_i V_0)^2}{2 (C_i/3)} = \frac{3}{2} C_i V_0^2 = 3 U_i$.
Work done $W = U_f - U_i = 3 U_i - U_i = 2 U_i$.
$W = 2 \times (\frac{1}{2} \frac{\varepsilon_0 A}{d} V_0^2) = \frac{\varepsilon_0 A V_0^2}{d}$.

(D) The energy stored in a capacitor is given by the formula $U = \frac{1}{2}CV^2$.
Given: $C = 2\,\mu F = 2 \times 10^{-6}\,F$ and $V = 50\,V$.
Substituting the values: $U = \frac{1}{2} \times (2 \times 10^{-6}) \times (50)^2$.
$U = 10^{-6} \times 2500 = 2500 \times 10^{-6} = 2.5 \times 10^{-3}\,J$.
Since $1\,J = 10^7\,erg$,we convert the energy to $erg$:
$U = 2.5 \times 10^{-3} \times 10^7\,erg = 2.5 \times 10^4\,erg$.
Note: Based on the provided options,the intended calculation likely assumes $1\,J = 10^6\,erg$ or a specific unit conversion context. Recalculating with the provided option $25 \times 10^3\,erg$ as the target: $2.5 \times 10^{-3}\,J = 25 \times 10^{-4}\,J$. Converting to $erg$ using $1\,J = 10^7\,erg$ gives $25 \times 10^3\,erg$.

(D) The energy $U$ stored in a capacitor is given by the formula $U = \frac{1}{2}CV^2$.
Given:
Capacitance $C = 5\,\mu F = 5 \times 10^{-6}\,F$.
Potential difference $V = 20\,kV = 20 \times 10^3\,V$.
Substituting these values into the formula:
$U = \frac{1}{2} \times (5 \times 10^{-6}) \times (20 \times 10^3)^2$
$U = \frac{1}{2} \times 5 \times 10^{-6} \times 400 \times 10^6$
$U = \frac{1}{2} \times 5 \times 400 \times 10^0$
$U = 1000\,J = 1\,kJ$.
Therefore,the correct option is $D$.

(C) The energy stored in a capacitor is given by the formula $U = \frac{1}{2} C V^2$.
Given:
Capacitance $C = 6 \, \mu F = 6 \times 10^{-6} \, F$
Potential difference $V = 100 \, V$
Substituting the values into the formula:
$U = \frac{1}{2} \times (6 \times 10^{-6}) \times (100)^2$
$U = 3 \times 10^{-6} \times 10,000$
$U = 3 \times 10^{-6} \times 10^4$
$U = 3 \times 10^{-2} \, J$
$U = 0.03 \, J$
Therefore,the energy stored in the capacitor is $0.03 \, J$.

(D) When a charged capacitor is discharged through a resistance, the entire electrostatic potential energy stored in the capacitor is dissipated as heat in the resistance.
The energy stored in a capacitor is given by the formula $U = \frac{1}{2}CV^2$.
Given:
Capacitance $C = 10\, \mu F = 10 \times 10^{-6}\, F = 10^{-5}\, F$.
Potential difference $V = 500\, V$.
Substituting the values into the formula:
$U = \frac{1}{2} \times (10^{-5}) \times (500)^2$
$U = \frac{1}{2} \times 10^{-5} \times 250,000$
$U = 0.5 \times 2.5 = 1.25\, J$.
Therefore, the heat produced in the resistance is $1.25\, J$.

(A) The energy stored in a capacitor,which is equal to the work done in charging it,is given by the formula $W = \frac{Q^2}{2C}$.
Given:
Charge $Q = 8 \times 10^{-18} \, C$
Capacitance $C = 100 \, \mu F = 100 \times 10^{-6} \, F = 10^{-4} \, F$
Substituting the values into the formula:
$W = \frac{(8 \times 10^{-18})^2}{2 \times 10^{-4}}$
$W = \frac{64 \times 10^{-36}}{2 \times 10^{-4}}$
$W = 32 \times 10^{-32} \, J$
Thus,the correct option is $A$.

(C) The initial capacitance of the parallel plate capacitor is $C = \frac{\epsilon_0 A}{d}$. The initial energy stored is $U_i = \frac{q^2}{2C}$.
When the distance between the plates is doubled $(d' = 2d)$,the new capacitance becomes $C' = \frac{\epsilon_0 A}{2d} = \frac{C}{2}$.
Since the capacitor is isolated,the charge $q$ remains constant. The new energy stored is $U_f = \frac{q^2}{2C'} = \frac{q^2}{2(C/2)} = \frac{q^2}{C}$.
The work done by the external force is equal to the change in potential energy: $W = U_f - U_i$.
$W = \frac{q^2}{C} - \frac{q^2}{2C} = \frac{q^2}{2C}$.

(C) Given: Potential difference $V = 200\,V$ and charge $Q = 0.1\,C$.
The energy $U$ stored in a capacitor is given by the formula $U = \frac{1}{2}QV$.
Substituting the given values:
$U = \frac{0.1 \times 200}{2}$
$U = \frac{20}{2}$
$U = 10\,J$.
Therefore,the energy released upon discharge is $10\,J$.

(B) The energy $U$ stored in a capacitor is given by the formula $U = \frac{1}{2} C V^2$.
Given: $C = 40 \, \mu F = 40 \times 10^{-6} \, F$,$V = 3000 \, V$,and $t = 2 \, ms = 2 \times 10^{-3} \, s$.
First,calculate the energy stored:
$U = \frac{1}{2} \times (40 \times 10^{-6}) \times (3000)^2$
$U = 20 \times 10^{-6} \times 9 \times 10^6 = 180 \, J$.
Power $P$ is defined as the energy delivered per unit time:
$P = \frac{U}{t} = \frac{180 \, J}{2 \times 10^{-3} \, s} = 90 \times 10^3 \, W = 90 \, kW$.

(A) The energy stored in a capacitor is given by the formula $U = \frac{Q^2}{2C}$.
When a charged capacitor is disconnected from the battery,the charge $Q$ on its plates remains constant.
The capacitance $C$ of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$,where $d$ is the separation between the plates.
As the plate separation $d$ increases,the capacitance $C$ decreases because $C \propto \frac{1}{d}$.
Since $U = \frac{Q^2}{2C}$ and $C$ is in the denominator,a decrease in $C$ leads to an increase in the stored energy $U$.
Therefore,the energy increases.

(B) When a capacitor of capacitance $C$ is charged to a potential $V$ by giving it a charge $q$,the work done in the process is stored as energy.
The potential $V$ is given by $V = \frac{q}{C}$.
The work done $dW$ to bring an additional charge $dq$ is $dW = V dq = \frac{q}{C} dq$.
Integrating this from $0$ to $Q$,the total energy $U$ stored is $U = \int_{0}^{Q} \frac{q}{C} dq = \frac{1}{2} \frac{Q^2}{C} = \frac{1}{2} CV^2$.
This energy is stored in the form of electrostatic potential energy within the electric field between the plates of the capacitor.

(B) The energy stored in a capacitor is given by $U = \frac{1}{2}CV^2$.
Given: Capacitance $C = 6 \times 10^{-6} \,F$,initial potential $V_1 = 10 \,V$,and final potential $V_2 = 20 \,V$.
The increase in energy $\Delta U$ is given by $\Delta U = U_2 - U_1 = \frac{1}{2}C(V_2^2 - V_1^2)$.
Substituting the values: $\Delta U = \frac{1}{2} \times (6 \times 10^{-6}) \times (20^2 - 10^2)$.
$\Delta U = 3 \times 10^{-6} \times (400 - 100)$.
$\Delta U = 3 \times 10^{-6} \times 300 = 900 \times 10^{-6} \,J$.
$\Delta U = 9 \times 10^{-4} \,J$.

(B) When a charged capacitor is connected through a resistance,the energy stored in the capacitor is dissipated as heat in the resistance.
The energy stored in a capacitor is given by the formula $U = \frac{1}{2} C V^2$.
Given:
Capacitance $C = 4\, \mu F = 4 \times 10^{-6}\, F$
Potential difference $V = 400\, V$
Substituting the values:
$U = \frac{1}{2} \times (4 \times 10^{-6}) \times (400)^2$
$U = 2 \times 10^{-6} \times 160000$
$U = 2 \times 10^{-6} \times 1.6 \times 10^5$
$U = 3.2 \times 10^{-1} = 0.32\, J$.
Thus,the heat produced in the resistance is $0.32\, J$.

(B) The energy density $u$ of a parallel plate capacitor is given by the formula: $u = \frac{1}{2} \varepsilon_0 E^2$.
Since the electric field $E = \frac{V}{d}$,we can write the energy density as $u = \frac{1}{2} \varepsilon_0 \left( \frac{V}{d} \right)^2$.
Given values: $\varepsilon_0 = 8.85 \times 10^{-12} \, F/m$,$V = 300 \, V$,and $d = 2 \times 10^{-3} \, m$.
Substituting these values: $u = \frac{1}{2} \times 8.85 \times 10^{-12} \times \left( \frac{300}{2 \times 10^{-3}} \right)^2$.
$u = 0.5 \times 8.85 \times 10^{-12} \times (1.5 \times 10^5)^2$.
$u = 4.425 \times 10^{-12} \times 2.25 \times 10^{10}$.
$u \approx 9.956 \times 10^{-2} \approx 0.1 \, J/m^3$.

(C) The energy stored in a capacitor is given by the formula $U = \frac{1}{2} C_{eq} V^2$.
Since the voltage $V$ is constant for all combinations,the energy $U$ is directly proportional to the equivalent capacitance $C_{eq}$.
To maximize the energy $U$,we must maximize the equivalent capacitance $C_{eq}$.
For $n$ identical capacitors of capacitance $C$:
$1$. In series: $C_{eq} = \frac{C}{n} = \frac{C}{3}$.
$2$. In parallel: $C_{eq} = nC = 3C$.
$3$. Two in parallel and one in series: $C_{eq} = \frac{(2C) \cdot C}{2C + C} = \frac{2}{3}C$.
$4$. Two in series and one in parallel: $C_{eq} = \frac{C}{2} + C = 1.5C$.
Comparing these values,the parallel combination $(3C)$ provides the maximum equivalent capacitance. Therefore,the combination with three capacitors in parallel stores the greatest energy.

(B) From the figure, plates $A$ and $C$ are connected together, and plate $B$ is connected to the other terminal of the battery. This forms two capacitors in parallel, one between $A$ and $B$, and another between $B$ and $C$.
Given: Area $A = 50 \, cm^2 = 50 \times 10^{-4} \, m^2$, separation $d = 3 \, mm = 3 \times 10^{-3} \, m$, and voltage $V = 12 \, V$.
The capacitance of each capacitor is $C_0 = \frac{\varepsilon_0 A}{d}$.
Since they are in parallel, the total capacitance is $C_{eq} = C_0 + C_0 = \frac{2 \varepsilon_0 A}{d}$.
Substituting the values:
$C_{eq} = \frac{2 \times 8.854 \times 10^{-12} \times 50 \times 10^{-4}}{3 \times 10^{-3}} \approx 2.95 \times 10^{-11} \, F$.
The energy stored $U = \frac{1}{2} C_{eq} V^2 = \frac{1}{2} \times (2.95 \times 10^{-11}) \times (12)^2$.
$U = 0.5 \times 2.95 \times 10^{-11} \times 144 = 2.124 \times 10^{-9} \, J \approx 2.1 \times 10^{-9} \, J$.

(D) The energy density of an electric field is given by $u = \frac{1}{2} \varepsilon_0 E^2$.
The electric field $E$ between the cloud and the earth is $E = \frac{V}{d}$,where $V = 10^5 \, V$ and $d = 0.75 \, km = 750 \, m$.
The volume of the space between the cloud and the earth is $Volume = A \times d$,where $A = 25 \times 10^6 \, m^2$.
The total energy $U$ is $U = u \times Volume = \frac{1}{2} \varepsilon_0 \left( \frac{V}{d} \right)^2 \times (A \times d) = \frac{1}{2} \varepsilon_0 \frac{V^2 A}{d}$.
Substituting the values: $U = \frac{1}{2} \times (8.85 \times 10^{-12}) \times \frac{(10^5)^2 \times 25 \times 10^6}{750}$.
$U = \frac{1}{2} \times 8.85 \times 10^{-12} \times \frac{10^{10} \times 25 \times 10^6}{750} = \frac{8.85 \times 25 \times 10^4}{1500} = \frac{221.25 \times 10^4}{1500} \approx 1475 \, J$.

(A) The electrostatic energy $U$ stored in a parallel plate capacitor connected to a battery of voltage $V$ is given by $U = \frac{1}{2}CV^2$.
Since the capacitance $C = \frac{\varepsilon_0 A}{x}$,we have $U = \frac{1}{2} \left( \frac{\varepsilon_0 A}{x} \right) V^2$.
To find the time rate of change of energy,we differentiate $U$ with respect to time $t$:
$\frac{dU}{dt} = \frac{d}{dt} \left( \frac{\varepsilon_0 A V^2}{2x} \right) = \frac{\varepsilon_0 A V^2}{2} \frac{d}{dt} (x^{-1})$.
Using the chain rule,$\frac{dU}{dt} = \frac{\varepsilon_0 A V^2}{2} (-x^{-2}) \frac{dx}{dt}$.
Since the plates are pulled apart at a uniform speed,$\frac{dx}{dt} = v$ (a constant).
Therefore,$\frac{dU}{dt} = -\frac{\varepsilon_0 A V^2 v}{2} x^{-2}$.
This shows that $\frac{dU}{dt} \propto x^{-2}$.

(D) The energy stored in a capacitor is given by the formula $U = \frac{1}{2}QV$.
In the given graph,$Q$ is plotted on the $X$-axis and $V$ is plotted on the $Y$-axis.
The area of the triangle $OAB$ is given by $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
Here,the base is $OB = Q$ and the height is $AB = V$.
Therefore,$\text{Area} = \frac{1}{2} \times Q \times V = \frac{1}{2}QV$.
This expression represents the energy stored in the capacitor.

(D) The energy stored in a capacitor is given by the formula $U = \frac{1}{2} C V^2$.
Substituting the given values: $C = 200 \, \mu F = 200 \times 10^{-6} \, F$ and $V = 200 \, V$.
$U = \frac{1}{2} \times (200 \times 10^{-6}) \times (200)^2 = 100 \times 10^{-6} \times 40000 = 4 \, J$.
When a capacitor discharges through a resistor, the entire stored energy is dissipated as heat in the resistor, regardless of the resistance value.
Therefore, the heat generated in both cases is $4 \, J$.

(B) The work done in charging a capacitor is stored as potential energy $U$ in the electric field.
For a capacitor with capacitance $C$ charged to a potential $V$,the energy $U$ is given by the integral of the work done to bring small charge $dq$ to the plates:
$U = \int_0^V q \, dV = \int_0^V (CV) \, dV$
$U = C \int_0^V V \, dV$
$U = C \left[ \frac{V^2}{2} \right]_0^V$
$U = \frac{1}{2}CV^2$
Therefore,the correct option is $B$.

(D) The area of $\triangle OAB$ is given by $Area = \frac{1}{2} \times \text{base} \times \text{height}$.
$Area = \frac{1}{2} \times OB \times AB = \frac{1}{2} \times Q \times V$.
Since the energy stored in a capacitor is $U = \frac{1}{2} QV$, the area under the $V-Q$ graph represents the energy stored in the capacitor.

(A) The energy stored in a capacitor is given by $U = \frac{1}{2}CV^2$.
Since the charge $q = CV$,we can substitute $V = \frac{q}{C}$ into the energy formula.
$U = \frac{1}{2} C \left( \frac{q}{C} \right)^2 = \frac{1}{2} C \left( \frac{q^2}{C^2} \right) = \frac{q^2}{2C}$.
Therefore,the correct formula is $\frac{q^2}{2C}$.

(B) Let the capacitance of the capacitor be $C$ and the electromotive force of the battery be $V$.
When the capacitor is connected to the battery,the total work done by the battery is $W = QV = (CV)V = CV^2$.
The energy stored in the capacitor is $U = \frac{1}{2} CV^2$.
Comparing the two,we see that the energy stored in the capacitor is exactly half of the total energy supplied by the battery $(U = \frac{1}{2} W)$.
The remaining half of the energy is dissipated as heat in the connecting wires or during the charging process.

(A) The electric field $E$ produced by a single charged plate with surface charge density $\sigma = q/A$ is given by $E = \frac{\sigma}{2 \epsilon_0} = \frac{q}{2 A \epsilon_0}$.
In a parallel plate capacitor,the electric field between the plates is the sum of the fields due to both plates: $E_{total} = \frac{q}{2 A \epsilon_0} + \frac{q}{2 A \epsilon_0} = \frac{q}{A \epsilon_0}$.
The electric energy density $u$ is defined as $u = \frac{1}{2} \epsilon_0 E^2$.
Substituting the value of $E_{total}$ into the energy density formula:
$u = \frac{1}{2} \epsilon_0 \left( \frac{q}{A \epsilon_0} \right)^2 = \frac{1}{2} \epsilon_0 \left( \frac{q^2}{A^2 \epsilon_0^2} \right) = \frac{q^2}{2 \epsilon_0 A^2}$.

(D) The energy stored in a capacitor is given by the formula $U = \frac{1}{2} C V^2$.
Given:
Capacitance $C = 2 \times 10^{-4} \, \mu F = 2 \times 10^{-4} \times 10^{-6} \, F = 2 \times 10^{-10} \, F$.
Potential $V = 20 \, kV = 20 \times 10^3 \, V = 2 \times 10^4 \, V$.
Substituting the values into the formula:
$U = \frac{1}{2} \times (2 \times 10^{-10}) \times (2 \times 10^4)^2$
$U = 10^{-10} \times (4 \times 10^8)$
$U = 4 \times 10^{-2} \, J$.

(A) The energy stored in the capacitor is completely converted into heat energy in the block.
The energy stored in a capacitor is given by $U = \frac{1}{2}CV^2$.
The heat energy absorbed by the block is given by $Q = mS\Delta T$.
Equating the two energies: $\frac{1}{2}CV^2 = mS\Delta T$.
Solving for the potential difference $V$: $V^2 = \frac{2mS\Delta T}{C}$.
Therefore,$V = \sqrt{\frac{2mS\Delta T}{C}}$.