Energy Stored in a Capacitor Questions in English

(A) The energy density (energy per unit volume) $u$ of an electric field $E$ is given by the formula: $u = \frac{1}{2}{\varepsilon _0}{E^2}$.
For a parallel plate capacitor,the electric field $E$ between the plates is related to the potential difference $V$ and the distance $d$ by the relation: $E = \frac{V}{d}$.
Substituting this value of $E$ into the energy density formula,we get: $u = \frac{1}{2}{\varepsilon _0}{\left( \frac{V}{d} \right)^2} = \frac{1}{2}{\varepsilon _0}\frac{{{V^2}}}{{{d^2}}}$.

(D) The work done by the battery $(W_b)$ in charging a capacitor to a potential $V$ is given by $W_b = Q \cdot V = (CV) \cdot V = CV^2$.
The energy stored in the capacitor $(U)$ is given by $U = \frac{1}{2} CV^2$.
Therefore,the ratio of the energy stored in the capacitor to the work done by the battery is:
$\frac{U}{W_b} = \frac{\frac{1}{2} CV^2}{CV^2} = \frac{1}{2} = 0.5$.

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

(A) When capacitors are connected in series,the charge $Q$ stored on each capacitor is the same.
The energy stored in a capacitor is given by the formula $U = \frac{Q^2}{2C}$.
Since $Q$ is constant for both capacitors in a series connection,the energy $U$ is inversely proportional to the capacitance $C$ $(U \propto \frac{1}{C})$.
Therefore,the ratio of the energy stored in the two capacitors is:
$\frac{U_1}{U_2} = \frac{C_2}{C_1}$
Given $C_1 = 2 \ \mu F$ and $C_2 = 4 \ \mu F$:
$\frac{U_1}{U_2} = \frac{4 \ \mu F}{2 \ \mu F} = \frac{2}{1}$
Thus,the ratio of the energy stored is $2 : 1$.

(B) The energy density $u$ of an electric field is given by the formula: $u = \frac{1}{2} \epsilon_0 E^2$.
Given: $u = 1.8 \times 10^{-9} \ J/m^3$ and $\epsilon_0 = 9 \times 10^{-12} \ C^2/N \cdot m^2$.
Substituting the values into the formula:
$1.8 \times 10^{-9} = \frac{1}{2} \times (9 \times 10^{-12}) \times E^2$.
$1.8 \times 10^{-9} = 4.5 \times 10^{-12} \times E^2$.
$E^2 = \frac{1.8 \times 10^{-9}}{4.5 \times 10^{-12}}$.
$E^2 = \frac{1.8}{4.5} \times 10^3 = 0.4 \times 1000 = 400$.
$E = \sqrt{400} = 20 \ NC^{-1}$.
Therefore,the correct option is $B$.

(D) The cloud and the earth form a parallel plate capacitor.
The energy stored in the electric field is given by $U = \frac{1}{2} \epsilon_0 E^2 \times \text{Volume}$.
Here, $E = \frac{V}{d}$, where $V = 10^5\ V$ and $d = 0.75\ km = 750\ m$.
The volume of the space between the cloud and the earth is $V_{ol} = A \times d$, where $A = 25 \times 10^6\ m^2$.
Substituting these values into the energy formula:
$U = \frac{1}{2} \epsilon_0 \left( \frac{V}{d} \right)^2 \times (A \times d) = \frac{1}{2} \epsilon_0 \frac{V^2 A}{d}$.
$U = \frac{1}{2} \times (8.85 \times 10^{-12}) \times \frac{(10^5)^2 \times 25 \times 10^6}{750}$.
$U = \frac{1}{2} \times 8.85 \times 10^{-12} \times \frac{10^{10} \times 25 \times 10^6}{750} = \frac{8.85 \times 25 \times 10^4}{1500} \approx 1475\ J$.

(B) When $n$ capacitors,each of capacitance $C$,are connected in parallel to a voltage source $V$,the potential difference across each capacitor is $V$.
The energy stored in a single capacitor is given by $U_i = \frac{1}{2}CV^2$.
Since the capacitors are in parallel,the total energy $U$ is the sum of the energies stored in each individual capacitor:
$U = U_1 + U_2 + ... + U_n$
$U = \frac{1}{2}CV^2 + \frac{1}{2}CV^2 + ... + \frac{1}{2}CV^2$ ($n$ times)
$U = n \times (\frac{1}{2}CV^2)$
$U = \frac{1}{2}nCV^2$.

(C) The energy stored in a capacitor is given by the formula $W = \frac{Q^2}{2C}$.
When the charge is increased from $Q$ to $Q' = 2Q$,the new energy $W'$ becomes:
$W' = \frac{(Q')^2}{2C} = \frac{(2Q)^2}{2C} = \frac{4Q^2}{2C}$.
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 new stored energy is $4W$.

(B) The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$. If the distance $d$ is doubled $(d' = 2d)$,the new capacitance becomes $C' = \frac{C}{2}$.
Since the capacitor is disconnected from the battery,the charge $Q$ remains constant.
The initial energy stored is $U_1 = \frac{Q^2}{2C}$.
The final energy stored is $U_2 = \frac{Q^2}{2C'} = \frac{Q^2}{2(C/2)} = \frac{Q^2}{C}$.
The work done $W$ is equal to the change in potential energy: $W = U_2 - U_1 = \frac{Q^2}{C} - \frac{Q^2}{2C} = \frac{Q^2}{2C}$.

(D) The energy stored in a capacitor,which is equal to the work done to charge it,is given by the formula: $W = \frac{Q^2}{2C}$.
Given:
Charge $Q = 8 \times 10^{-18} \ C$
Capacitance $C = 800 \ \mu F = 800 \times 10^{-6} \ F = 8 \times 10^{-4} \ F$.
Substituting the values into the formula:
$W = \frac{(8 \times 10^{-18})^2}{2 \times (8 \times 10^{-4})}$
$W = \frac{64 \times 10^{-36}}{16 \times 10^{-4}}$
$W = 4 \times 10^{-32} \ J$.

(B) The energy stored in a fully charged capacitor is given by $U = \frac{1}{2}CV^2$,where $V$ is the potential difference across the capacitor.
Since the capacitor is discharged through a resistance coil in a thermally insulated block,the entire electrical energy stored in the capacitor is converted into heat energy absorbed by the block.
The heat energy absorbed by the block is given by $Q = ms\Delta T$.
Equating the energy stored to the heat energy absorbed:
$\frac{1}{2}CV^2 = ms\Delta T$
$V^2 = \frac{2ms\Delta T}{C}$
$V = \sqrt{\frac{2ms\Delta T}{C}}$

(B) The energy stored in a capacitor is given by $U = \frac{1}{2}CV^2$.
Given: $C = 40\ \mu F = 40 \times 10^{-6}\ F$,$V = 3000\ V$,and $t = 2\ ms = 2 \times 10^{-3}\ s$.
Substituting the values:
$U = \frac{1}{2} \times (40 \times 10^{-6}) \times (3000)^2 = 20 \times 10^{-6} \times 9 \times 10^6 = 180\ J$.
The power delivered is $P = \frac{U}{t} = \frac{180\ J}{2 \times 10^{-3}\ s} = 90 \times 10^3\ W = 90\ kW$.

(C) For capacitors connected in parallel,the equivalent capacitance $C_{eq}$ is given by $C_{eq} = C_1 + C_2$.
Given $C_1 = C_2 = 10 \ \mu F$,the equivalent capacitance is $C_{eq} = 10 \ \mu F + 10 \ \mu F = 20 \ \mu F = 20 \times 10^{-6} \ F$.
The energy $U$ stored in a capacitor is given by the formula $U = \frac{1}{2} C_{eq} V^2$.
Substituting the values $C_{eq} = 20 \times 10^{-6} \ F$ and $V = 200 \ V$:
$U = \frac{1}{2} \times (20 \times 10^{-6}) \times (200)^2$
$U = 10 \times 10^{-6} \times 40000$
$U = 10 \times 10^{-6} \times 4 \times 10^4$
$U = 40 \times 10^{-2} = 0.4 \ J$.

(C) The initial capacitance is $C = \frac{\varepsilon_0 A}{d}$.
Since the capacitor is disconnected from the battery,the charge $Q$ on the plates remains constant.
The initial energy stored is $U_i = \frac{Q^2}{2C}$.
When the separation is increased to $2d$,the new capacitance becomes $C' = \frac{\varepsilon_0 A}{2d} = \frac{C}{2}$.
The final energy stored is $U_f = \frac{Q^2}{2C'} = \frac{Q^2}{2(C/2)} = \frac{Q^2}{C}$.
The work done $W$ is the change in potential energy: $W = U_f - U_i = \frac{Q^2}{C} - \frac{Q^2}{2C} = \frac{Q^2}{2C}$.
Substituting $Q = CV$,we get $W = \frac{(CV)^2}{2C} = \frac{1}{2} C V^2$.
Substituting $C = \frac{\varepsilon_0 A}{d}$,we get $W = \frac{1}{2} \left( \frac{\varepsilon_0 A}{d} \right) V^2 = \frac{\varepsilon_0 A V^2}{2d}$.

(B) The area of the plates is $A = \pi r^2 = \pi \times (8 \times 10^{-2})^2 \approx 0.0201\,m^2$ and the distance is $d = 1\,mm = 1 \times 10^{-3}\,m$.
The capacitance of the capacitor with the dielectric is given by $C = \frac{K \varepsilon_0 A}{d}$.
Substituting the values: $C = \frac{6 \times 8.854 \times 10^{-12} \times 0.0201}{1 \times 10^{-3}} \approx 1.068 \times 10^{-9}\,F$.
The potential difference is $V = 150\,V$.
The energy stored in the capacitor is $U = \frac{1}{2}CV^2$.
$U = \frac{1}{2} \times (1.068 \times 10^{-9}) \times (150)^2$.
$U = 0.5 \times 1.068 \times 10^{-9} \times 22500$.
$U \approx 1.2 \times 10^{-5}\,J$.

(B) The system consists of two capacitors connected in parallel.
Each capacitor has capacitance $C = \frac{\varepsilon_0 A}{d}$.
Since there are two such capacitors in parallel,the total capacitance is $C_{eq} = 2 \times \frac{\varepsilon_0 A}{d} = \frac{2 \varepsilon_0 A}{d}$.
The energy stored in the system is given by $U = \frac{1}{2} C_{eq} V^2 = \frac{1}{2} \left( \frac{2 \varepsilon_0 A}{d} \right) V^2 = \frac{\varepsilon_0 A V^2}{d}$.
Given: $\varepsilon_0 = 8.854 \times 10^{-12} \, F/m$,$A = 50 \times 10^{-4} \, m^2$,$d = 3 \times 10^{-3} \, m$,$V = 12 \, V$.
$U = \frac{8.854 \times 10^{-12} \times 50 \times 10^{-4} \times (12)^2}{3 \times 10^{-3}}$.
$U = \frac{8.854 \times 50 \times 144 \times 10^{-16}}{3 \times 10^{-3}} = \frac{63748.8 \times 10^{-16}}{3 \times 10^{-3}} \approx 2.125 \times 10^{-9} \, J$.
Thus,the energy stored is approximately $2.1 \times 10^{-9} \, J$.

(B) The capacitance of a cylindrical capacitor is given by $C = \frac{2\pi \varepsilon_0 L}{\ln(b/a)}$.
The energy stored in the capacitor is $U = \frac{1}{2} \frac{Q^2}{C}$.
Substituting the expression for $C$,we get $U = \frac{1}{2} \frac{Q^2 \ln(b/a)}{2\pi \varepsilon_0 L}$.
Since $\ln(b/a)$,$\pi$,and $\varepsilon_0$ are constants,the energy $U$ is proportional to $\frac{Q^2}{L}$,i.e.,$U \propto \frac{Q^2}{L}$.
Let the new charge be $Q' = 2Q$ and the new length be $L' = 2L$.
The new energy $U'$ is proportional to $\frac{(Q')^2}{L'} = \frac{(2Q)^2}{2L} = \frac{4Q^2}{2L} = 2 \left( \frac{Q^2}{L} \right)$.
Therefore,the energy becomes $2$ times the original energy.

(A) The energy stored in a capacitor is given by the formula $U = \frac{1}{2} CV^2$.
Given:
Capacitance $C = 4 \times 10^{-6} \ F$
Potential difference $V = 100 \ V$
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 required to discharge the capacitor is $0.02 \ J$.

(D) The work done to change the potential energy of a capacitor is equal to the change in its stored energy.
Initial energy $U_i = \frac{1}{2} C V_1^2 = \frac{1}{2} \times (6 \times 10^{-6}\ F) \times (10\ V)^2 = 3 \times 10^{-4}\ J$.
Final energy $U_f = \frac{1}{2} C V_2^2 = \frac{1}{2} \times (6 \times 10^{-6}\ F) \times (20\ V)^2 = 12 \times 10^{-4}\ J$.
Work done $W = \Delta U = U_f - U_i = (12 - 3) \times 10^{-4}\ J = 9 \times 10^{-4}\ J$.

(D) When a charged capacitor is connected to a resistor,the total energy stored in the capacitor is dissipated as heat in the resistor.
The energy stored in a capacitor is given by the formula $U = \frac{1}{2}CV^2$.
Given:
Capacitance $C = 4\ \mu F = 4 \times 10^{-6}\ F$
Voltage $V = 400\ V$
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 total heat energy produced is $0.32\ J$.

(D) The work done $W$ to change the potential of a capacitor from $V_1$ to $V_2$ is given by the change in stored energy: $W = U_f - U_i = \frac{1}{2} C (V_2^2 - V_1^2)$.
For the first case,$W = \frac{1}{2} C (10^2 - 5^2) = \frac{1}{2} C (100 - 25) = \frac{1}{2} C (75) = 37.5 C$.
For the second case,$W' = \frac{1}{2} C (15^2 - 10^2) = \frac{1}{2} C (225 - 100) = \frac{1}{2} C (125) = 62.5 C$.
Taking the ratio: $\frac{W'}{W} = \frac{62.5 C}{37.5 C} = \frac{62.5}{37.5} = \frac{625}{375} = \frac{5}{3} \approx 1.67$.
Therefore,$W' = 1.67\ W$.

(C) The two capacitors are connected in parallel.
The equivalent capacitance is given by $C_{eq} = C_1 + C_2 = 1 \ \mu F + 1 \ \mu F = 2 \ \mu F = 2 \times 10^{-6} \ F$.
The potential difference across the capacitors is $V = 200 \ V$.
The total energy stored in the system is given by the formula $U = \frac{1}{2} C_{eq} V^2$.
Substituting the values,we get $U = \frac{1}{2} \times (2 \times 10^{-6} \ F) \times (200 \ V)^2$.
$U = 10^{-6} \times 40000 = 0.04 \ J$.

(B) From the circuit diagram,$C_1$ and $C_2$ are in series,and their combination is in parallel with $C_3$.
First,calculate the equivalent capacitance of $C_1$ and $C_2$ in series:
$C_{12} = \frac{C_1 C_2}{C_1 + C_2} = \frac{2 \times 6}{2 + 6} = \frac{12}{8} = 1.5\ \mu F$.
Now,$C_{12}$ is in parallel with $C_3$,so the total equivalent capacitance $C_{eq}$ is:
$C_{eq} = C_{12} + C_3 = 1.5 + 4 = 5.5\ \mu F$.
The total energy $U$ stored in the system is given by:
$U = \frac{1}{2} C_{eq} V^2 = \frac{1}{2} \times 5.5 \times 10^{-6} \times (2)^2$
$U = \frac{1}{2} \times 5.5 \times 10^{-6} \times 4 = 11 \times 10^{-6}\ J$.

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

(C) The potential difference $V$ between the plates of a parallel plate capacitor in a uniform electric field $E$ is given by $V = Ed$.
The capacitance $C$ of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$.
The energy $U$ stored in a capacitor is given by the formula $U = \frac{1}{2} CV^2$.
Substituting the expressions for $C$ and $V$ into the energy formula:
$U = \frac{1}{2} \left( \frac{\varepsilon_0 A}{d} \right) (Ed)^2$
$U = \frac{1}{2} \left( \frac{\varepsilon_0 A}{d} \right) E^2 d^2$
$U = \frac{1}{2} \varepsilon_0 E^2 Ad$.

(C) The force of attraction $F$ between the plates of a parallel plate capacitor is given by the formula:
$F = \frac{Q^2}{2 \varepsilon_0 A}$
where $Q$ is the charge on the capacitor,$\varepsilon_0$ is the permittivity of free space,and $A$ is the area of each plate.
We know that the charge $Q$ is related to capacitance $C$ and potential difference $V$ by:
$Q = CV$
Also,the capacitance of a parallel plate capacitor is given by:
$C = \frac{\varepsilon_0 A}{d} \implies \varepsilon_0 A = Cd$
Substituting these expressions into the force formula:
$F = \frac{(CV)^2}{2(Cd)}$
$F = \frac{C^2 V^2}{2Cd}$
$F = \frac{CV^2}{2d}$

(B) The energy stored in a capacitor is given by $U = \frac{1}{2} CV^2 = \frac{Q^2}{2C}$.
Let the potential at $A$ be $0 \ V$ and at $C$ be $10 \ V$. Let the potential at node $B$ be $V_B$.
Using Kirchhoff's current law at node $B$,the sum of charges on the plates connected to $B$ must be zero (assuming initial charge is zero): $10(V_B - 0) + 5(V_B - 0) + 4(V_B - 10) + 6(V_B - 10) = 0$.
$15V_B + 10V_B - 40 - 60 = 0 \implies 25V_B = 100 \implies V_B = 4 \ V$.
Now,calculate the energy $U = \frac{1}{2} C(V_{diff})^2$ for each capacitor:
For $10 \ \mu F$: $U = \frac{1}{2} \times 10 \times (4 - 0)^2 = 5 \times 16 = 80 \ \mu J$.
For $5 \ \mu F$: $U = \frac{1}{2} \times 5 \times (4 - 0)^2 = 2.5 \times 16 = 40 \ \mu J$.
For $4 \ \mu F$: $U = \frac{1}{2} \times 4 \times (4 - 10)^2 = 2 \times 36 = 72 \ \mu J$.
For $6 \ \mu F$: $U = \frac{1}{2} \times 6 \times (4 - 10)^2 = 3 \times 36 = 108 \ \mu J$.
The maximum energy is stored in the $6 \ \mu F$ capacitor.

(A) Let the capacitance of the capacitor be $C$ and the $EMF$ of the battery be $V$.
When the capacitor is fully charged,the charge stored on the plates is $Q = CV$.
The energy stored in the capacitor is given by $U = \frac{1}{2} CV^2$.
The work done by the battery in charging the capacitor is $W = Q \times V = (CV) \times V = CV^2$.
The ratio of the energy stored in the capacitor to the work done by the battery is $\frac{U}{W} = \frac{\frac{1}{2} CV^2}{CV^2} = \frac{1}{2} = 0.5$.

(A) The energy density $u$ (energy per unit volume) in a region with an electric field $E$ is given by the formula: $u = \frac{1}{2} \varepsilon_{0} E^{2}$.
The volume $V$ of the space between the plates of the 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:
$U = u \times V$
$U = (\frac{1}{2} \varepsilon_{0} E^{2}) \times (Ad)$
$U = \frac{1}{2} \varepsilon_{0} E^{2} Ad$.

(D) The graph shows the variation of voltage $V$ (on the y-axis) with charge $q$ (on the x-axis) for a capacitor.
The area under the $V-q$ graph represents the work done in charging the capacitor,which is stored as potential energy.
The area of the triangle $AOB$ is given by:
$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$
$\text{Area} = \frac{1}{2} \times q \times V$
Since the energy stored in a capacitor is $U = \frac{1}{2} qV$,the area of the triangle $AOB$ represents the energy stored by the capacitor.

(D) In steady state,the capacitor acts as an open circuit,so no current flows through it.
The circuit consists of two parallel branches connected across the $6 \, V$ battery.
Branch $1$ (left side) consists of resistors $3 \, \Omega$ and $7 \, \Omega$ in series. The potential at $D$ relative to $A$ is given by the voltage divider rule: $V_{A} - V_{D} = \frac{3}{3+7} \times 6 = 1.8 \, V$.
Branch $2$ (right side) consists of resistors $2 \, \Omega$ and $4 \, \Omega$ in series. The potential at $B$ relative to $A$ is: $V_{A} - V_{B} = \frac{2}{2+4} \times 6 = 2 \, V$.
The potential difference across the capacitor (between points $D$ and $B$) is $V_{DB} = |V_{A} - V_{B} - (V_{A} - V_{D})| = |2 - 1.8| = 0.2 \, V$.
The energy stored in the capacitor is $U = \frac{1}{2} CV^{2}$.
Given $C = 3 \, \mu F = 3 \times 10^{-6} \, F$ and $V = 0.2 \, V$.
$U = \frac{1}{2} \times (3 \times 10^{-6}) \times (0.2)^{2} = \frac{1}{2} \times 3 \times 10^{-6} \times 0.04 = 6 \times 10^{-8} \, J = 60 \times 10^{-9} \, J = 60 \, nJ$.

(B) The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$.
If the distance $d$ is doubled $(d' = 2d)$,the new capacitance becomes $C' = \frac{\epsilon_0 A}{2d} = \frac{C}{2}$.
The charge $Q$ remains constant as the capacitor is isolated.
The initial energy stored is $U_i = \frac{Q^2}{2C}$.
The final 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 agent is equal to the change in potential energy: $W = U_f - U_i$.
$W = \frac{Q^2}{C} - \frac{Q^2}{2C} = \frac{Q^2}{2C}$.

(D) The energy stored in a charged spherical conductor is given by $U = \frac{Q^2}{2C}$.
Given: $U = 4.5\, J$ and $Q = 4\,\mu C = 4 \times 10^{-6}\, C$.
Substituting the values: $4.5 = \frac{(4 \times 10^{-6})^2}{2C} = \frac{16 \times 10^{-12}}{2C}$.
$C = \frac{16 \times 10^{-12}}{9} = 1.77 \times 10^{-12}\, F$.
The capacitance of a spherical conductor is $C = 4\pi\varepsilon_0 R$.
Therefore,$R = \frac{C}{4\pi\varepsilon_0} = C \times (9 \times 10^9)$.
$R = \frac{16 \times 10^{-12}}{9} \times 9 \times 10^9 = 16 \times 10^{-3}\, m$.
Converting to millimeters: $R = 16\, mm$.

(A) The charge $q$ on the capacitor remains constant because the capacitor is isolated after charging.
Initial capacitance $C_i = 5\,\mu F = 5 \times 10^{-6}\,F$.
Final capacitance $C_f = 2\,\mu F = 2 \times 10^{-6}\,F$.
Charge $q = 5\,\mu C = 5 \times 10^{-6}\,C$.
The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$.
Work done $W = \Delta U = U_f - U_i = \frac{q^2}{2C_f} - \frac{q^2}{2C_i} = \frac{q^2}{2} \left( \frac{1}{C_f} - \frac{1}{C_i} \right)$.
Substituting the values:
$W = \frac{(5 \times 10^{-6})^2}{2} \left( \frac{1}{2 \times 10^{-6}} - \frac{1}{5 \times 10^{-6}} \right)$.
$W = \frac{25 \times 10^{-12}}{2} \left( \frac{5 - 2}{10 \times 10^{-6}} \right) = \frac{25 \times 10^{-12}}{2} \left( \frac{3}{10 \times 10^{-6}} \right)$.
$W = \frac{75 \times 10^{-12}}{20 \times 10^{-6}} = 3.75 \times 10^{-6}\,J$.

(D) The energy stored in a capacitor is given by the work done in charging it. For a capacitor,the charge $q$ is directly proportional to the potential difference $V$,i.e.,$q = CV$.
The graph shows a linear relationship between $V$ and $q$.
The area under the $V-q$ graph represents the work done,which is stored as potential energy $U$ in the capacitor.
Area of $\Delta OPM = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times q \times V$.
Since $U = \frac{1}{2} qV$,the area of $\Delta OPM$ represents the energy stored in the capacitor.

(D) The two capacitors are connected in parallel to the battery of potential $V$.
The equivalent capacitance $C_{eq}$ for capacitors in parallel is given by $C_{eq} = C_1 + C_2$.
Here,$C_{eq} = C + \frac{C}{2} = \frac{3C}{2}$.
The work done $W$ in charging a capacitor is equal to the energy stored in the capacitor,which is given by $W = \frac{1}{2} C_{eq} V^2$.
Substituting the value of $C_{eq}$,we get $W = \frac{1}{2} \left( \frac{3C}{2} \right) V^2 = \frac{3}{4} CV^2$.

(D) The area of the triangle $AOB$ in the $V-q$ graph is given by:
Area $= \frac{1}{2} \times \text{base} \times \text{height}$
Area $= \frac{1}{2} \times q \times V$
We know that the energy stored in a capacitor is given by $U = \frac{1}{2} qV$.
Therefore,the area of the triangle $AOB$ is equal to the energy stored by the capacitor.
Thus,the area is proportional to the energy stored by the capacitor.

(D) The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$.
The potential difference between the plates is $V = Ed$.
The energy stored in the capacitor is $U = \frac{1}{2} C V^2$.
Substituting the values of $C$ and $V$:
$U = \frac{1}{2} \left( \frac{\epsilon_0 A}{d} \right) (Ed)^2$.
$U = \frac{1}{2} \left( \frac{\epsilon_0 A}{d} \right) E^2 d^2$.
$U = \frac{1}{2} \epsilon_0 E^2 Ad$.

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

(N/A) Given:
Area $A = 90 \,cm^{2} = 90 \times 10^{-4} \,m^{2}$
Distance $d = 2.5 \,mm = 2.5 \times 10^{-3} \,m$
Potential $V = 400 \,V$
Permittivity $\epsilon_{0} = 8.85 \times 10^{-12} \,C^{2} \,N^{-1} \,m^{-2}$
$(a)$ Capacitance $C = \frac{\epsilon_{0} A}{d} = \frac{8.85 \times 10^{-12} \times 90 \times 10^{-4}}{2.5 \times 10^{-3}} = 3.186 \times 10^{-11} \,F$
Energy $U = \frac{1}{2} C V^{2} = \frac{1}{2} \times 3.186 \times 10^{-11} \times (400)^{2} = 2.55 \times 10^{-6} \,J$
$(b)$ Volume $V' = A \times d = 90 \times 10^{-4} \times 2.5 \times 10^{-3} = 2.25 \times 10^{-5} \,m^{3}$
Energy density $u = \frac{U}{V'} = \frac{2.55 \times 10^{-6}}{2.25 \times 10^{-5}} = 0.113 \,J \,m^{-3}$
Relation: $u = \frac{U}{Ad} = \frac{\frac{1}{2} (\frac{\epsilon_{0} A}{d}) V^{2}}{Ad} = \frac{1}{2} \epsilon_{0} (\frac{V}{d})^{2} = \frac{1}{2} \epsilon_{0} E^{2}$

(N/A) capacitor stores energy in the form of an electrostatic field between its plates. When charging a capacitor,work is done by an external agent to move charge from one plate to the other against the existing potential difference.
Consider two uncharged conductors $1$ and $2$ as shown in the figure.
Imagine a process of transferring charge from conductor $2$ to conductor $1$ bit by bit. As charge is transferred,conductor $1$ becomes positively charged and conductor $2$ becomes negatively charged.
In transferring positive charge from conductor $2$ to conductor $1$,external work must be done because conductor $1$ is at a higher potential than conductor $2$.
Consider the situation when the conductors $1$ and $2$ have charges $Q'$ and $-Q'$ respectively.
The potential difference $V'$ between conductors $1$ and $2$ is $V' = \frac{Q'}{C}$,where $C$ is the capacitance of the system.
If a small charge $\delta Q'$ is transferred from conductor $2$ to $1$,the work done is $\delta W = V' \delta Q'$.
Substituting $V'$,we get $\delta W = \frac{Q' \delta Q'}{C}$.
The total work done $W$ to bring a total charge $Q$ from conductor $2$ to $1$ is obtained by integrating the work done for small charge transfers from $0$ to $Q$:
$W = \int_{0}^{Q} \frac{Q'}{C} dQ' = \frac{1}{C} \left[ \frac{(Q')^2}{2} \right]_{0}^{Q} = \frac{Q^2}{2C}$.
Since $Q = CV$,the energy stored can also be expressed as $U = \frac{1}{2}CV^2$ or $U = \frac{1}{2}QV$.

Energy stored per unit volume is known as energy density. The energy stored in a capacitor is given by $U = \frac{1}{2} \frac{Q^2}{C}$.
Substituting $Q = \sigma A$ and $C = \frac{\epsilon_0 A}{d}$,we get:
$U = \frac{1}{2} \frac{(\sigma A)^2}{\epsilon_0 A / d} = \frac{1}{2} \frac{\sigma^2 A^2 d}{\epsilon_0 A} = \frac{1}{2} \frac{\sigma^2 A d}{\epsilon_0}$.
Since the electric field between the plates is $E = \frac{\sigma}{\epsilon_0}$,we have $\sigma = \epsilon_0 E$.
Substituting this into the expression for $U$:
$U = \frac{1}{2} \frac{(\epsilon_0 E)^2 A d}{\epsilon_0} = \frac{1}{2} \epsilon_0 E^2 (A d)$.
Here,$A d$ represents the volume $V$ of the region between the plates.
Therefore,the energy stored per unit volume $(u)$ is:
$u = \frac{U}{V} = \frac{U}{A d} = \frac{1}{2} \epsilon_0 E^2$.

(B) capacitor is a device that stores electrical energy in an electric field.
When a potential difference is applied across the plates of a capacitor,charges are separated,creating an electric field between the plates.
The energy stored in the capacitor is given by the formula $U = \frac{1}{2} CV^2 = \frac{Q^2}{2C}$,where $C$ is the capacitance,$V$ is the potential difference,and $Q$ is the charge.
This energy is stored in the space between the plates in the form of an electric field.

(N/A) The energy stored $(U)$ in a capacitor with capacitance $C$,charge $Q$,and potential difference $V$ is given by the following three formulas:
$1$. $U = \frac{1}{2} CV^2$
$2$. $U = \frac{Q^2}{2C}$
$3$. $U = \frac{1}{2} QV$
These formulas are derived using the relationship $Q = CV$.

(N/A) Energy density is defined as the energy stored per unit volume in an electric field.
For a parallel plate capacitor,the energy stored $U$ is given by $U = \frac{1}{2} CV^2$.
Since $C = \frac{\epsilon_0 A}{d}$ and $V = Ed$,we have $U = \frac{1}{2} (\frac{\epsilon_0 A}{d}) (Ed)^2 = \frac{1}{2} \epsilon_0 A d E^2$.
The volume of the space between the plates is $V_{vol} = Ad$.
Therefore,the energy density $u$ is $u = \frac{U}{V_{vol}} = \frac{\frac{1}{2} \epsilon_0 A d E^2}{Ad} = \frac{1}{2} \epsilon_0 E^2$.

(C) The energy $U$ stored in a capacitor is given by the formula:
$U = \frac{1}{2} C V^2$
Given:
Capacitance $C = 5.0 \mu F = 5.0 \times 10^{-6} F$
Potential difference $V = 800 V$
Substituting the values into the formula:
$U = \frac{1}{2} \times (5.0 \times 10^{-6} F) \times (800 V)^2$
$U = \frac{1}{2} \times 5.0 \times 10^{-6} \times 640000$
$U = 2.5 \times 10^{-6} \times 6.4 \times 10^5$
$U = 16 \times 10^{-1} = 1.6 J$
Thus,the energy given to the conductor is $1.6 J$.

(C) The energy density $u$ (energy per unit volume) in a parallel plate capacitor is given by the formula: $u = \frac{1}{2} \varepsilon_{0} E^{2}$.
The volume $V$ of the space between the plates 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:
$U = u \times V$
$U = (\frac{1}{2} \varepsilon_{0} E^{2}) \times (Ad)$
$U = \frac{1}{2} \varepsilon_{0} E^{2} Ad$.

(A) The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$,where $q$ is the charge and $C$ is the capacitance.
Since $C$ is constant,$U \propto q^2$.
Let the initial charge be $q$ and the initial energy be $U_i = \frac{q^2}{2C}$.
When the charge is increased by $2 \ C$,the new charge is $q' = q + 2$.
The new energy is $U_f = \frac{(q+2)^2}{2C}$.
Given that the energy increases by $44 \%$,the new energy is $U_f = U_i + 0.44 \, U_i = 1.44 \, U_i$.
Substituting the expressions for $U_f$ and $U_i$:
$\frac{(q+2)^2}{2C} = 1.44 \times \frac{q^2}{2C}$.
$(q+2)^2 = 1.44 \, q^2$.
Taking the square root on both sides:
$q + 2 = \sqrt{1.44} \, q$.
$q + 2 = 1.2 \, q$.
$0.2 \, q = 2$.
$q = \frac{2}{0.2} = 10 \ C$.