Parallel Plate Capacitor Questions in English

(B) The formula for the capacitance of a parallel plate capacitor is $C = \frac{\epsilon A}{d}$,where $\epsilon$ is the absolute permittivity,$A$ is the area,and $d$ is the distance between plates.
Rearranging for $\epsilon$,we get $\epsilon = \frac{C \cdot d}{A}$.
The unit of capacitance $C$ is Farad $(F)$,the unit of distance $d$ is meter $(m)$,and the unit of area $A$ is square meter $(m^2)$.
Substituting these units: $\text{Unit of } \epsilon = \frac{F \cdot m}{m^2} = F \cdot m^{-1}$.
Therefore,the unit of absolute permittivity is $F \cdot m^{-1}$.

(A) The electric field $E$ between two parallel plates is given by $E = \frac{V}{d}$,where $V$ is the potential difference and $d$ is the distance between the plates.
Given: $V = 50\,V$,$d = 5\,mm = 5 \times 10^{-3}\,m$.
$E = \frac{50}{5 \times 10^{-3}} = 10^4\,V/m$.
The force $F$ on the particle is $F = qE$.
The acceleration $a$ is given by $a = \frac{F}{m} = \frac{qE}{m}$.
Given: $q = 10^{-11}\,C$,$m = 10^{-15}\,kg$.
$a = \frac{10^{-11} \times 10^4}{10^{-15}} = \frac{10^{-7}}{10^{-15}} = 10^8\,m/s^2$.

(D) The capacitance $C$ of a parallel plate capacitor is given by the formula $C = \frac{K\varepsilon_0 A}{d}$,where $K$ is the dielectric constant of the medium between the plates,$\varepsilon_0$ is the permittivity of free space,$A$ is the area of each plate,and $d$ is the separation distance between the plates.
From this formula,it is clear that the capacitance depends on the area of the plates,the distance between them,and the dielectric medium,but it does not depend on the thickness of the plates,the material of the plates,or the potential difference applied across them.
Therefore,among the given options,the capacity depends on the separation between the plates.

(B) The capacitance of a parallel plate capacitor is given by the formula $C = \frac{\varepsilon_0 A}{d}$,where $\varepsilon_0$ is the permittivity of free space,$A$ is the area of the plates,and $d$ is the separation between them.
When the separation between the plates is halved,the new separation becomes $d' = d/2$.
The new capacitance $C'$ is given by $C' = \frac{\varepsilon_0 A}{d'} = \frac{\varepsilon_0 A}{d/2}$.
This simplifies to $C' = 2 \times \frac{\varepsilon_0 A}{d} = 2C$.
Therefore,the new capacity is $2C$.

(C) The capacitance $C$ of a parallel plate capacitor is given by the formula $C = \frac{\varepsilon_0 A}{d}$,where $\varepsilon_0$ is the permittivity of free space,$A$ is the area of the plates,and $d$ is the distance between the plates.
From this relation,it is clear that the capacitance $C$ is directly proportional to the area $A$ of the plates $(C \propto A)$.
Therefore,if the area $A$ of the plates increases,the capacity $C$ of the capacitor also increases.

(B) The capacitance of a parallel plate capacitor filled with a dielectric is given by $C = \frac{K \varepsilon_0 A}{d}$.
For a spherical conductor of radius $r$,the capacitance is $C = 4 \pi \varepsilon_0 r$.
Equating the two capacitances: $4 \pi \varepsilon_0 r = \frac{K \varepsilon_0 A}{d}$.
Solving for $r$: $r = \frac{KA}{4 \pi d}$.
Given: $A = 100 \, cm^2 = 100 \times 10^{-4} \, m^2 = 10^{-2} \, m^2$,$d = 1 \, mm = 10^{-3} \, m$,and $K = 6$.
Substituting the values: $r = \frac{6 \times 10^{-2}}{4 \times 3.14 \times 10^{-3}} = \frac{6 \times 10}{12.56} = \frac{60}{12.56} \approx 4.77 \, m$.

(A) For a parallel plate capacitor,the capacitance is given by $C = \frac{\epsilon_0 A}{d}$.
Assuming the capacitor is disconnected from the battery,the charge $Q$ remains constant.
Since $Q = CV$,we have $V = \frac{Q}{C} = \frac{Qd}{\epsilon_0 A}$.
This implies $V \propto d$.
Therefore,$\frac{V_1}{V_2} = \frac{d_1}{d_2}$.
Given $V_1 = 60 \ V$,$d_1 = 4 \ mm$,and $d_2 = 12 \ mm$.
$V_2 = \frac{V_1 \times d_2}{d_1} = \frac{60 \times 12}{4} = 180 \ V$.

(C) The electric field $E$ between the plates of a parallel plate capacitor is given by the formula $E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{A\varepsilon_0}$,where $Q$ is the charge on the plates,$A$ is the area of the plates,and $\varepsilon_0$ is the permittivity of free space.
Since the electric field $E$ depends only on the charge density $\sigma$ (or charge $Q$ and area $A$) and is independent of the distance $d$ between the plates,the electric intensity remains unchanged when the distance is increased,assuming the charge $Q$ remains constant.
Therefore,$E \propto d^0$.

(A) The electric field $E$ produced by one plate of a parallel plate capacitor at the location of the other plate is given by $E = \frac{\sigma}{2\varepsilon_0 K}$,where $\sigma = \frac{q}{A}$ is the surface charge density.
Substituting $\sigma$,we get $E = \frac{q}{2A\varepsilon_0 K}$.
The force $F$ on a charge $q$ placed in this electric field is $F = qE$.
Therefore,$F = q \times \left( \frac{q}{2A\varepsilon_0 K} \right) = \frac{q^2}{2\varepsilon_0 AK}$.

(C) In a parallel plate capacitor,the charge on the plates is induced by the electric field between them.
According to the principle of electrostatic induction,the magnitude of the induced charge on the plates must be equal and opposite to maintain the charge neutrality of the capacitor system as a whole.
Even if the physical dimensions (area) of the plates are different,the magnitude of the charge $q$ on both plates remains equal in magnitude ($+q$ and $-q$).
Therefore,the charge on the smaller plate is equal to the charge on the larger plate.

(B) Initially,the electric field between the plates of a parallel plate capacitor is $E = \frac{\sigma}{\varepsilon_0}$,where $\sigma$ is the surface charge density. The force on the particle is $F = qE = \frac{q\sigma}{\varepsilon_0}$.
When one plate is removed,the capacitor effectively becomes a single charged sheet. The electric field due to a single charged sheet is given by $E' = \frac{\sigma}{2\varepsilon_0}$.
Therefore,the new force acting on the particle is $F' = qE' = q \left( \frac{\sigma}{2\varepsilon_0} \right) = \frac{1}{2} \left( \frac{q\sigma}{\varepsilon_0} \right) = \frac{F}{2}$.

(B) The capacitance of a sphere of radius $R$ is given by $C_{sphere} = 4\pi \varepsilon_0 R$.
Given $R = 1\,m$,so $C_{sphere} = 4\pi \varepsilon_0 \times 1 = 4\pi \varepsilon_0$.
The capacitance of a parallel plate capacitor is given by $C_{parallel} = \frac{\varepsilon_0 A}{d}$,where $A = \pi r^2$ is the area of the plates and $d$ is the distance between them.
The diameter of the plate is $40\,mm$,so the radius $r = 20\,mm = 20 \times 10^{-3}\,m$.
Equating the two capacitances: $4\pi \varepsilon_0 = \frac{\varepsilon_0 \pi r^2}{d}$.
Canceling $\pi \varepsilon_0$ from both sides,we get $4 = \frac{r^2}{d}$,which implies $d = \frac{r^2}{4}$.
Substituting the value of $r$: $d = \frac{(20 \times 10^{-3})^2}{4} = \frac{400 \times 10^{-6}}{4} = 100 \times 10^{-6}\,m = 0.1 \times 10^{-3}\,m = 0.1\,mm$.

(D) The capacitance $C$ of a parallel plate capacitor is given by the formula:
$C = \frac{K \varepsilon_0 A}{d}$
Where:
$K$ is the dielectric constant of the medium between the plates.
$\varepsilon_0$ is the permittivity of free space.
$A$ is the area of each plate.
$d$ is the distance between the plates.
From this formula,it is evident that the capacitance depends on the area of the plates $(A)$,the medium between the plates $(K)$,and the distance between the plates $(d)$.
It does not depend on the material (metal) of the plates,provided the plates are good conductors.

(B) The electric field $E$ between the plates of a capacitor is given by the formula $E = \frac{V}{d}$,where $V$ is the potential difference and $d$ is the distance between the plates.
Given:
Potential difference $V = 100\,V$
Distance $d = 1\,mm = 1 \times 10^{-3}\,m$
Substituting the values:
$E = \frac{100}{1 \times 10^{-3}} = 100 \times 10^3 = 100,000\,V/m$.
Note: The dielectric constant does not affect the electric field calculation when the potential difference $V$ is maintained constant across the plates.

(A) The capacitance of a spherical conductor of radius $R$ is given by $C_1 = 4\pi \varepsilon_0 R$. Given diameter $D = 20\,cm$,so radius $R = 10\,cm = 0.1\,m$. Thus,$C_1 = 4\pi \varepsilon_0 (0.1)$.
The capacitance of a parallel plate capacitor is given by $C_2 = \frac{\varepsilon_0 A}{d}$,where $A = \pi r^2$ and $r$ is the radius of the plate. Given diameter of the plate is $4\,cm$,so $r = 2\,cm = 0.02\,m$. Thus,$A = \pi (0.02)^2 = 4\pi \times 10^{-4}\,m^2$.
Equating the two capacitances,$C_1 = C_2$:
$4\pi \varepsilon_0 (0.1) = \frac{\varepsilon_0 (4\pi \times 10^{-4})}{d}$
$0.1 = \frac{4 \times 10^{-4}}{d}$
$d = \frac{4 \times 10^{-4}}{0.1} = 4 \times 10^{-3}\,m$.

(D) The electric field $E$ between the plates of a parallel plate capacitor is given by the formula $E = \frac{\sigma}{\epsilon_0}$,where $\sigma$ is the surface charge density and $\epsilon_0$ is the permittivity of free space.
Since $\sigma = \frac{Q}{A}$,the electric field can be expressed as $E = \frac{Q}{A\epsilon_0}$.
In this expression,$Q$ is the charge on the plates,$A$ is the area of the plates,and $\epsilon_0$ is a constant.
As the formula does not contain the distance $d$ between the plates,the electric field is uniform and independent of the distance between the plates.

(A) Initial capacitance $C = 50\,\mu F = 50 \times 10^{-6}\,F$ and initial potential $V = 100\;V$.
Initial charge $Q = CV = 50 \times 10^{-6} \times 100 = 5 \times 10^{-3}\,C$.
Since the capacitor is disconnected from the battery,the charge $Q$ remains constant.
When the distance between the plates is doubled,the new capacitance $C' = C/2 = 25\,\mu F$.
The new potential $V' = Q/C' = Q/(C/2) = 2V = 200\;V$.
The work done by the external agent is the change in stored energy: $W = U_f - U_i$.
$U_i = \frac{1}{2}CV^2 = \frac{1}{2} \times 50 \times 10^{-6} \times (100)^2 = 0.25\;J$.
$U_f = \frac{1}{2}C'V'^2 = \frac{1}{2} \times (C/2) \times (2V)^2 = \frac{1}{2} \times (C/2) \times 4V^2 = CV^2 = 0.50\;J$.
$W = 0.50 - 0.25 = 0.25\;J = 25 \times 10^{-2}\,J$.

(A) The capacitance of a parallel plate capacitor is given by the formula $C = \frac{\varepsilon_0 A}{d}$.
Rearranging the formula to solve for the area $A$,we get $A = \frac{Cd}{\varepsilon_0}$.
Given values are $C = 3\,F$,$d = 5\,mm = 5 \times 10^{-3}\,m$,and $\varepsilon_0 = 8.854 \times 10^{-12}\,F/m$.
Substituting these values into the formula:
$A = \frac{3 \times 5 \times 10^{-3}}{8.854 \times 10^{-12}}$
$A = \frac{15 \times 10^{-3}}{8.854 \times 10^{-12}}$
$A \approx 1.694 \times 10^9\,m^2$.

(B) The capacitance $C$ of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$,where $A = \pi r^2$.
Given: $r = 0.08\,m$,$d = 1.0 \times 10^{-3}\,m$,$V = 100\,V$,and $\varepsilon_0 = 8.854 \times 10^{-12}\,F/m$.
First,calculate the area $A = \pi (0.08)^2 = \pi \times 0.0064\,m^2$.
Now,calculate the capacitance $C = \frac{8.854 \times 10^{-12} \times \pi \times 0.0064}{1.0 \times 10^{-3}} \approx 1.78 \times 10^{-10}\,F$.
The charge $Q$ is given by $Q = CV = (1.78 \times 10^{-10}) \times 100 = 1.78 \times 10^{-8}\,C \approx 1.8 \times 10^{-8}\,C$.

(C) The electric field between the plates of a parallel plate capacitor is uniform,meaning its magnitude and direction are the same at every point between the plates.
The force $F$ experienced by a charge $q$ in an electric field $E$ is given by $F = qE$.
Since both $A$ and $B$ are protons,they have the same charge $q = e$.
Because the electric field $E$ is uniform throughout the region between the plates,both protons experience the same force.
Therefore,$F_A = F_B$.

(C) The capacitance of a parallel plate capacitor is given by the formula $C = \frac{\epsilon_0 A}{d}$,where $A$ is the area of the plates and $d$ is the distance between them.
From this formula,we can see that $C \propto \frac{1}{d}$.
Given initial capacitance $C_1 = 15\,\mu F$ at distance $d_1 = 6\,cm$.
New distance $d_2 = 2\,cm$.
Using the ratio $\frac{C_2}{C_1} = \frac{d_1}{d_2}$,we get:
$\frac{C_2}{15} = \frac{6}{2}$
$\frac{C_2}{15} = 3$
$C_2 = 15 \times 3 = 45\,\mu F$.
Therefore,the new capacity is $45\,\mu F$.

(C) The capacitance $C$ of a parallel plate capacitor is given by the formula $C = \frac{\epsilon_0 A}{d}$,where $A$ is the area of the plates and $d$ is the distance between them.
From this relation,we can see that $C \propto \frac{1}{d}$.
Therefore,the ratio of the capacitances is $\frac{C_1}{C_2} = \frac{d_2}{d_1}$.
Given: $C_1 = 10\,\mu F$,$d_1 = 8\,cm$,and $d_2 = 4\,cm$.
Substituting these values into the ratio formula: $\frac{10}{C_2} = \frac{4}{8}$.
This simplifies to $\frac{10}{C_2} = \frac{1}{2}$.
Solving for $C_2$,we get $C_2 = 10 \times 2 = 20\,\mu F$.

(B) The force $F$ on one plate of a parallel plate capacitor due to the electric field $E$ created by the other plate is given by $F = Q \times E_{plate}$.
Since the total electric field between the plates is $E = E_{plate} + E_{plate} = 2 E_{plate}$,the field due to one plate is $E_{plate} = E/2$.
Therefore,the force on each plate is $F = Q \times (E/2) = (Q \times E) / 2$.
Given $Q = 1 \, \mu C = 10^{-6} \, C$ and $E = 10^5 \, V/m$.
Substituting the values: $F = (10^{-6} \times 10^5) / 2 = 10^{-1} / 2 = 0.1 / 2 = 0.05 \, N$.

(D) The capacitance of a parallel plate capacitor is given by the formula $C = \frac{\varepsilon_0 A}{d}$.
Given initial capacitance $C = 12\,\mu F$.
When the distance between the plates is doubled,$d' = 2d$.
When the area of the plates is halved,$A' = \frac{A}{2}$.
The new capacitance $C'$ is given by $C' = \frac{\varepsilon_0 A'}{d'} = \frac{\varepsilon_0 (A/2)}{2d} = \frac{1}{4} \left( \frac{\varepsilon_0 A}{d} \right) = \frac{1}{4} C$.
Substituting the value of $C$,we get $C' = \frac{1}{4} \times 12\,\mu F = 3\,\mu F$.

(C) In a stack of $n$ plates connected alternately, the plates form $(n - 1)$ capacitors connected in parallel.
Each pair of adjacent plates acts as a single capacitor with capacitance $C$.
Since these $(n - 1)$ capacitors are connected in parallel, the equivalent capacitance $C_R$ is the sum of individual capacitances.
Therefore, $C_R = C + C + ... + (n - 1) \text{ times} = (n - 1)C$.

(D) The capacitance of a parallel plate capacitor is given by the formula $C = \frac{K \varepsilon_0 A}{d}$,where $A = l \times b$.
Given:
Capacitance $C = 2 \times 10^{-6}\,F$
Dielectric constant $K = 2.5$
Thickness $d = 0.15 \times 10^{-3}\,m$
Width $b = 400 \times 10^{-3}\,m = 0.4\,m$
Permittivity of free space $\varepsilon_0 = 8.85 \times 10^{-12}\,F/m$
Substituting the values into the formula:
$2 \times 10^{-6} = \frac{2.5 \times 8.85 \times 10^{-12} \times (l \times 0.4)}{0.15 \times 10^{-3}}$
$2 \times 10^{-6} = \frac{2.5 \times 8.85 \times 10^{-12} \times 0.4 \times l}{0.15 \times 10^{-3}}$
$2 \times 10^{-6} = \frac{8.85 \times 10^{-12} \times l}{0.15 \times 10^{-3}}$
$l = \frac{2 \times 10^{-6} \times 0.15 \times 10^{-3}}{8.85 \times 10^{-12}}$
$l = \frac{0.3 \times 10^{-9}}{8.85 \times 10^{-12}} = \frac{300}{8.85} \approx 33.9\,m$.

(C) The given circuit consists of $4$ capacitors connected in parallel. Each capacitor is formed by two adjacent plates with area $A$ and separation $d$. The capacitance of each capacitor is $C = \frac{\varepsilon_0 A}{d}$.
Plate $1$ is connected to the positive terminal of the battery. It forms one capacitor with plate $2$. The charge on plate $1$ is $q_1 = +CV = +\frac{\varepsilon_0 AV}{d}$.
Plate $4$ is connected to the negative terminal of the battery. It acts as one plate for two capacitors: one with plate $3$ and one with plate $5$. Since plate $4$ is connected to the negative terminal,the charge on both sides of plate $4$ will be negative. The total charge on plate $4$ is $q_4 = -CV - CV = -2CV = -\frac{2\varepsilon_0 AV}{d}$.
Thus,the charges on plates $1$ and $4$ are $\frac{\varepsilon_0 AV}{d}$ and $-\frac{2\varepsilon_0 AV}{d}$ respectively.

(D) Let the capacitance of each pair of adjacent plates be $C = \frac{{\varepsilon _0}A}{d}$.
From the given circuit diagram,we can identify the arrangement of capacitors.
The top plate and the second plate form one capacitor $C_1$.
The second plate and the third plate form another capacitor $C_2$.
The third plate and the fourth plate form a third capacitor $C_3$.
Based on the connections,the capacitor $C_1$ is in parallel with the series combination of $C_2$ and $C_3$.
Thus,the equivalent capacitance is $C_{eq} = C_1 + \frac{C_2 \cdot C_3}{C_2 + C_3} = C + \frac{C \cdot C}{C + C} = C + \frac{C}{2} = \frac{3C}{2}$.
Substituting $C = \frac{{\varepsilon _0}A}{d}$,we get $C_{eq} = \frac{3{\varepsilon _0}A}{2d}$.

(C) For a parallel plate capacitor,the charges on the inner surfaces facing each other must be equal in magnitude and opposite in sign. Therefore,$Q_2 = -Q_3$.
The potential difference $V$ between the plates of a capacitor is given by the ratio of the charge on the inner surface of the positive plate to the capacitance $C$.
$V = \frac{Q_2}{C}$
Since $Q_3 = -Q_2$,we can write $Q_2 = \frac{Q_2 - Q_3}{2}$.
Substituting this into the potential difference formula:
$V = \frac{Q_2 - Q_3}{2C}$

(B) The capacitance $C$ of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$,where $\varepsilon_0$ is the permittivity of free space,$A$ is the area of the plates,and $d$ is the separation distance.
To find the rate of change of capacitance with respect to time $t$,we differentiate $C$ with respect to $t$:
$\frac{dC}{dt} = \frac{d}{dt} \left( \frac{\varepsilon_0 A}{d} \right) = \varepsilon_0 A \frac{d}{dt} (d^{-1})$.
Using the chain rule,$\frac{dC}{dt} = \varepsilon_0 A (-d^{-2}) \frac{dd}{dt}$.
Since the plates are pulled apart with velocity $v$,the rate of change of separation is $\frac{dd}{dt} = v$.
Substituting this,we get $\frac{dC}{dt} = -\frac{\varepsilon_0 A}{d^2} v$.
The magnitude of the rate of change is $\left| \frac{dC}{dt} \right| = \frac{\varepsilon_0 A v}{d^2}$.
Thus,$\left| \frac{dC}{dt} \right| \propto \frac{1}{d^2}$.

(D) The capacitance of a parallel plate capacitor with a dielectric slab of thickness $t$ is given by $C = \frac{\epsilon_0 A}{d - t + (t/K)}$.
For a metal sheet (aluminum),the dielectric constant $K$ is infinite $(K = \infty)$.
Given that the thickness $t$ is negligible $(t \approx 0)$,we analyze the effect.
However,if we place a conducting sheet of thickness $t$ inside,the effective distance between the plates becomes $(d - t)$.
The new capacitance is $C' = \frac{\epsilon_0 A}{d - t}$.
Since $(d - t) < d$,the denominator decreases,which means the capacitance $C'$ increases.
Even if the thickness is negligible,the presence of a conducting plate effectively reduces the separation distance between the plates,leading to an increase in capacitance.

(A) The potential difference $V$ across the plates remains constant because the capacitor is connected to a battery.
The electric field $E$ between the plates is given by $E = V/d$,where $d$ is the distance between the plates.
As the plates are moved further apart,$d$ increases,so $E = V/d$ decreases.
The capacitance $C$ of a parallel plate capacitor is given by $C = \epsilon_0 A / d$.
As $d$ increases,the capacitance $C$ decreases.
The charge $Q$ on the plates is given by $Q = CV$.
Since $C$ decreases and $V$ is constant,the charge $Q$ on the plates also decreases.
Therefore,both the electric field intensity and the charge on the plates decrease.

(B) The electric field $E$ between the plates of a parallel plate capacitor is given by $E = \frac{\sigma}{\epsilon_0}$,where $\sigma$ is the surface charge density.
This field is the superposition of the fields produced by the two individual plates,each contributing $E_{plate} = \frac{\sigma}{2\epsilon_0}$.
Since $F = qE$,the force $F$ experienced by a test charge $q$ is $F = q \left( \frac{\sigma}{\epsilon_0} \right)$.
When one plate is removed,the electric field becomes $E' = \frac{\sigma}{2\epsilon_0}$,which is half of the original field.
Therefore,the new force $F'$ experienced by the test charge is $F' = qE' = q \left( \frac{\sigma}{2\epsilon_0} \right) = \frac{F}{2}$.

(B) The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$.
Since the battery is disconnected,the charge $Q$ on the plates remains constant.
When the distance $d$ between the plates is increased,the capacitance $C$ decreases because $C \propto \frac{1}{d}$.
The potential difference $V$ between the plates is given by $V = \frac{Q}{C}$.
Since $Q$ is constant and $C$ decreases,the potential difference $V$ must increase.
Therefore,the correct statement is that the potential difference between the plates increases.

(D) The capacitance $C$ of a parallel plate capacitor is given by the formula:
$C = \frac{K \varepsilon_0 A}{d}$
Where:
$A$ is the area of the plates,
$d$ is the distance between the plates,
$K$ is the dielectric constant of the medium between the plates,
$\varepsilon_0$ is the permittivity of free space.
From this formula,it is evident that the capacitance depends on the area of the plates,the distance between them,and the dielectric medium,but it does not depend on the material (metal) of the plates.

(B) Given: Area $A = 400 \, cm^2 = 400 \times 10^{-4} \, m^2 = 4 \times 10^{-2} \, m^2$.
Distance $d = 2 \, mm = 2 \times 10^{-3} \, m$.
Potential difference $V = 200 \, V$.
Permittivity of free space $\epsilon_0 = 8.854 \times 10^{-12} \, F/m$.
The capacitance $C$ is given by $C = \frac{\epsilon_0 A}{d}$.
Substituting the values: $C = \frac{8.854 \times 10^{-12} \times 4 \times 10^{-2}}{2 \times 10^{-3}} = 17.708 \times 10^{-11} \, F$.
The charge $Q$ is given by $Q = CV$.
$Q = (17.708 \times 10^{-11}) \times 200 = 3541.6 \times 10^{-11} \, C = 3.5416 \times 10^{-8} \, C$.
Thus,the charge on the plates is approximately $3.54 \times 10^{-8} \, C$.

(B) The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$.
Let the initial capacitance be $C_1 = \frac{\varepsilon_0 A_1}{d_1}$.
Given: New area $A_2 = 2A_1$ and new distance $d_2 = \frac{d_1}{2}$.
The new capacitance $C_2$ is given by $C_2 = \frac{\varepsilon_0 A_2}{d_2}$.
Substituting the values: $C_2 = \frac{\varepsilon_0 (2A_1)}{(d_1/2)} = 4 \times \frac{\varepsilon_0 A_1}{d_1} = 4C_1$.

(C) The capacitance of a parallel plate capacitor is given by the formula $C = \frac{\epsilon_0 A}{d}$.
From this,we can see that $C \propto \frac{1}{d}$.
Therefore,the ratio of the capacitances is $\frac{C_1}{C_2} = \frac{d_2}{d_1}$.
Given: $C_1 = 15\ \mu F$,$d_1 = 6\, cm$,and $d_2 = 2\, cm$.
Substituting these values into the ratio formula:
$\frac{15}{C_2} = \frac{2}{6}$
$\frac{15}{C_2} = \frac{1}{3}$
$C_2 = 15 \times 3 = 45\ \mu F$.
Thus,the new capacitance is $45\ \mu F$.

(B) The capacitance of a parallel plate capacitor is given by $C_p = \frac{\epsilon_0 A}{d}$,where $A = \pi r^2$ is the area of the plates and $r$ is the radius of the plates.
Given diameter $= 40\, mm$,so radius $r = 20\, mm = 20 \times 10^{-3}\, m$.
The capacitance of an isolated metallic sphere is given by $C_s = 4\pi \epsilon_0 R$,where $R = 1\, mm = 1 \times 10^{-3}\, m$.
Equating the two capacitances: $\frac{\epsilon_0 A}{d} = 4\pi \epsilon_0 R$.
Substituting $A = \pi r^2$: $\frac{\pi r^2}{d} = 4\pi R$.
Solving for $d$: $d = \frac{r^2}{4R}$.
Substituting the values: $d = \frac{(20 \times 10^{-3})^2}{4 \times 1 \times 10^{-3}} = \frac{400 \times 10^{-6}}{4 \times 10^{-3}} = 100 \times 10^{-3}\, m = 0.1\, mm$.

(D) The capacitance of a parallel plate capacitor is given by the formula $C = \frac{\varepsilon_0 A}{d}$.
Given initial capacitance $C = 12 \ \mu F$.
When the area $A$ is halved $(A' = A/2)$ and the distance $d$ is doubled $(d' = 2d)$,the new capacitance $C'$ becomes:
$C' = \frac{\varepsilon_0 A'}{d'} = \frac{\varepsilon_0 (A/2)}{2d} = \frac{1}{4} \left( \frac{\varepsilon_0 A}{d} \right)$.
Substituting the initial value of $C$:
$C' = \frac{1}{4} C = \frac{12 \ \mu F}{4} = 3 \ \mu F$.

(D) The dielectric strength $E$ is the maximum electric field the dielectric can withstand. The separation $d$ between the plates is given by $d = \frac{V}{E}$.
Substituting the given values,$V = 12 \times 10^3 \; V$ and $E = 10^9 \; Vm^{-1}$,we get $d = \frac{12 \times 10^3}{10^9} = 12 \times 10^{-6} \; m$.
The capacitance of a parallel plate capacitor is $C = \frac{K \varepsilon_0 A}{d}$.
Rearranging for area $A$,we get $A = \frac{C d}{K \varepsilon_0} = \frac{C V}{K E \varepsilon_0}$.
Substituting the values $C = 80 \times 10^{-12} \; F$,$V = 12 \times 10^3 \; V$,$K = 5$,$E = 10^9 \; Vm^{-1}$,and $\varepsilon_0 = 8.85 \times 10^{-12} \; Fm^{-1}$:
$A = \frac{80 \times 10^{-12} \times 12 \times 10^3}{5 \times 10^9 \times 8.85 \times 10^{-12}} = \frac{960 \times 10^{-9}}{44.25 \times 10^{-3}} \approx 21.69 \times 10^{-6} \; m^2$.
Thus,the minimum area is approximately $21.7 \times 10^{-6} \; m^2$.

(B) The energy density $(u)$ of a parallel plate capacitor is given by the formula:
$u = \frac{1}{2} \epsilon_0 E^2 = \frac{1}{2} \epsilon_0 \left( \frac{V}{d} \right)^2$
Given:
$V = 300 \ V$
$d = 2 \ mm = 2 \times 10^{-3} \ m$
$\epsilon_0 = 8.85 \times 10^{-12} \ F/m$
Substituting the values:
$u = \frac{1}{2} \times 8.85 \times 10^{-12} \times \left( \frac{300}{2 \times 10^{-3}} \right)^2$
$u = 0.5 \times 8.85 \times 10^{-12} \times (1.5 \times 10^5)^2$
$u = 0.5 \times 8.85 \times 10^{-12} \times 2.25 \times 10^{10}$
$u \approx 0.1 \ J/m^3$

(A) The capacitance of a parallel plate capacitor is given by the formula $C = \frac{\varepsilon_0 A}{d}$.
Rearranging the formula to solve for the area $A$,we get $A = \frac{Cd}{\varepsilon_0}$.
Given values are $C = 3 \ F$,$d = 5 \ mm = 5 \times 10^{-3} \ m$,and $\varepsilon_0 = 8.85 \times 10^{-12} \ F/m$.
Substituting these values into the equation:
$A = \frac{3 \times 5 \times 10^{-3}}{8.85 \times 10^{-12}}$
$A = \frac{15 \times 10^{-3}}{8.85 \times 10^{-12}}$
$A \approx 1.694 \times 10^9 \ m^2$.

(C) For a parallel plate capacitor,the charges on the facing surfaces are equal in magnitude and opposite in sign. Thus,$Q_2 = -Q_3$.
The potential difference $V$ between the plates of a capacitor is given by $V = \frac{Q}{C}$,where $Q$ is the charge on the inner surface of the positive plate.
Here,the charge on the inner surface of the first plate is $Q_2$. Therefore,the potential difference is $V = \frac{Q_2}{C}$.
Since $Q_2 = -Q_3$,we can write $Q_2 = \frac{Q_2 - Q_3}{2}$.
Substituting this into the potential formula:
$V = \frac{Q_2 - Q_3}{2C}$.

(B) Let $C = \frac{{\varepsilon _0}A}{d}$ be the capacitance of a single pair of adjacent plates.
From the circuit diagram,we identify the capacitors formed between the plates:
$1$. Capacitor between plate $2$ and $1$ (connected to $A$ and $B$ respectively).
$2$. Capacitor between plate $2$ and $3$ (connected to $A$ and $B$ respectively).
$3$. Capacitor between plate $4$ and $3$ (connected to $A$ and $B$ respectively).
$4$. Capacitor between plate $4$ and $5$ (connected to $A$ and $B$ respectively).
Analyzing the connections: Plate $2$ is at potential $A$. Plates $1, 3, 5$ are connected to $B$. Plate $4$ is connected to $1$,so it is also at potential $B$.
Thus,we have capacitors between $(2,1)$,$(2,3)$,$(4,3)$,and $(4,5)$.
Since $1, 3, 5$ are all at potential $B$ and $2$ is at $A$,we have capacitors $C_{21}, C_{23}$ in parallel. Also,$4$ is at $B$,so $C_{43}$ and $C_{45}$ are in parallel.
Wait,looking at the diagram: $2$ is connected to $A$. $1, 3, 4, 5$ are connected to $B$.
Actually,the plates form capacitors: $C_{21}$ (between $2$ and $1$),$C_{23}$ (between $2$ and $3$),$C_{43}$ (between $4$ and $3$),$C_{45}$ (between $4$ and $5$).
Since $1, 3, 5$ are connected to $B$ and $4$ is connected to $1$,all plates $1, 3, 4, 5$ are at potential $B$. Plate $2$ is at $A$.
This results in $4$ capacitors in parallel: $C_{eq} = C_{21} + C_{23} + C_{43} + C_{45} = 4C = \frac{4{\varepsilon _0}A}{d}$.

(A) In a system of $n$ plates connected alternately,we form $(n - 1)$ capacitors in parallel.
Each capacitor is formed by two adjacent plates,and the capacitance of each such pair is given as $C$.
Since the plates are connected alternately,all these $(n - 1)$ capacitors are connected in parallel.
The equivalent capacitance $C_{eq}$ for capacitors in parallel is the sum of individual capacitances:
$C_{eq} = C_1 + C_2 + ... + C_{n-1}$
Since each $C_i = C$,we have:
$C_{eq} = (n - 1)C$.

(C) The capacitance $C$ of a parallel plate capacitor is given by the formula $C = \frac{\epsilon_0 A}{d}$.
Here,the area $A = 5 \, km \times 5 \, km = (5 \times 10^3 \, m) \times (5 \times 10^3 \, m) = 25 \times 10^6 \, m^2$.
The distance $d = 1 \, km = 10^3 \, m$.
The permittivity of free space $\epsilon_0 = 8.854 \times 10^{-12} \, F/m$.
Substituting these values into the formula:
$C = \frac{8.854 \times 10^{-12} \times 25 \times 10^6}{10^3}$
$C = 8.854 \times 25 \times 10^{-12+6-3}$
$C = 221.35 \times 10^{-9} \, F$
$C = 0.22135 \times 10^{-6} \, F = 0.22 \, \mu F$.

(B) The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$,where $A = \pi r^2$ is the area of the plate.
Given diameter of the plate is $4 \ cm$,so radius $r = 2 \ cm = 2 \times 10^{-2} \ m$.
Area $A = \pi (2 \times 10^{-2})^2 = 4\pi \times 10^{-4} \ m^2$.
The capacitance of a sphere of radius $R$ is $C = 4\pi \varepsilon_0 R$.
Given diameter of the sphere is $20 \ cm$,so radius $R = 10 \ cm = 0.1 \ m$.
Equating the two capacitances: $\frac{\varepsilon_0 A}{d} = 4\pi \varepsilon_0 R$.
$\frac{A}{d} = 4\pi R$.
$d = \frac{A}{4\pi R} = \frac{4\pi \times 10^{-4}}{4\pi \times 0.1} = \frac{10^{-4}}{10^{-1}} = 10^{-3} \ m$.

(A) The energy density $u$ of an electric field is given by $u = \frac{1}{2} \varepsilon_0 E^2$.
For a parallel plate capacitor,the electric field $E$ between the plates is $E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{A \varepsilon_0}$.
Substituting this value of $E$ into the energy density formula:
$u = \frac{1}{2} \varepsilon_0 \left( \frac{Q}{A \varepsilon_0} \right)^2$
$u = \frac{1}{2} \varepsilon_0 \left( \frac{Q^2}{A^2 \varepsilon_0^2} \right)$
$u = \frac{Q^2}{2 \varepsilon_0 A^2}$.

(A) The force $F$ between the plates of a parallel plate capacitor is given by $F = \frac{Q^2}{2\varepsilon_0 A}$.
Since the charge $Q = CV$,we substitute this into the formula:
$F = \frac{(CV)^2}{2\varepsilon_0 A} = \frac{C^2V^2}{2\varepsilon_0 A}$.
We know that the capacitance $C = \frac{\varepsilon_0 A}{d}$,which implies $\varepsilon_0 A = Cd$.
Substituting $\varepsilon_0 A = Cd$ into the force equation:
$F = \frac{C^2V^2}{2(Cd)} = \frac{CV^2}{2d}$.
Thus,the force acting between the plates is $\frac{CV^2}{2d}$.