Parallel Plate Capacitor Questions in English

(B) The initial capacitance of a parallel plate capacitor with plate separation $d$ and area $A$ is given by $C = \frac{\varepsilon_0 A}{d}$.
When a metal slab of thickness $b$ is inserted between the plates,the effective separation between the plates becomes $(d - b)$.
The new capacitance $C'$ is given by $C' = \frac{\varepsilon_0 A}{d - b}$.
Given $b = d/2$,we substitute this into the formula:
$C' = \frac{\varepsilon_0 A}{d - d/2} = \frac{\varepsilon_0 A}{d/2} = \frac{2\varepsilon_0 A}{d}$.
Therefore,the ratio of the new capacitance to the original capacitance is:
$\frac{C'}{C} = \frac{2\varepsilon_0 A / d}{\varepsilon_0 A / d} = \frac{2}{1}$ or $2 : 1$.

(B) Let the potential of the positive terminal of the battery be $V$ and the negative terminal be $0$.
Plates $1, 3,$ and $5$ are connected to the positive terminal,so their potential is $V$.
Plates $2$ and $4$ are connected to the negative terminal,so their potential is $0$.
This arrangement forms capacitors between adjacent plates.
Capacitor $C_1$ is formed by plates $1$ and $2$,$C_2$ by plates $2$ and $3$,$C_3$ by plates $3$ and $4$,and $C_4$ by plates $4$ and $5$.
Each capacitor has capacitance $C = \frac{{{\varepsilon _0}A}}{d}$.
Plate $1$ has only one face forming a capacitor with plate $2$. The charge on plate $1$ is $Q_1 = C \times (V - 0) = \frac{{{\varepsilon _0}AV}}{d}$.
Plate $4$ has two faces: one forming a capacitor with plate $3$ and another with plate $5$. Both plates $3$ and $5$ are at potential $V$,while plate $4$ is at potential $0$.
The charge on plate $4$ is $Q_4 = C \times (0 - V) + C \times (0 - V) = -\frac{{{\varepsilon _0}AV}}{d} - \frac{{{\varepsilon _0}AV}}{d} = -\frac{{2{\varepsilon _0}AV}}{d}$.

(B) The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_{0} A}{d}$,where $A$ is the area of the plates and $d$ is the distance between them.
When the distance $d$ is increased,the capacitance $C$ decreases.
Since the battery is disconnected,the charge $Q$ on the plates remains constant.
The relationship between charge,capacitance,and potential difference is $V = \frac{Q}{C}$.
Since $Q$ is constant and $C$ decreases,the potential difference $V$ must increase.

(A) For an isolated parallel plate capacitor,the charge $Q$ on the plates remains constant.
The electric field $E$ produced by one plate at the location of the other plate is given by $E = \frac{\sigma}{2\varepsilon_0}$,where $\sigma = \frac{Q}{A}$ is the surface charge density.
The electrostatic force $F$ experienced by one plate due to the electric field of the other is $F = Q \times E$.
Substituting the expression for $E$,we get $F = Q \times \frac{Q}{2A\varepsilon_0} = \frac{Q^2}{2A\varepsilon_0}$.
Since $Q$,$A$,and $\varepsilon_0$ are constants,the force $F$ is independent of the distance $d$ between the plates.

(B) Let the potential of plate $1$ be $V_A$ and plate $3$ be $V_B$. The plates $2$ and $4$ are connected together,let their potential be $V_x$. Plate $5$ is isolated.
The capacitors are formed between adjacent plates:
$C_1$ between $1$ and $2$ with capacitance $C = \frac{\varepsilon_0 A}{d}$.
$C_2$ between $2$ and $3$ with capacitance $C = \frac{\varepsilon_0 A}{d}$.
$C_3$ between $3$ and $4$ with capacitance $C = \frac{\varepsilon_0 A}{d}$.
$C_4$ between $4$ and $5$ with capacitance $C = \frac{\varepsilon_0 A}{d}$.
Since plate $5$ is isolated,it does not participate in the circuit. The circuit consists of $C_1$ in series with the parallel combination of $C_2$ and $C_3$.
Equivalent capacitance $C_{eq} = \frac{C_1 \times (C_2 + C_3)}{C_1 + C_2 + C_3} = \frac{C \times (2C)}{C + 2C} = \frac{2}{3}C = \frac{2\varepsilon_0 A}{3d}$.
The charge flown through the cell is $Q = C_{eq} E = \frac{2\varepsilon_0 AE}{3d}$.

(B) In the given arrangement,the plates $A, C, E$ are connected to the positive terminal of the battery,and plates $B, D$ are connected to the negative terminal.
This configuration forms four capacitors in parallel,each with capacitance $C = 2 \mu F$.
The capacitors are formed between the pairs of plates: $(A, B)$,$(B, C)$,$(C, D)$,and $(D, E)$.
Plate $C$ acts as a common plate for two capacitors: one formed with plate $B$ and one formed with plate $D$.
For the capacitor formed by plates $B$ and $C$,the charge on plate $C$ is $q_1 = +CV$.
For the capacitor formed by plates $C$ and $D$,the charge on plate $C$ is $q_2 = +CV$.
Therefore,the total charge on plate $C$ is $q = q_1 + q_2 = 2CV$.
Given $C = 2 \mu F$ and $V = 10 \ V$,we have $q = 2 \times 2 \mu F \times 10 \ V = 40 \mu C$.

(A) Consider a small elemental strip of thickness $dy$ at a distance $y$ from one plate. This strip acts as a capacitor with capacitance $dC = \frac{K \varepsilon_0 A}{dy}$.
Since these elemental capacitors are in series,the equivalent capacitance $C$ is given by $\frac{1}{C} = \int_0^d \frac{dy}{K \varepsilon_0 A}$.
Substituting $K = \lambda \sec(\pi y / 2d)$,we get $\frac{1}{C} = \frac{1}{\varepsilon_0 A \lambda} \int_0^d \cos(\pi y / 2d) dy$.
Evaluating the integral: $\int_0^d \cos(\pi y / 2d) dy = [\frac{2d}{\pi} \sin(\pi y / 2d)]_0^d = \frac{2d}{\pi} \sin(\pi/2) = \frac{2d}{\pi}$.
Thus,$\frac{1}{C} = \frac{1}{\varepsilon_0 A \lambda} \cdot \frac{2d}{\pi} = \frac{2d}{\pi \varepsilon_0 \lambda A}$.
Therefore,$C = \frac{\pi \varepsilon_0 \lambda A}{2d}$.

(B) The system consists of three plates $A, B,$ and $C$. Based on the circuit diagram, plates $A$ and $C$ are connected together to one terminal of the battery, and plate $B$ is connected to the other terminal. This configuration forms two capacitors in parallel, each with plates separated by distance $d = 0.885 \ mm = 0.885 \times 10^{-3} \ m$.
The capacitance of each capacitor is given by $C = \frac{\epsilon_0 A}{d}$.
Substituting the values: $C = \frac{8.854 \times 10^{-12} \times 0.1}{0.885 \times 10^{-3}} = 10^{-9} \ F = 1 \ nF$.
Since the two capacitors are in parallel, the equivalent capacitance is $C_{eq} = C + C = 2 \ nF = 2 \times 10^{-9} \ F$.
The energy stored in the system is $U = \frac{1}{2} C_{eq} V^2$.
Substituting $C_{eq} = 2 \times 10^{-9} \ F$ and $V = 10 \ V$:
$U = \frac{1}{2} \times (2 \times 10^{-9}) \times (10)^2 = 10^{-9} \times 100 = 10^{-7} \ J$.
Converting to microjoules $(\mu J)$:
$U = 10^{-7} \ J = 10^{-1} \times 10^{-6} \ J = 10^{-1} \ \mu J$.

(C) In the given arrangement, there are four plates. Let the plates be numbered $1$, $2$, $3$, and $4$ from top to bottom.
Plate $1$ and $4$ are connected to point $A$.
Plate $2$ and $3$ are connected to point $B$.
This creates three capacitors in parallel between points $A$ and $B$.
Specifically, the space between plate $1$ and $2$ forms a capacitor, the space between plate $2$ and $3$ forms a capacitor, and the space between plate $3$ and $4$ forms a capacitor.
However, looking at the connections: Plates $1$ and $4$ are connected to $A$, and plates $2$ and $3$ are connected to $B$. This forms three parallel capacitors, each with capacitance $C = \frac{\varepsilon_0 A}{d}$.
Thus, the total equivalent capacitance is $C_{eq} = C + C + C = 3C = \frac{3\varepsilon_0 A}{d}$.

(C) Let the initial capacitance be $C = \frac{\epsilon_0 A}{d}$,charge be $Q = CV$,and electric field be $E = \frac{V}{d}$.
When the separation $d$ is halved to $d' = \frac{d}{2}$,the new capacitance becomes $C' = \frac{\epsilon_0 A}{d'} = 2C$.
The new charge becomes $Q' = C'V = 2CV = 2Q$.
The new electric field becomes $E' = \frac{V}{d'} = \frac{2V}{d} = 2E$.
The force of attraction between the plates of a capacitor is given by $F = \frac{QE}{2}$.
Substituting the new values,the new force $F'$ is:
$F' = \frac{Q' E'}{2} = \frac{(2Q)(2E)}{2} = 4 \left( \frac{QE}{2} \right) = 4F$.
Therefore,the force of attraction becomes four times the initial force.

(D) The electric field produced by one plate at the location of the other plate is $E = \frac{\sigma}{2 \varepsilon_0}$,where $\sigma = \frac{Q}{A}$ is the surface charge density.
The force of attraction $F$ on a charge $Q$ due to this electric field is $F = Q E = Q \left( \frac{Q}{2 \varepsilon_0 A} \right) = \frac{Q^2}{2 \varepsilon_0 A}$.
Since the capacitance of a parallel plate capacitor is $C = \frac{\varepsilon_0 A}{d}$,we can write $\varepsilon_0 A = C d$.
Substituting this into the force expression: $F = \frac{Q^2}{2 C d}$.
Using the relation $Q = CV$,we get $F = \frac{(CV)^2}{2 C d} = \frac{C^2 V^2}{2 C d} = \frac{1}{2} \frac{C V^2}{d}$.
Thus,both expressions $\frac{Q^2}{2 \varepsilon_0 A}$ and $\frac{C V^2}{2 d}$ are correct. Therefore,the correct option is $D$.

(B) The electrostatic force of attraction between the plates of a parallel plate capacitor is given by $F = \frac{Q^2}{2A\epsilon_0}$.
Since $Q = CV$ and $C = \frac{\epsilon_0 A}{d}$,we have $Q = \frac{\epsilon_0 A V}{d}$.
Substituting $Q$ into the force equation: $F = \frac{(\epsilon_0 A V / d)^2}{2A\epsilon_0} = \frac{\epsilon_0 A V^2}{2d^2}$.
To maintain balance,this force must be balanced by the weight of the additional mass $m$,so $mg = F$.
$m = \frac{\epsilon_0 A V^2}{2d^2g}$.
Given: $\epsilon_0 = 8.85 \times 10^{-12} \, F/m$,$A = 100 \, cm^2 = 10^{-2} \, m^2$,$V = 5000 \, V$,$d = 5 \, mm = 5 \times 10^{-3} \, m$,$g = 9.8 \, m/s^2$.
$m = \frac{8.85 \times 10^{-12} \times 10^{-2} \times (5000)^2}{2 \times (5 \times 10^{-3})^2 \times 9.8}$.
$m = \frac{8.85 \times 10^{-14} \times 25 \times 10^6}{2 \times 25 \times 10^{-6} \times 9.8} = \frac{8.85 \times 10^{-8}}{19.6 \times 10^{-6}} \approx 0.004515 \, kg$.
$m \approx 4.515 \, g$. Rounding to the nearest option,the correct answer is $4.4 \, g$.

(C) Let the potential of the positive terminal be $V$ and the negative terminal be $0$.
Plates $1, 3, 5$ are connected to the positive terminal (potential $V$).
Plates $2, 4$ are connected to the negative terminal (potential $0$).
The potential difference across each adjacent pair of plates is $V$.
Charge on plate $1$: It has only one face (inner) facing plate $2$. The charge is $q_1 = C_{eff} V = (\epsilon_0 A/d) V$.
Charge on plate $4$: It has two faces (both sides) facing plates $3$ and $5$. Both faces contribute to the charge.
Charge on left face of plate $4$ (facing plate $3$) is $-(\epsilon_0 A/d) V$.
Charge on right face of plate $4$ (facing plate $5$) is $-(\epsilon_0 A/d) V$.
Total charge on plate $4$ is $q_4 = -(\epsilon_0 A/d) V - (\epsilon_0 A/d) V = -2\epsilon_0 AV/d$.

(A) The energy stored in a capacitor is given by $U = \frac{Q^2}{2C}$.
Since the plates are isolated,the charge $Q$ remains constant.
The force $F$ on the movable plate is given by the negative gradient of the potential energy with respect to the displacement $x$:
$F = -\frac{dU}{dx} = -\frac{d}{dx} \left( \frac{Q^2}{2C} \right) = \frac{Q^2}{2C^2} \frac{dC}{dx}$.
The capacitance of a parallel-plate capacitor with overlap length $x$ and width $b$ is $C = \frac{\epsilon_0 A}{d} = \frac{\epsilon_0 (bx)}{d}$,where $d$ is the fixed separation between the plates.
Thus,$\frac{dC}{dx} = \frac{\epsilon_0 b}{d}$.
Substituting this into the force equation:
$F = \frac{Q^2}{2 \left( \frac{\epsilon_0 bx}{d} \right)^2} \left( \frac{\epsilon_0 b}{d} \right) = \frac{Q^2 d}{2 \epsilon_0 b x^2}$.
Therefore,the force $F$ is proportional to $x^{-2}$.

(C) The dielectric strength of air is $E_{m} = 3 \times 10^{6} \, V/m$.
The maximum potential difference $(V_{m})$ that can be applied across the plates with separation $d = 2 \, mm = 2 \times 10^{-3} \, m$ is given by:
$V_{m} = E_{m} \cdot d = (3 \times 10^{6} \, V/m) \times (2 \times 10^{-3} \, m) = 6000 \, V$.
The capacitance $C$ of the parallel plate capacitor is given by:
$C = \frac{\varepsilon_{0} A}{d} = \frac{8.854 \times 10^{-12} \, F/m \times 20 \times 10^{-4} \, m^2}{2 \times 10^{-3} \, m} = 8.854 \times 10^{-12} \times 10^{-3} \, F = 8.854 \times 10^{-12} \, F$.
The maximum charge $q_{m}$ on the plates is:
$q_{m} = C \cdot V_{m} = (8.854 \times 10^{-12} \, F) \times (6000 \, V) = 53.124 \times 10^{-9} \, C \approx 53 \, nC$.
Thus,the maximum emf is $6000 \, V$ and the charge is $53 \, nC$.

(A) Let $F$ be the force applied to separate the plates of a parallel plate capacitor by a small distance $\Delta x$. The work done by the force is $W = F \Delta x$.
This work done is stored as an increase in the potential energy of the capacitor. The energy density $u$ between the plates is given by $u = \frac{1}{2} \epsilon_0 E^2$.
The total energy stored in the volume $V = A \Delta x$ is $\Delta U = u \times (A \Delta x) = (\frac{1}{2} \epsilon_0 E^2) A \Delta x$.
Equating work done to the change in potential energy: $F \Delta x = \frac{1}{2} \epsilon_0 E^2 A \Delta x$.
Thus,$F = \frac{1}{2} \epsilon_0 A E^2$.
Since the electric field between the plates is $E = \frac{\sigma}{\epsilon_0} = \frac{Q}{A \epsilon_0}$,we have $\epsilon_0 A = \frac{Q}{E}$.
Substituting this into the force equation: $F = \frac{1}{2} (\frac{Q}{E}) E^2 = \frac{1}{2} QE$.

(B) For a parallel plate capacitor with charges $Q_1$ and $Q_2$ on the plates,the charge on the inner surfaces is given by $q = \frac{Q_1 - Q_2}{2}$.
Here,$Q_1 = 2Q$ and $Q_2 = -Q$.
Therefore,the charge on the inner surface is $q = \frac{2Q - (-Q)}{2} = \frac{3Q}{2}$.
The potential difference $V$ between the plates is given by $V = \frac{q}{C}$.
Substituting the value of $q$,we get $V = \frac{3Q/2}{C} = \frac{3Q}{2C}$.

(D) The capacitance $C$ of a parallel plate capacitor is given by the formula $C = \frac{\epsilon_0 A}{d}$,where $\epsilon_0$ is the permittivity of free space,$A$ is the area of the plates,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 $(A)$ and the distance between them $(d)$.
Therefore,the capacity depends on the separation between the plates.

(C) Given:
Area of parallel plate capacitor,$A = 200\,cm^2 = 200 \times 10^{-4}\,m^2 = 2 \times 10^{-2}\,m^2$
Separation between the plates,$d = 1.5\,cm = 1.5 \times 10^{-2}\,m$
Force of attraction between the plates,$F = 25 \times 10^{-6}\,N$
Permittivity of free space,$\epsilon_0 = 8.85 \times 10^{-12}\,C^2/Nm^2$
The force of attraction between the plates of a capacitor is given by the formula:
$F = \frac{Q^2}{2A\epsilon_0}$
Since $Q = CV = \frac{\epsilon_0 A V}{d}$,we substitute $Q$ into the force equation:
$F = \frac{(\frac{\epsilon_0 A V}{d})^2}{2A\epsilon_0} = \frac{\epsilon_0^2 A^2 V^2}{d^2 \cdot 2A\epsilon_0} = \frac{\epsilon_0 A V^2}{2d^2}$
Rearranging for $V^2$:
$V^2 = \frac{2F d^2}{\epsilon_0 A}$
Substituting the values:
$V^2 = \frac{2 \times (25 \times 10^{-6}) \times (1.5 \times 10^{-2})^2}{(8.85 \times 10^{-12}) \times (200 \times 10^{-4})}$
$V^2 = \frac{50 \times 10^{-6} \times 2.25 \times 10^{-4}}{8.85 \times 10^{-12} \times 2 \times 10^{-2}}$
$V^2 = \frac{112.5 \times 10^{-10}}{17.7 \times 10^{-14}} \approx 6.356 \times 10^4 \approx 63560$
$V = \sqrt{63560} \approx 252.1\,V$
Thus,the value of $V$ is approximately $250\,V$.

(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 plate,$A$ is the area of the plate,and $\varepsilon_0$ is the permittivity of free space $(8.85 \times 10^{-12}\,F/m)$.
Rearranging the formula to solve for $q$:
$q = E \cdot A \cdot \varepsilon_0$
Substituting the given values ($E = 100\,N/C$,$A = 1\,m^2$,$\varepsilon_0 = 8.85 \times 10^{-12}\,F/m$):
$q = 100 \times 1 \times 8.85 \times 10^{-12}$
$q = 8.85 \times 10^{-10}\,C$
Therefore,the magnitude of the charge on each plate is $8.85 \times 10^{-10}\,C$.

(B) When charges $q_1$ and $q_2$ are given to the two plates of a parallel plate capacitor,the charge on the inner surfaces of the plates is given by $q_{inner} = \frac{q_1 - q_2}{2}$.
Here,$q_1 = +2\,\mu C$ and $q_2 = +4\,\mu C$.
Therefore,the charge on the inner surface is $q = \frac{2\,\mu C - 4\,\mu C}{2} = -1\,\mu C$ (in magnitude,$1\,\mu C$).
The potential difference $V$ across the capacitor is given by $V = \frac{q}{C}$.
Substituting the values,$V = \frac{1\,\mu C}{1\,\mu F} = 1\,V$.

(B) For a parallel plate capacitor with charges $Q_1$ and $Q_2$ on the two plates,the charge on the inner surfaces is given by $q = \frac{Q_1 - Q_2}{2}$.
Here,$Q_1 = 2Q$ and $Q_2 = -Q$.
Therefore,the charge on the inner surface of the first plate is $q = \frac{2Q - (-Q)}{2} = \frac{3Q}{2}$.
The charge on the inner surface of the second plate is $-q = -\frac{3Q}{2}$.
The potential difference $V$ between the plates is given by the relation $q = CV$,where $q$ is the magnitude of the charge on the inner surface.
Substituting the values,we get $\frac{3Q}{2} = CV$.
Thus,$V = \frac{3Q}{2C}$.

(A) The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$.
Given radius $r = 10\, cm = 0.1\, m$,so $A = \pi r^2 = \pi (0.1)^2 = 0.01\pi\, m^2$.
Initial distance $d_1 = 1\, mm = 10^{-3}\, m$. Initial capacitance $C_1 = \frac{\epsilon_0 A}{d_1}$.
Final distance $d_2 = 1\, cm = 10^{-2}\, m = 10 d_1$. Final capacitance $C_2 = \frac{\epsilon_0 A}{d_2} = \frac{C_1}{10}$.
Since the battery remains connected,the voltage $V = 100\, V$ remains constant.
Initial energy $U_1 = \frac{1}{2} C_1 V^2$.
Final energy $U_2 = \frac{1}{2} C_2 V^2 = \frac{1}{2} (\frac{C_1}{10}) V^2 = \frac{U_1}{10}$.
Change in energy $\Delta U = U_2 - U_1 = \frac{1}{2} V^2 (C_2 - C_1) = \frac{1}{2} (100)^2 (\frac{C_1}{10} - C_1) = -0.9 \times \frac{1}{2} C_1 (100)^2$.
Using $\epsilon_0 = 8.854 \times 10^{-12}\, F/m$,$C_1 = \frac{8.854 \times 10^{-12} \times 0.01\pi}{10^{-3}} \approx 2.78 \times 10^{-10}\, F$.
$\Delta U = -0.45 \times 2.78 \times 10^{-10} \times 10000 \approx -1.25 \times 10^{-6}\, J$.
Since $1\, J = 10^7\, ergs$,$\Delta U = -1.25 \times 10^{-6} \times 10^7 = -12.5\, ergs$.
The negative sign indicates a loss of energy.

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

(D) When capacitors are connected in series, the charge $Q$ on each capacitor is the same.
Since both capacitors have the same capacitance $C$, the potential difference across each capacitor, given by $V = Q/C$, is also the same.
The energy stored in a capacitor is given by $U = Q^2 / (2C)$. Since $Q$ and $C$ are the same for both, the stored energy is also the same.
The electric field between the plates of a parallel plate capacitor is given by $E = V/d$. Since the potential difference $V$ is the same but the plate separations $d$ are different, the electric field $E$ must be different for the two capacitors.

(A) The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_{0} A}{d}$,where $A$ is the area of the plates and $d$ is the distance between them.
When the distance $d$ is increased,the capacitance $C$ decreases.
Since the battery is disconnected,the charge $Q$ on the plates remains constant.
The relationship between charge,capacitance,and potential difference is $V = \frac{Q}{C}$.
Since $Q$ is constant and $C$ decreases,the potential difference $V$ must increase.

(B) In the given arrangement,there are $6$ metallic plates.
Plates $1, 3, 5$ are connected to point $P$,and plates $2, 4, 6$ are connected to point $Q$.
This configuration creates $5$ capacitors connected in parallel,where each capacitor is formed between two adjacent plates.
The capacitance of each individual capacitor is $C = \frac{\varepsilon_0 A}{d}$.
Since there are $5$ such capacitors in parallel,the equivalent capacitance $C_{eq}$ is given by:
$C_{eq} = C + C + C + C + C = 5C = \frac{5\varepsilon_0 A}{d}$.

(B) The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$.
When the plates are pulled apart,the separation distance $d$ increases.
Since $C \propto \frac{1}{d}$,the capacitance $C$ decreases.
Because the battery is removed,the charge $Q$ on the plates remains constant.
Using the relation $V = \frac{Q}{C}$,since $Q$ is constant and $C$ decreases,the potential difference $V$ must increase.

(C) The charge stored on the two plates of a capacitor are $+Q$ and $-Q$. The total charge is $Q + (-Q) = 0$. Thus,the Assertion is correct.
The electric field outside a parallel plate capacitor is zero because the fields produced by the two plates (each of magnitude $\frac{\sigma}{2\varepsilon_0}$) are equal and opposite. The field between the plates is $\frac{\sigma}{\varepsilon_0}$.
As shown in the figure,by drawing a Gaussian surface $ABCD$,we can apply Gauss's Law: $\oint \vec{E} \cdot d\vec{s} = \frac{q_{enclosed}}{\varepsilon_0}$. Since the net charge enclosed by the surface is $Q - Q = 0$,the net flux is zero,which implies the electric field outside the plates is zero. Therefore,the Reason is incorrect.

(D) Consider a small strip of width $dx$ at a distance $x$ from the left end. The separation between the plates at this distance $x$ is $d' = d + x\alpha$.
The capacitance of this small strip is $dC = \frac{\varepsilon_0 a dx}{d + x\alpha}$.
To find the total capacitance $C$,we integrate from $x = 0$ to $x = a$:
$C = \int_{0}^{a} \frac{\varepsilon_0 a dx}{d + x\alpha} = \frac{\varepsilon_0 a}{\alpha} [\ln(d + x\alpha)]_{0}^{a} = \frac{\varepsilon_0 a}{\alpha} \ln\left(\frac{d + a\alpha}{d}\right) = \frac{\varepsilon_0 a}{\alpha} \ln\left(1 + \frac{a\alpha}{d}\right)$.
Using the Taylor series expansion $\ln(1 + y) \approx y - \frac{y^2}{2}$ for small $y = \frac{a\alpha}{d}$:
$C \approx \frac{\varepsilon_0 a}{\alpha} \left(\frac{a\alpha}{d} - \frac{1}{2} \left(\frac{a\alpha}{d}\right)^2\right) = \frac{\varepsilon_0 a^2}{d} \left(1 - \frac{a\alpha}{2d}\right)$.

(N/A) Given:
Area of each plate,$A = 6 \times 10^{-3} \, m^{2}$
Distance between the plates,$d = 3 \, mm = 3 \times 10^{-3} \, m$
Supply voltage,$V = 100 \, V$
Permittivity of free space,$\varepsilon_{0} = 8.854 \times 10^{-12} \, F/m$
$1$. Calculation of Capacitance $(C)$:
The formula for the capacitance of a parallel plate capacitor is $C = \frac{\varepsilon_{0} A}{d}$.
Substituting the values:
$C = \frac{8.854 \times 10^{-12} \times 6 \times 10^{-3}}{3 \times 10^{-3}}$
$C = 8.854 \times 10^{-12} \times 2 = 17.708 \times 10^{-12} \, F \approx 17.71 \, pF$.
$2$. Calculation of Charge $(q)$:
The charge on each plate is given by $q = CV$.
$q = 17.71 \times 10^{-12} \, F \times 100 \, V$
$q = 1.771 \times 10^{-9} \, C$.
Thus,the capacitance is $17.71 \, pF$ and the charge on each plate is $1.771 \times 10^{-9} \, C$.

(N/A) The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$.
Given values are $C = 2\,\mu F = 2 \times 10^{-6}\, F$ and $d = 0.5\, cm = 0.5 \times 10^{-2}\, m$.
The permittivity of free space is $\epsilon_0 = 8.854 \times 10^{-12}\, F/m$.
Rearranging the formula for area $A$,we get $A = \frac{C d}{\epsilon_0}$.
Substituting the values: $A = \frac{(2 \times 10^{-6}) \times (0.5 \times 10^{-2})}{8.854 \times 10^{-12}}$.
$A = \frac{1 \times 10^{-8}}{8.854 \times 10^{-12}} \approx 1129.43\, m^2$.
Thus,the area of the plates is approximately $1130\, m^2$.

(N/A) Let $F$ be the force applied to separate the plates of a parallel plate capacitor by a distance of $\Delta x$.
The work done by the external force is $W = F \Delta x$.
This work increases the potential energy of the capacitor by an amount equal to the change in energy density multiplied by the change in volume: $\Delta U = u A \Delta x$,where $u = \frac{1}{2} \varepsilon_0 E^2$ is the energy density,$A$ is the area of the plates,and $\Delta x$ is the change in separation.
Equating work done to the change in potential energy:
$F \Delta x = (\frac{1}{2} \varepsilon_0 E^2) A \Delta x$
$F = \frac{1}{2} \varepsilon_0 E^2 A$
Since the electric field $E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{A \varepsilon_0}$,we can write $\varepsilon_0 A = \frac{Q}{E}$.
Substituting this into the force equation:
$F = \frac{1}{2} (\frac{Q}{E}) E^2 = \frac{1}{2} Q E$.
The factor $\frac{1}{2}$ arises because the electric field $E$ between the plates is the sum of the fields produced by both plates. Each plate experiences a force only due to the field produced by the other plate,which is $\frac{E}{2}$. Thus,the force on a plate with charge $Q$ is $F = Q \times (\frac{E}{2}) = \frac{1}{2} Q E$.

(C) Given:
Voltage rating,$V = 1\; kV = 1000\; V$
Dielectric constant,$\varepsilon_{r} = 3$
Dielectric strength,$E_{max} = 10^{7}\; V/m$
Safety limit for electric field,$E = 10\% \text{ of } 10^{7} = 10^{6}\; V/m$
Capacitance,$C = 50\; pF = 50 \times 10^{-12}\; F$
The distance $d$ between the plates is given by $d = V/E = 1000 / 10^{6} = 10^{-3}\; m$.
Using the formula for capacitance $C = \frac{\varepsilon_{0} \varepsilon_{r} A}{d}$,we can find the area $A$:
$A = \frac{C \cdot d}{\varepsilon_{0} \cdot \varepsilon_{r}}$
Substituting the values:
$A = \frac{50 \times 10^{-12} \times 10^{-3}}{8.854 \times 10^{-12} \times 3}$
$A = \frac{50 \times 10^{-15}}{26.562 \times 10^{-12}} \approx 1.882 \times 10^{-3}\; m^2$
Converting to $cm^2$ $(1\; m^2 = 10^{4}\; cm^2)$:
$A \approx 1.882 \times 10^{-3} \times 10^{4} = 18.82\; cm^2 \approx 19\; cm^2$.

(N/A) parallel plate capacitor consists of two large plane parallel conducting plates separated by a small distance $d$.
$A$ non-conducting medium (dielectric) is typically kept between the two plates.
Let the charges on plate $1$ and plate $2$ be $+Q$ and $-Q$ respectively,$A$ be the area of each plate,and $d$ be the separation between them.
Since $d$ is much smaller than the linear dimensions of the plates $(d^2 << A)$,we can use the result for the electric field $E = \frac{\sigma}{2\epsilon_0}$ produced by an infinite plane sheet of uniform surface charge density $\sigma = \frac{Q}{A}$.
In the region between the plates,the electric fields due to both plates point in the same direction (from positive to negative plate):
$E = E_1 + E_2 = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0} = \frac{Q}{A\epsilon_0}$.
The potential difference $V$ between the plates is given by $V = E \cdot d = \frac{Qd}{A\epsilon_0}$.
The capacitance $C$ is defined as $C = \frac{Q}{V} = \frac{Q}{(Qd / A\epsilon_0)} = \frac{\epsilon_0 A}{d}$.
The capacitance depends on:
$1$. The area of the plates $(A)$.
$2$. The distance between the plates $(d)$.
$3$. The permittivity of the medium between the plates ($\epsilon_0$ or $\epsilon = k\epsilon_0$).

(N/A) The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$.
For a capacitance of $C = 1$ $F$,with a plate separation of $d = 1$ $cm = 10^{-2}$ $m$,the required area $A$ is:
$A = \frac{C d}{\epsilon_0} = \frac{1 \times 10^{-2}}{8.854 \times 10^{-12}} \approx 1.13 \times 10^9$ $m^2$.
If the plates are square,the side length $L = \sqrt{A} \approx \sqrt{1.13 \times 10^9} \approx 33.6$ $km$.
Since it is physically impossible to construct a capacitor with such massive plates,$1$ $F$ is considered an extremely large unit for practical purposes.

(A) The capacitance $C$ of a parallel plate capacitor is given by the formula:
$C = \frac{\epsilon_0 A}{d}$
Given:
Area $A = 1 \ m^2$
Distance $d = 1 \ mm = 1 \times 10^{-3} \ m$
Permittivity of free space $\epsilon_0 = 8.85 \times 10^{-12} \ F/m$
Substituting the values:
$C = \frac{8.85 \times 10^{-12} \times 1}{1 \times 10^{-3}}$
$C = 8.85 \times 10^{-12} \times 10^3$
$C = 8.85 \times 10^{-9} \ F$

(A) parallel plate capacitor is a fundamental electrical component used to store electrical energy in an electric field.
It consists of two large plane parallel conducting plates,each of area $A$,separated by a small distance $d$.
The space between the plates is filled with an insulating material called a dielectric (or vacuum).
When a potential difference $V$ is applied across the plates,they acquire equal and opposite charges $+Q$ and $-Q$,creating a uniform electric field between them.

(A) For a parallel plate capacitor,the electric field due to one plate is $E_1 = \frac{\sigma}{2\epsilon_0}$,where $\sigma = \frac{Q}{A}$ is the surface charge density.
Since both plates produce electric fields in the same direction between them,the total electric field $E$ is the sum of the fields from both plates.
$E = E_1 + E_2 = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}$.
Substituting $\sigma = \frac{Q}{A}$,we get $E = \frac{Q}{\epsilon_0 A}$.

(C) The capacitance $(C)$ of a parallel plate capacitor is given by the formula $C = \frac{\epsilon_0 A}{d}$,where $\epsilon_0$ is the permittivity of free space,$A$ is the area of the plates,and $d$ is the distance between them.
From this formula,it is clear that the capacitance depends only on the physical dimensions of the capacitor (area and separation distance) and the properties of the dielectric material between the plates.
It does not depend on the charge $(Q)$ stored on the plates or the potential difference $(V)$ applied across them.
Therefore,the correct answer is that it does not depend on the potential difference.

(N/A) The capacitance of a parallel plate capacitor with vacuum or air between its plates is given by $C_0 = \frac{\epsilon_0 A}{d}$,where $\epsilon_0$ is the permittivity of free space,$A$ is the area of the plates,and $d$ is the distance between them.
When a dielectric medium of dielectric constant $K$ is introduced between the plates,the permittivity of the medium becomes $\epsilon = K \epsilon_0$.
Therefore,the new capacitance $C$ is given by $C = \frac{\epsilon A}{d} = \frac{K \epsilon_0 A}{d}$.
Substituting the expression for $C_0$,we get $C = K C_0$.

(C) The arrangement consists of three parallel plate capacitors connected in parallel.
The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$.
From the figure,the separation distances between the plates are $b$,$2b$,and $3b$.
Therefore,the individual capacitances are:
$C_1 = \frac{\varepsilon_0 A}{b}$
$C_2 = \frac{\varepsilon_0 A}{2b}$
$C_3 = \frac{\varepsilon_0 A}{3b}$
Since they are in parallel,the equivalent capacitance is:
$C_{eq} = C_1 + C_2 + C_3$
$C_{eq} = \frac{\varepsilon_0 A}{b} + \frac{\varepsilon_0 A}{2b} + \frac{\varepsilon_0 A}{3b}$
$C_{eq} = \frac{\varepsilon_0 A}{b} \left( 1 + \frac{1}{2} + \frac{1}{3} \right)$
$C_{eq} = \frac{\varepsilon_0 A}{b} \left( \frac{6 + 3 + 2}{6} \right) = \frac{11}{6} \frac{\varepsilon_0 A}{b}$
Wait,re-evaluating the figure: The gaps are $b$,$2b$,and $3b$ based on the visual structure. If the gaps are $b$,$3b$,and $5b$ (as implied by the provided solution's denominator $15b$):
$C_{eq} = \frac{\varepsilon_0 A}{b} + \frac{\varepsilon_0 A}{3b} + \frac{\varepsilon_0 A}{5b} = \frac{\varepsilon_0 A}{b} \left( 1 + \frac{1}{3} + \frac{1}{5} \right) = \frac{\varepsilon_0 A}{b} \left( \frac{15 + 5 + 3}{15} \right) = \frac{23}{15} \frac{\varepsilon_0 A}{b}$.
Thus,$x = 23$.

(C) When two large conducting plates with charges $q_{1}$ and $q_{2}$ are placed parallel to each other,the charge on the inner surfaces is given by $q_{inner} = \frac{q_{1}-q_{2}}{2}$.
The electric field $E$ between the plates is due to these inner charges: $E = \frac{\sigma}{\varepsilon_{0}} = \frac{q_{inner}}{A\varepsilon_{0}} = \frac{q_{1}-q_{2}}{2A\varepsilon_{0}}$.
The potential difference $V$ between the plates is given by $V = E \cdot d$,where $d$ is the separation between the plates.
Since the capacitance of a parallel plate capacitor is $C = \frac{A\varepsilon_{0}}{d}$,we can write $V = \frac{q_{1}-q_{2}}{2A\varepsilon_{0}} \cdot d = \frac{q_{1}-q_{2}}{2(A\varepsilon_{0}/d)}$.
Substituting $C$,we get $V = \frac{q_{1}-q_{2}}{2C}$.

(C) For a parallel plate capacitor,the capacitance is given by $C = \frac{\varepsilon_0 A}{d}$.
Given that the charge $q$ on the capacitor remains constant,the potential difference $V$ between the plates is given by $V = \frac{q}{C}$.
Substituting the expression for $C$,we get $V = \frac{q d}{\varepsilon_0 A}$.
Since $q$,$\varepsilon_0$,and $A$ are constants,we have $V \propto d$.
This implies that the graph of $V$ versus $d$ is a straight line passing through the origin. However,in a real physical capacitor,when the plates are in contact $(d=0)$,the capacitance is not infinite,and the potential difference is not zero due to the finite size of the plates and edge effects. Therefore,the graph starts from a small positive value of $d$ and is a straight line,which does not pass through the origin.

(D) Given that both capacitors have the same charge $Q$ and potential $V$,their capacitances $C = Q/V$ must be equal.
For a parallel plate capacitor,$C = \frac{\varepsilon_0 A}{d}$.
Since $C_1 = C_2$,we have $\frac{\varepsilon_0 A_1}{d_1} = \frac{\varepsilon_0 A_2}{d_2}$.
Given $A_1 = 100 \,cm^2$,$A_2 = 500 \,cm^2$,and $d_1 = 0.5 \,mm = 0.05 \,cm$.
Substituting the values: $\frac{100}{0.05} = \frac{500}{d_2}$.
$d_2 = \frac{500 \times 0.05}{100} = 5 \times 0.05 = 0.25 \,cm$.

(A) The force of attraction between the plates of a parallel plate capacitor is given by $F = \frac{q^2}{2 A \varepsilon_0}$.
In this system,the inner plates are attracted towards their respective fixed outer plates due to the electrostatic force of attraction.
At equilibrium,the spring force $F_s = kx$ must balance the electrostatic force of attraction $F$ acting on the inner plate.
Therefore,$kx = \frac{q^2}{2 A \varepsilon_0}$.
Solving for the extension $x$,we get $x = \frac{q^2}{2 A \varepsilon_0 k}$.

(D) Let the plates be numbered $1, 2, 3, 4, 5$ from top to bottom.
Plate $1$ and $4$ are connected together. Plate $3$ and $5$ are connected to terminal $Q$. Plate $2$ is connected to terminal $P$.
The capacitance of each pair of adjacent plates is $C = \frac{\varepsilon_0 A}{d}$.
There are $4$ capacitors formed between the $5$ plates: $C_{12}, C_{23}, C_{34}, C_{45}$.
- $C_{12}$ is between plate $1$ (connected to $4$) and plate $2$ $(P)$.
- $C_{23}$ is between plate $2$ $(P)$ and plate $3$ $(Q)$.
- $C_{34}$ is between plate $3$ $(Q)$ and plate $4$ (connected to $1$).
- $C_{45}$ is between plate $4$ (connected to $1$) and plate $5$ $(Q)$.
By analyzing the connections,we find that the equivalent circuit consists of capacitors in a combination of series and parallel. The net capacitance is calculated as $C_{\text{net}} = \frac{5}{3} C = \frac{5 \varepsilon_0 A}{3d}$.
Thus,the correct option is $(d)$.

(B) In the given arrangement,the plates are connected such that they form capacitors in parallel.
There are $n = 7$ plates,which form $n - 1 = 6$ capacitors.
Each capacitor has an area $A$ and separation $d$,so the capacitance of each is $C = \frac{\varepsilon_0 A}{d}$.
Since all $6$ capacitors are connected in parallel between points $P$ and $Q$,the equivalent capacitance is:
$C_{\text{net}} = C_1 + C_2 + C_3 + C_4 + C_5 + C_6 = 6C$
$C_{\text{net}} = \frac{6 \varepsilon_0 A}{d}$

(D) The initial capacitance of a parallel plate capacitor is given by the formula $C = \frac{\varepsilon_0 A}{d}$,where $A$ is the area of the plates and $d$ is the distance between them.
According to the problem,the new distance $d' = 2d$ and the new area $A' = A / 2$.
The final capacitance $C'$ is given by $C' = \frac{\varepsilon_0 A'}{d'}$.
Substituting the new values,we get $C' = \frac{\varepsilon_0 (A / 2)}{2d} = \frac{\varepsilon_0 A}{4d}$.
Since $C = \frac{\varepsilon_0 A}{d}$,we can write $C' = \frac{C}{4}$.

(C) From the figure,the total distance $d = 5 \, mm$ is composed of $a + c + b$,where $c = 1 \, mm$ is the thickness of the conducting slab between the two capacitors.
Thus,$a + b = d - c = 5 \, mm - 1 \, mm = 4 \, mm = 4 \times 10^{-3} \, m$.
The two capacitors are in series,with effective distance $d_{eff} = a + b = 4 \times 10^{-3} \, m$.
The equivalent capacitance $C_{eq}$ is given by $C_{eq} = \frac{\varepsilon_0 A}{d_{eff}}$.
Given $A = 200 \, cm^2 = 200 \times 10^{-4} \, m^2$.
$C_{eq} = \frac{\varepsilon_0 \times 200 \times 10^{-4}}{4 \times 10^{-3}} = \frac{200 \times 10^{-1}}{4} \varepsilon_0 = 5 \varepsilon_0 \, F$.
Comparing this with $x \varepsilon_0 \, F$,we get $x = 5$.