Parallel Plate Capacitor Questions in English

(A) The arrangement can be viewed as three capacitors connected in parallel,each having an area of $\frac{A}{3}$ but different plate separations.
For the first capacitor,the separation is $d$. Its capacitance is $C_1 = \frac{\varepsilon_0 (A/3)}{d} = \frac{\varepsilon_0 A}{3d}$.
For the second capacitor,the separation is $2d$. Its capacitance is $C_2 = \frac{\varepsilon_0 (A/3)}{2d} = \frac{\varepsilon_0 A}{6d}$.
For the third capacitor,the separation is $3d$. Its capacitance is $C_3 = \frac{\varepsilon_0 (A/3)}{3d} = \frac{\varepsilon_0 A}{9d}$.
Since they are in parallel,the equivalent capacitance is $C_{eq} = C_1 + C_2 + C_3$.
$C_{eq} = \frac{\varepsilon_0 A}{3d} + \frac{\varepsilon_0 A}{6d} + \frac{\varepsilon_0 A}{9d} = \frac{\varepsilon_0 A}{d} \left( \frac{1}{3} + \frac{1}{6} + \frac{1}{9} \right)$.
Taking the least common multiple of $3, 6, 9$,which is $18$:
$C_{eq} = \frac{\varepsilon_0 A}{d} \left( \frac{6 + 3 + 2}{18} \right) = \frac{11 \varepsilon_0 A}{18 d}$.

(A) Given that the capacitor is connected to a battery,the potential difference $V$ remains constant.
$(i)$ Capacitance $C = \frac{\varepsilon_0 A}{d}$. As the plates are moved closer,the distance $d$ decreases,so the capacitance $C$ increases. Thus,statement $C$ is correct.
$(ii)$ Charge $Q = CV$. Since $C$ increases and $V$ is constant,the charge $Q$ increases. Thus,statement $A$ is correct.
$(iii)$ Energy stored $U = \frac{1}{2}CV^2$. Since $C$ increases and $V$ is constant,the energy $U$ increases. Thus,statement $B$ is incorrect.
$(iv)$ The ratio of charge to potential is $\frac{Q}{V} = C$. Since $C$ increases,the ratio $\frac{Q}{V}$ changes. Thus,statement $D$ is incorrect.
$(v)$ The product of charge and voltage is $QV = CV^2$. Since $C$ increases and $V$ is constant,the product $QV$ increases. Thus,statement $E$ is correct.
Therefore,statements $A, C,$ and $E$ are correct.

(A) The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$,where $A = l \times b$ is the area of the plates and $d$ is the distance between them.
Initial capacitance $C_i = \frac{\epsilon_0 (3 \ cm \times 1 \ cm)}{3 \ \mu m} = \epsilon_0 \times 10^4 \ cm^2/m$.
We want the new capacitance $C_f = 10 C_i$.
For option $A$: $C_A = \frac{\epsilon_0 (30 \times 1)}{1} = 30 C_i$ (Incorrect).
For option $B$: $C_B = \frac{\epsilon_0 (3 \times 1)}{30} = 0.1 C_i$ (Incorrect).
For option $C$: $C_C = \frac{\epsilon_0 (6 \times 5)}{3} = 10 C_i$ (Correct).
For option $D$: $C_D = \frac{\epsilon_0 (1 \times 1)}{10} = 0.033 C_i$ (Incorrect).
For option $E$: $C_E = \frac{\epsilon_0 (5 \times 2)}{1} = 10 C_i$ (Correct).
Thus,options $C$ and $E$ increase the capacitance by a factor of $10$.

(B) The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$,where $A = \pi r^2$.
Given $V = \frac{Q}{C}$,we have $V = \frac{Q d}{\epsilon_0 A}$.
Differentiating with respect to time $t$,we get $\frac{dV}{dt} = \frac{d}{dt} \left( \frac{Q d}{\epsilon_0 A} \right) = \frac{d}{\epsilon_0 A} \frac{dQ}{dt}$.
Since $\frac{dQ}{dt} = I$,the rate of change of potential difference is $\frac{dV}{dt} = \frac{I d}{\epsilon_0 A}$.
Rearranging for $d$,we get $d = \frac{\epsilon_0 A (dV/dt)}{I} = \frac{\epsilon_0 (\pi r^2) (dV/dt)}{I}$.
Substituting the given values: $r = 0.1 \ m$,$I = 0.15 \ A$,$\frac{dV}{dt} = 7 \times 10^8 \ V/s$,$\epsilon_0 = 9 \times 10^{-12} \ F/m$,and $\pi = 22/7$.
$d = \frac{(9 \times 10^{-12}) \times (22/7) \times (0.1)^2 \times (7 \times 10^8)}{0.15} \ m$.
$d = \frac{9 \times 10^{-12} \times 22 \times 0.01 \times 10^8}{0.15} \ m = \frac{9 \times 22 \times 10^{-6}}{0.15} \ m = \frac{198 \times 10^{-6}}{0.15} \ m = 1320 \times 10^{-6} \ m$.
Since $1 \ \mu m = 10^{-6} \ m$,the distance $d = 1320 \ \mu m$.

(C) The force of attraction between the plates of a parallel plate capacitor is given by the formula $F = \frac{Q^2}{2 \varepsilon_0 A}$.
Since the charge $Q = CV$ and the capacitance $C = \frac{\varepsilon_0 A}{d}$,we can write $\varepsilon_0 A = Cd$.
Substituting these values into the force equation:
$F = \frac{(CV)^2}{2(Cd)} = \frac{C^2 V^2}{2 Cd} = \frac{CV^2}{2 d}$.

(B) The relationship between charge $(Q)$,voltage $(V)$,and capacitance $(C)$ is given by $Q = CV$,which can be rearranged as $V = Q/C$.
In the given graph,voltage $(V)$ is plotted on the $y$-axis and charge $(Q)$ is plotted on the $x$-axis.
The slope of this graph is $V/Q = 1/C$.
From the graph,for a fixed charge $Q$,the voltage $V_B > V_A$.
Since the slope is inversely proportional to capacitance $(C)$,a higher slope corresponds to a lower capacitance.
Therefore,the slope of line $B$ is greater than the slope of line $A$,which implies that the capacitance of $A$ is greater than the capacitance of $B$ $(C_A > C_B)$.

(C) When two charged plates with charges $q_1$ and $q_2$ are brought close to form a parallel plate capacitor,the charge on the inner faces of the plates becomes $q_{inner} = \frac{q_1 - q_2}{2}$.
This is because the total charge on the inner surfaces must be equal and opposite to create a uniform electric field between the plates.
The potential difference $V$ across a capacitor is given by the formula $V = \frac{Q}{C}$,where $Q$ is the magnitude of the charge on the inner face of the capacitor plates.
Substituting $Q = \frac{q_1 - q_2}{2}$,we get $V = \frac{q_1 - q_2}{2C}$.
Therefore,the correct option is $C$.

(C) For the $1^{\text{st}}$ capacitor,the capacitance is given by $C_1 = \frac{\varepsilon_0 A_1}{d} = \frac{\varepsilon_0 \pi r^2}{d}$.
For the $2^{\text{nd}}$ capacitor,the new radius is $r' = \sqrt{2}r$ and the new distance is $d' = \frac{d}{2}$.
The new area is $A_2 = \pi (r')^2 = \pi (\sqrt{2}r)^2 = 2\pi r^2$.
The new capacitance is $C_2 = \frac{\varepsilon_0 A_2}{d'} = \frac{\varepsilon_0 (2\pi r^2)}{d/2} = \frac{4\varepsilon_0 \pi r^2}{d} = 4C_1$.
Therefore,the ratio $\frac{C_1}{C_2} = \frac{C_1}{4C_1} = \frac{1}{4}$.

(C) Given that the two parallel plate capacitors have the same charge $q$ and the same potential $V$, their capacitances must be equal because $C = q/V$.
Therefore, $C_1 = C_2$.
The formula for the capacitance of a parallel plate capacitor is $C = \frac{\varepsilon_0 A}{d}$.
Equating the two capacitances: $\frac{\varepsilon_0 A_1}{d_1} = \frac{\varepsilon_0 A_2}{d_2}$.
This simplifies to $d_2 = \frac{A_2}{A_1} d_1$.
Given $A_1 = 100 \,cm^2$, $A_2 = 500 \,cm^2$, and $d_1 = 0.5 \,mm = 0.05 \,cm$.
Substituting these values: $d_2 = \frac{500}{100} \times 0.05 \,cm = 5 \times 0.05 \,cm = 0.25 \,cm$.

(C) The capacitance of a parallel plate capacitor is given by the formula $C = \frac{k A \varepsilon_0}{d}$, where $k$ is the dielectric constant, $A$ is the area of the plates, $\varepsilon_0$ is the permittivity of free space, and $d$ is the separation between the plates.
Initially, the capacitance is $C_1 = \frac{k A \varepsilon_0}{d}$.
When the plates are moved closer such that the new separation $d' = \frac{d}{2}$, the new capacitance $C_2$ becomes:
$C_2 = \frac{k A \varepsilon_0}{d'} = \frac{k A \varepsilon_0}{d/2} = 2 \left( \frac{k A \varepsilon_0}{d} \right)$.
Substituting $C_1$ into the equation, we get $C_2 = 2C_1$.

(A) The capacitance $C$ 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.
Given the initial capacity $C = 10 \mu F$.
If the distance is doubled,the new distance $d' = 2d$.
The new capacitance $C'$ is:
$C' = \frac{\varepsilon_{0} A}{d'} = \frac{\varepsilon_{0} A}{2d}$
Substituting the initial capacitance expression:
$C' = \frac{C}{2}$
$C' = \frac{10 \mu F}{2} = 5 \mu F$
Therefore,the new capacity is $5 \mu F$.

(C) 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,$\varepsilon_{0}$ is the permittivity of free space,$A$ is the area of the plates,and $d$ is the distance between the plates.
From the formula,we can see that $C \propto A$.
Therefore,the capacity of the capacitor increases if the area of the plate is increased.

(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 distance between the plates.
From this formula,it is clear that the capacitance $C$ is inversely proportional to the distance $d$ between the plates $(C \propto \frac{1}{d})$.
Therefore,by decreasing the distance $d$ between the plates,the capacitance $C$ of the capacitor increases.

(B) When the capacitor is charged and then disconnected from the battery,the charge $Q$ on the plates remains constant because there is no path for the charge to flow.
The capacitance $C$ of a parallel plate capacitor is given by $C = \frac{A \varepsilon_0}{d}$.
When the plates are moved further apart,the distance $d$ increases,which causes the capacitance $C$ to decrease.
Since the charge $Q$ is constant and $Q = CV$,the voltage $V = \frac{Q}{C}$ must increase as $C$ decreases.
Therefore,the voltage across the capacitor increases.

(B) The electric field intensity $E$ between the plates of a parallel plate capacitor is given by the formula:
$E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{A \varepsilon_0}$ --- $(i)$
The capacitance $C$ of a parallel plate capacitor with plate area $A$ and separation distance $t$ is given by:
$C = \frac{A \varepsilon_0}{t}$
From this,we can write $A \varepsilon_0 = Ct$ --- $(ii)$
Substituting the value of $A \varepsilon_0$ from equation $(ii)$ into equation $(i)$:
$E = \frac{Q}{Ct}$
Therefore,the electric field intensity is $\frac{Q}{Ct}$.

(A) The total electric field $E$ between the plates of a parallel plate capacitor is given by $E = \frac{V}{d}$.
Since the electric field due to one plate is half of the total field,the field due to one plate is $E_1 = \frac{E}{2} = \frac{V}{2 d}$.
The charge on the plates is $Q = C V$.
The force $F$ exerted by one plate on the other is given by $F = Q \times E_1$.
Substituting the values,we get $F = (C V) \times \left( \frac{V}{2 d} \right) = \frac{C V^2}{2 d}$.

(D) The total electric field '$E$' between the plates of a parallel plate capacitor is the sum of the fields produced by each plate individually.
Since the plates are identical,each plate produces an electric field of magnitude $\frac{E}{2}$.
The force '$F$' on one plate is due to the electric field produced by the other plate.
Therefore,$F = Q \times (\text{Field produced by the other plate}) = Q \times \frac{E}{2} = \frac{QE}{2}$.

(D) The electric field $E$ between two parallel plates separated by a distance $d$ with a potential difference $V$ is given by $E = \frac{V}{d}$.
The force $F$ acting on a particle of charge $q$ in this electric field is $F = qE = \frac{qV}{d}$.
According to Newton's second law of motion,the acceleration $a$ is given by $a = \frac{F}{m}$.
Substituting the value of $F$,we get $a = \frac{qV}{md}$.

(B) In an ideal parallel plate capacitor,we assume the plates are infinite in extent,resulting in a uniform electric field between them.
However,for real capacitors with finite plate dimensions,the electric field lines do not remain perfectly parallel near the edges.
Instead,they bulge or bend outward at the edges of the plates.
This phenomenon is known as the fringing of the field.

(D) Given that the parallel plate capacitor is charged and then isolated. Therefore,the charge $Q$ remains constant.
Capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_{0} A}{d}$,which implies $C \propto \frac{1}{d}$.
If the separation $d$ between the plates is increased,the capacitance $C$ decreases.
From the relation $Q = CV$,we have $V = \frac{Q}{C}$. Since $Q$ is constant and $C$ decreases,the potential $V$ must increase.
Thus,on increasing the plate separation,the charge remains constant,the potential increases,and the capacitance decreases.

(C) The system consists of two capacitors in parallel connected to plate '$Q$'.
One capacitor is formed by plates '$P$' and '$Q$' with separation $d$,and the other by plates '$Q$' and '$R$' with separation $2d$.
The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$.
Let $C_1$ be the capacitance between '$P$' and '$Q$': $C_1 = \frac{\epsilon_0 A}{d}$.
Let $C_2$ be the capacitance between '$Q$' and '$R$': $C_2 = \frac{\epsilon_0 A}{2d} = \frac{C_1}{2}$.
Since both plates '$P$' and '$R$' are grounded,the potential of both is $0 \,V$. Thus,the capacitors are in parallel with respect to plate '$Q$'.
The equivalent capacitance is $C_{eq} = C_1 + C_2 = C_1 + \frac{C_1}{2} = \frac{3}{2} C_1$.
Substituting the values: $C_1 = \frac{8.85 \times 10^{-12} \times 2}{2 \times 10^{-3}} = 8.85 \times 10^{-9} \,F$.
$C_{eq} = \frac{3}{2} \times 8.85 \times 10^{-9} = 1.3275 \times 10^{-8} \,F$.
The potential of plate '$Q$' is $V = \frac{q}{C_{eq}} = \frac{8.85 \times 10^{-8}}{1.3275 \times 10^{-8}} = \frac{8.85}{1.3275} = \frac{8.85}{1.5 \times 8.85} \times 10 = \frac{20}{3} \,V$.

(D) In equilibrium,the attractive force between the plates of the capacitor is balanced by the spring force.
The force between the plates of a parallel plate capacitor is given by $F = \frac{q^2}{2 \varepsilon_0 A} = \frac{C^2 V^2}{2 \varepsilon_0 A}$.
Since $C = \frac{\varepsilon_0 A}{d'}$,where $d'$ is the new distance between the plates,we have $F = \frac{(\varepsilon_0 A / d')^2 V^2}{2 \varepsilon_0 A} = \frac{\varepsilon_0 A V^2}{2 d'^2}$.
Given that the initial distance is $d$ and the final distance is $d' = 0.75 d = \frac{3}{4} d$,the extension in the spring is $x = d - d' = d - \frac{3}{4} d = \frac{1}{4} d$.
The spring force is $F_s = kx = k \left( \frac{1}{4} d \right)$.
Equating the forces: $\frac{\varepsilon_0 A V^2}{2 (\frac{3}{4} d)^2} = k \left( \frac{1}{4} d \right)$.
$\frac{\varepsilon_0 A V^2}{2 (\frac{9}{16} d^2)} = k \left( \frac{d}{4} \right)$.
$\frac{\varepsilon_0 A V^2}{\frac{9}{8} d^2} = k \left( \frac{d}{4} \right)$.
$\frac{8 \varepsilon_0 A V^2}{9 d^2} = k \left( \frac{d}{4} \right)$.
$k = \frac{32 \varepsilon_0 A V^2}{9 d^3}$.

(D) Let the capacitance of each pair of plates be $C = \frac{\varepsilon_0 A}{d}$.
Based on the arrangement,the system can be modeled as two capacitors in series ($C_1$ and $C_2$) which are then in parallel with a third capacitor $(C_3)$.
Here,$C_1 = C_2 = C_3 = C = \frac{\varepsilon_0 A}{d}$.
The equivalent capacitance of the series combination of $C_1$ and $C_2$ is $C_{12} = \frac{C_1 C_2}{C_1 + C_2} = \frac{C^2}{2C} = \frac{C}{2}$.
Now,$C_{12}$ and $C_3$ are in parallel,so the total capacitance $C_T$ is:
$C_T = C_{12} + C_3 = \frac{C}{2} + C = \frac{3C}{2}$.
Substituting $C = \frac{\varepsilon_0 A}{d}$,we get:
$C_T = \frac{3 \varepsilon_0 A}{2 d}$.

(B) The magnitude of the charge transferred between the plates is given by $Q = N \cdot e$.
Given $N = 10^{12}$ electrons and $e = 1.6 \times 10^{-19} \,C$.
So,$Q = 10^{12} \times 1.6 \times 10^{-19} = 1.6 \times 10^{-7} \,C$.
The potential difference developed is $V = 10 \,V$.
The capacitance $C$ is defined as $C = \frac{Q}{V}$.
Substituting the values,$C = \frac{1.6 \times 10^{-7}}{10} = 1.6 \times 10^{-8} \,F$.

(D) The arrangement can be considered as three parallel plate capacitors connected in parallel,each having an area of $\frac{A}{3}$.
Let the distance between the flat plate and the first,second,and third stairs be $d$,$2d$,and $3d$ respectively.
The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$.
For the first part,$C_1 = \frac{\varepsilon_0 (A/3)}{d} = \frac{\varepsilon_0 A}{3d}$.
For the second part,$C_2 = \frac{\varepsilon_0 (A/3)}{2d} = \frac{\varepsilon_0 A}{6d}$.
For the third part,$C_3 = \frac{\varepsilon_0 (A/3)}{3d} = \frac{\varepsilon_0 A}{9d}$.
Since these capacitors are connected in parallel,the equivalent capacitance $C_{eq}$ is the sum of individual capacitances:
$C_{eq} = C_1 + C_2 + C_3$
$C_{eq} = \frac{\varepsilon_0 A}{3d} + \frac{\varepsilon_0 A}{6d} + \frac{\varepsilon_0 A}{9d}$
Taking the common denominator as $18d$:
$C_{eq} = \frac{6\varepsilon_0 A + 3\varepsilon_0 A + 2\varepsilon_0 A}{18d} = \frac{11\varepsilon_0 A}{18d}$

(B) Let the four plates be numbered $1, 2, 3, 4$ from top to bottom. The outer plates $1$ and $4$ are connected together. Points $A$ and $B$ are connected to plates $2$ and $3$ respectively.
Between plates $1$ and $2$, there is a capacitor $C_1 = \frac{\varepsilon_0 S}{d}$.
Between plates $2$ and $3$, there is a capacitor $C_2 = \frac{\varepsilon_0 S}{d}$.
Between plates $3$ and $4$, there is a capacitor $C_3 = \frac{\varepsilon_0 S}{d}$.
Since plates $1$ and $4$ are connected, they are at the same potential. Let this potential be $V_0$. Plate $2$ is at potential $V_A$ and plate $3$ is at potential $V_B$.
The capacitor $C_1$ is connected between $V_A$ and $V_0$. The capacitor $C_3$ is connected between $V_B$ and $V_0$. The capacitor $C_2$ is connected between $V_A$ and $V_B$.
This is equivalent to two capacitors $C_1$ and $C_3$ in series, which are then in parallel with $C_2$.
Since $C_1 = C_2 = C_3 = C = \frac{\varepsilon_0 S}{d}$, the equivalent capacitance $C_{eq}$ is:
$C_{eq} = C_2 + (C_1 \text{ in series with } C_3)$
$C_{eq} = C + \frac{C \times C}{C + C} = C + \frac{C}{2} = \frac{3C}{2}$
Substituting $C = \frac{\varepsilon_0 S}{d}$, we get:
$C_{eq} = \frac{3}{2} \frac{\varepsilon_0 S}{d}$

(A) For an isolated charged parallel plate capacitor,the charge $q$ on the plates remains constant.
The electric field $E$ between the plates is given by $E = \frac{\sigma}{\varepsilon_{0}} = \frac{q}{A \varepsilon_{0}}$,which is constant since $q$,$A$,and $\varepsilon_{0}$ are constant.
The electrostatic force of attraction between the plates is $F = qE = \frac{q^2}{2A \varepsilon_{0}}$.
Since the force is constant,the acceleration $a$ of the plates is constant $(a = F/m)$.
The separation $d$ between the plates at time $t$ is given by $d(t) = d_{0} - \frac{1}{2} a t^2$,where $d_{0}$ is the initial separation.
The potential difference $V$ is given by $V = E \cdot d(t) = E (d_{0} - \frac{1}{2} a t^2)$.
This equation $V(t) = E d_{0} - (\frac{E a}{2}) t^2$ represents a downward-opening parabola,which corresponds to Graph $A$.

(A) The given arrangement consists of four plates. Let the plates connected to $P$ be the positive plates and the plates connected to $Q$ be the negative plates.
From the figure,we can see that there are two capacitors formed between the plates.
Each capacitor is formed by two adjacent plates separated by a distance $d$ with an area $a$.
The capacitance of a single parallel plate capacitor is given by $C = \frac{\varepsilon_{0} a}{d}$.
Since the two capacitors are connected in parallel,the equivalent capacitance $C_{eq}$ is the sum of the individual capacitances.
$C_{eq} = C_{1} + C_{2} = \frac{\varepsilon_{0} a}{d} + \frac{\varepsilon_{0} a}{d} = \frac{2 \varepsilon_{0} a}{d}$.

(A) In a parallel plate capacitor,the electric field between the plates is uniform.
Since the force $F = eE$,where $e$ is the charge of the electron and $E$ is the uniform electric field,the force experienced by an electron at any point between the plates is constant.
Thus,the forces acting on the electrons at both points $P$ and $Q$ are the same.
Therefore,option $(A)$ is correct.