CAPACITOR

 

Electric Capacitance:$\\$

• Capacitors are charge storing device.
• Capacitance (C) of a capacitor is the ratio of charge (Q) given and the potential ( V ) to which it is raised i.e $C=\frac{Q}{V}$
• Its S.I unit is Farad and its cgs unit is Statfarad. Stat-farad =$\frac{1}{9}×10^{-11}$ farad
• The dimensional formula of electrical capacitance is $[M^{-1} L^{-2} T^{-4} A^{-2}]$
• The capacitance of a spherical conductor is C=$4πε_0$ R=$\frac{R}{9×10^9}$

where R is the radius of the conductor.

• The capacitance of the earth is $(6.37×10^6 )/(9×10^9 )=7.12×10^{-4}$ farad

Capacitor and Capacitance

Capacitor and Capacitance of Parallel Plate Capacitor\$

A capacitor consists of two conductors separated by an insulator or dielectric. The conductors carry equal and opposite charge±Q. It is an electrical device that stores electric charge.

Capacitance of parallel plate capacitor with air as dielectric is$\\$$C_0=\frac{ε_0.A}{d}$ Where A is the area of each plate and 'd’ is the distance between them.

With dielectric of dielectric constant $ε_r$ or K. $C_0=\frac{K.ε_0.A}{d}$

The capacitance does not depend on the material of plates, potential difference between the plates and the charge given to the plates.

The value of C depends on size, shape and relative position of two coatings of the capacitor. It also depends on nature of medium separating the two coatings.

A parallel plate capacitor is used to produce uniform electric field between the plates. Outside the plates no electric field is produced. A material used as a dielectric in capacitor must have high dielectric constant and high dielectric strength

Capacitance of a parallel plate capacitor with a dielectric slab of dielectric constant K of thickness t(t<d) is given by$\\$C=$\frac{ε_0 .A}{ d-t(1-\frac{1}{K})}$

Capacitance of a parallel plate capacitor with a conducting slab of thickness (t<d) is given by C=$\frac{ε_0 .A}{ d-t(1-\frac{1}{∞})}$ [∵K" for " Conductor = ∞] Hence C=$\frac{ε_0.A}{d-t}$

If dielectric slab fills the entire space between the plates (instead of filling it partially) i.e. t=d, then C=$\frac{ε_0.A}{\frac{d}{K}}$=K$\frac{ε_0.A}{d-t}$= K$C_0$ If the conducting slab entirely fills the space between plate i.e. t=d then C=$\frac{ε_0.A}{d-t}$=C=$\frac{ε_0.A}{d-d}$ = ∞

The capacitance of a parallel plate capacitor having a number of slabs of thickness $t_1,t_2,t_3,……….$ and dielectric constants $K_1,K_2,K_3,………$ respectively in

$$$$d=t_{1}+t_{2}+t_{3}+\cdots \ldots \ldots

When two dielectrics of equal thickness are arranged in series then:

$$$$\begin{array}{r} C_{\mathrm{eq}}=\frac{\varepsilon_{0} A}{\frac{d / 2}{K_{1}}+\frac{d / 2}{K_{2}}}=\frac{\varepsilon_{o} A}{d}\left(\frac{2 K_{1} K_{2}}{K_{1}+K_{2}}\right) \\ \text { Similarly, } K_{\text {eq }}=\frac{d}{\frac{t_{1}}{K_{1}}+\frac{t_{2}}{K_{2}}+\cdots \cdot \frac{t_{0}}{K_{n}}} \end{array}

For two dielectrics of equal thickness $\left(t_{1}=t_{2}=d / 2\right)$ arranged in series

$$$$K_{e q}=\frac{2 K_{1} K_{2}}{K_{I}+K_{2}}

$\textbf{Dielectrics in parallel}$$\\$When a number of dielectric slabs of same thickness (d) and different areas of cross-section $A_{1}, A_{2}, A_{3} \ldots \ldots$ having dielectric constants $K_{1}, K_{2}, K_{3 \ldots \ldots}$ respectively are placed between the plates of a parallel plate capacitor. Its capacitance is given by $\mathrm{C}=\frac{\varepsilon_{0}\left(\mathrm{~K}_{1} \mathrm{~A}_{1}+\mathrm{K}_{2} \mathrm{~A}_{2}+\mathrm{K}_{3} \mathrm{~A}_{3}+\cdots_{\ldots} \ldots\right)}{\mathrm{d}} \\$

(1) When two dielectrics of equal areas are arranged in parallel i.e. $A_{1}=A_{2}$

$$$$\begin{aligned} C_{e q} &=\frac{\varepsilon_{0} A}{d}\left(\frac{K_{1}+K_{2}}{2}\right) \\ \text { Similarly, } \mathbf{K}_{e q} &=\frac{K_{1} A_{1}+K_{2} A_{2}+\cdots . . .}{A} \end{aligned}

(2) For two dielectrics of equal areas $\left(A_{1}=A_{2}=\frac{A}{2}\right)$ arranged in parallel

$$$$\mathbf{K}_{\mathrm{eq}}=\frac{K_{1}+\mathrm{K}_{2}}{2}

Combination of Capacitors

Series Combination: When a potential difference V is applied across several capacitors connected in series, the capacitors have identical charge q. The sum of the potential differences across all the capacitors is equal to the applied potential difference V.

Q=$c_1.v_1=c_2.v_2= c_3.v_3$$\\$The total potential difference across combination is:

$V=V_{1}+V_{2}+V_{3}$

$$$$\begin{array}{l} V=\frac{Q}{C_{1}}+\frac{Q}{C_{2}}+\frac{Q}{C_{3}} \\ \frac{V}{Q}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}} \end{array}

The ratio $\mathrm{Q} \mathrm{N}$ is called as the equivalent capacitance $\mathrm{C}$ between point $\mathrm{a}$ and $\mathrm{b}$. The equivalent capacitance $\mathrm{C}$ is given by: $\frac{1}{c}=\frac{1}{c_{1}}+\frac{1}{c_{2}}+\frac{1}{c_{3}}$ The potential difference across $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ is $\mathrm{V}_{1}$ and $\mathrm{V}_{2}$ respectively, given as follows:

$$$$V_{1}=\frac{C_{2}}{C_{1}+C_{2}} ; V_{2}=\frac{C_{1}}{C_{1}+C_{2}} V

In case of more than two capacitors, the relation is: $\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}+\frac{1}{C_{4}}+\cdots . \ldots$

$\textbf{Parallel Combination:}\\$When a potential difference V is applied across several capacitors connected in parallel, that potential difference V is applied across each capacitor. The total charge q stored on the capacitors is the sum of the charges stored on all the capacitors.$\\$$\begin{array}{c} C_{\mathrm{eq}}=\frac{q}{V}=C_{1}+C_{2}+C_{3} \\ C_{\mathrm{eq}}=\sum_{j=1}^{n} C_{j}(n \text { capacitors in parallel }) \end{array} \\$

The total charge is Q given as: $Q=Q_{1}+Q_{2}+Q_{3}$,$\\ Q=C_{1} V+C_{2} V+C_{3} V$,$\\ \frac{Q}{V}=C_{1}+C_{2}+ \mathrm{C}_{3}$$\\$Equivalent capacitance between a and b is:$\\$$\mathrm{C}=\mathrm{C}_{1}+\mathrm{C}_{2}+C_{3}$ The charges on capacitors is given as:$\\$$Q_1=\frac{C_1}{C_1+C_2}$Q$\\$$Q_2=\frac{C_2}{C_1+C_2}$Q$\\$In case of more than two capacitors, $C=C_1+C_2+C_3+C_4+C_5$+...……

$\textbf{Spherical Capacitors}\\$Two concentric spherical capacitors having radii $r_{a}$ and $r_{b}$ of inner sphere and outer sphere respectively$\\$(i) If outer sphere is earthed, $\mathrm{C}=\frac{4 \pi \varepsilon_{0} \mathrm{Kr}_{\mathrm{a}} \mathrm{r}_{\mathrm{b}}}{\left(\mathrm{r}_{\mathrm{b}}-\mathrm{r}_{\mathrm{a}}\right)}$$\\$(ii) If inner sphere is earthed, $\mathrm{C}=\frac{4 \pi \varepsilon_{0} \mathrm{Kr}_{5}^{2}}{\left(\mathrm{r}_{\mathrm{b}}-\mathrm{r}_{\mathrm{a}}\right)} \\$

$\textbf{Electric Potential Energy Stored in a Charged Capacitor}\\$The capacitor stores charge as well as electric potential energy, which is given by$\\$$U=\frac{Q^{2}}{2 C}=\frac{1}{2} Q V=\frac{1}{2} C V^{2} \\$The energy is stored in the electric field between the plates.$\\$The energy stored per unit volume in the electric field between the plates is called energy density. $\left(U_{d}\right)$. It is given by$\\ U_{d}=\frac{1}{2} \varepsilon_{0} E^{2}=\frac{\rho^{2}}{2 \varepsilon_{0}}=\frac{Q^{2}}{2 \varepsilon_{0} A^{2}} \\$Where $U$ is surface charge density on plates of capacitor. Force between the plates of capacitor given by$\\$$\mathrm{F}=\frac{-\mathrm{d} \mathrm{U}}{\mathrm{dr}}$ (i.e. Energy gradient)$\\$$F=\frac{q^{2}}{2 C d}=\frac{q^{2}}{2 \frac{\varepsilon_{0} A}{d} \cdot d} \\$$\therefore F=\frac{q^{2}}{2 \varepsilon_{0} A} \\$Again, $F=\frac{d U}{d r}=\frac{C V^{2}}{2 d} \\$$\therefore F=\frac{C V^{2}}{2 d} \\$Also, $\frac{F}{A}=\frac{\rho^{2}}{2 \varepsilon_{0}}=\frac{1}{2} \varepsilon_{0} E^{2} \\$$\therefore F=\frac{\rho^{2} A}{2 \varepsilon_{0}}=\frac{1}{2} \varepsilon_{0} E^{2} \\$Thus,$\\$$\mathrm{F}=\frac{\mathrm{q}^{2}}{2 \varepsilon_{0} \mathrm{~A}}=\frac{\mathrm{CV}^{2}}{2 \mathrm{~d}}=\frac{\rho^{2} \mathrm{~A}}{2 \varepsilon_{0}}=\frac{1}{2} \varepsilon_{0} \mathrm{E}^{2} \mathrm{~A} \\$

$\textbf{Regrouping of Capacitors:}\\$When two capacitors of capacities $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ are charged with different potentials $\mathrm{V}_{1}$ and $\mathrm{V}_{2}$ are connected by a conducting wire, charge flows from higher potential to lower potential until they attain common potential (V). (i) Total charge before sharing $=$ Total charge after sharing

$$$$\mathrm{C}_{1} \mathrm{~V}_{1}+\mathrm{C}_{2} \mathrm{~V}_{2}=\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right) \mathrm{V}

(ii) Common potential (V)

$$$$\begin{array}{l} =\frac{C_{1} V_{1}+C_{2} V_{2}}{C_{1}+C_{2}} \\ \text { i.e } V=\frac{\text { total charge }}{\text { total capacity }} \end{array}

(iii) Initial Energy Stored,

$$$$U_{i}=\frac{1}{2} C_{1} V_{1}^{2}+\frac{1}{2} C_{2} V_{2}^{2}

(iv) Final Energy stored,

$$$$U_{f}=\frac{1}{2}\left(C_{1}+C_{2}\right) V^{2}

(v) Loss in Energy $(\Delta U)=U_{f}-U_{i}$

$$$$=\frac{1}{2} \frac{C_{1} C_{2}}{C_{1}+C_{2}}\left(V_{1}-V_{2}\right)^{2}

(vi) Final charges in two capacitors, $q_{1}^{\prime}=\mathrm{C}_{1} \mathrm{~V}$ and $\mathrm{q}_{2}^{\prime}=\mathrm{C}_{2} \mathrm{~V} \\$Note: Two capacitors $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ charged with $V_{1}$ and $V_{2}$ are connected with unlike plates joined together, then$\\$(i) Common potential $\mathrm{V}=\frac{c_{1} V_{1}-C_{2} V_{2}}{C_{1}-C_{2}}$$\\$(ii) Loss in Energy $=\frac{1}{2} \frac{C_{1} C_{2}}{C_{1}+C_{2}}\left(V_{1}+V_{2}\right)^{2}$

Inserting Dielectric

$\text{Effect of Inserting Dielectric:}$If Battery is disconnected charge remains constant and if the battery is connected potential difference remains constant

$\textbf{When battery is connected:}\\$(i) V= Constant$\\$ii) $\mathrm{E}=\frac{d v}{d x}=$ constant $[\because V$ is constant $]$

(iii) $\mathrm{C}=\frac{\mathrm{K} ε_{0} \mathrm{~A}}{\mathrm{~d}}=\mathrm{KC}_{0} \Rightarrow \mathrm{C}$ increases by $\mathrm{K}$ times.$\\$

(iv) $\mathrm{U}=\frac{1}{2} \mathrm{CV}^{2}$. As $\mathrm '{V}^{\prime}$ is constant and ' $\mathrm{C}^{\prime}$' increases $\mathrm{K}^{\prime}$ times. So $\mathrm{U}^{\prime}$ also increases $\mathrm{K}$ times.$\\$

(v) Charge (q) $=C V$. As $^{\prime} V’$ is constant and $C$ increases, so q also increases.$\\$

$(\mathrm{vi}) \mathrm{F}=\frac{\mathrm{CV}^{2}}{2 \mathrm{~d}}$. As 'C' increases, ' 'F' also increases.$\\$

(Vii) $\rho=\frac{Q}{A^{-}}$. As ' $Q$ ' increases, ' $\rho$ ' also increases. Thus, above relations can be concluded as Potential difference remains constant$\\$

• Electric field E remains constant.
• Capacitance, potential energy, charge, force, surface charge density increases (becomes K times)

$\textbf{When battery is disconnected :}\\$(i) $Q=$ constant$\\$(ii) $\mathrm{C}=\frac{\mathrm{K} e_{\mathrm{e}} \mathrm{A}}{d} \Rightarrow \mathrm{C}^{\prime}$ increases $^{\prime} \mathrm{K}^{\prime}$ times i.e. ' $\mathrm{C}^{\prime}=\mathrm{C}_{0} \mathrm{~K} \\$(iii) $U=\frac{1}{2} \frac{Q^{2}}{C}$. As $Q$ is constant, $C$ increases, $U$ decreases.$\\$$(i v) V=\frac{q}{c} \Rightarrow V \propto \frac{1}{c}$, As ' $C$ ' increases, $V$ decreases$\\ (\mathrm{v}) \mathrm{E}=\frac{\mathrm{dV}}{\mathrm{dx}}$. As ' $\mathrm{V}$ ' decreases, 'E' also decreases.$\\$(vi) $\mathrm{F}=\frac{\mathrm{Q}^{2}}{2 \mathrm{Ae}_{\mathrm{O}} \mathrm{K}} \Rightarrow \mathrm{F}^{\prime}$ decreases.$\\$(vii) $\rho=\frac{Q}{A}=$ constant $[\because Q$ is constant $]$ The above relations can be concluded as.$\\$

• Charge remains constant.$\\$
• Surface charge density [\rho) remains constant. Capacitance [C) Increases $\mathrm{K}$ times. Potential, Potential energy, Electric field intensity, force decreases $1 / K$ times.$\\$

$\textbf{When the plates of a parallel plate capacitor are moved apart}\\$(i,e, 'd' increases):$\\$

$\textbf{When battery is connected :}\\$(i) $V=$ constant$\\$

(iii) $Q=C . V=\mathrm{Q} \propto C .$ As $C$ decreases, $Q$ also decreases.$\\$

(iv) $E=\frac{d V}{d x}$. As $'V^{\prime}$ is constant and 'd' is increased, 'E decreases.$\\$

(v) $U=\frac{1}{2} C V^{2} \Rightarrow U \propto$ C. As 'C decreases, 'U' also decreases.$\\$

(vi) $F=\frac{c V^{2}}{2 d} \Rightarrow F \propto \frac{c}{d} .$ As $C$ decreases and d increases, $F$ decreases.$\\$

(vil) $\rho=\frac{4}{A} \cdot$ As $^{\prime} Q^{\prime}$ decreases, $\rho$ decreases. The above relations can be concluded as, Potential remains constant.$\\$

(viii) Capacitance, Charge, Electric field intensity, Potential energy, Force, surface charge density all decreases.$\\$

$\textbf{When battery is disconnected :}\\$(i) Q= Constant$\\$

(ii) C=$\frac{ε_0.A}{d}$ ⇒ C∝ $\frac{A}{d}$. As ‘d’ increases, ‘C ’ decreases$\\$

(iii) V=$\frac{Q}{C}$⇒V∝$\frac{1}{C}$⋅ As ' C' decreases, ' V' increases.$\\$

(iv) E=$\frac{V}{d}$. As ‘V' and 'd' both are increased, ‘E’ remains constant. Also, E=$\frac{σ}{ε_o}$ =$\frac{Q}{Aε_o}$. As ‘Q’ is constant, ‘E' is constant.$\\$

(v) U=1/2 QV. As ‘V ’ increases, ‘U’ also increases.$\\$

(vi) F= $\frac{Q^2}{2Aε_0}$ . As ' Q' is constant, ‘F' is also constant.$\\$

(vii) ρ=Q/A= Constant [ As Q is constant ].$\\$

The above relations can be concluded as, Charge, Electric field intensity, Force, Surface charge density are constant.

Charging of a Capacitor

Charging of a Capacitor:$\\$

Here, When the key is pressed, the capacitor begins to store charge. If at any time during charging, I is the current through the circuit and Q is the charge on the capacitor, then$\\$Potential difference across resistor =IR, and$\\$Potential difference between the plates of the capacitor =Q/C$\\$Since the sum of both these potentials is equal to ε,$\\$RI+Q/C=ε….............................(1)$\\$

As the current stops flowing when the capacitor is fully charged,$\\$

When Q=$Q_0$ (the maximum value of the charge on the capacitor), I=0$\\$

From equation. (1),$\\$

$\frac{Q_0}{C}$=ε…...........................(2)$\\$From equations (1) and (2),$\\$

RI+$\frac{Q}{C}=\frac{Q_0}{C} \\$or$\\$

$\frac{Q_0}{C}$-$\frac{Q}{C}$=RI$\\$

Or$\\$

$\frac{Q_0-Q}{CR}$=I….............(3)$\\$

Since I=$\frac{dq}{dt}$, from equation …....(3),$\\$

$\frac{Q_0-Q}{CR}$=$\frac{dq}{dt}$ or $\frac{dQ}{Q_0}$ = $\frac{dt}{CR} \\$

When t=0, Q=0 and when t=t, Q=Q.$\\$

Integrating both sides within proper limits, we get$\\$

$∫_0^0\frac{dQ}{(Q_0-Q) }=∫_0^t \frac{dt}{CR}=\frac{1}{CR} ∫_0^t dt \\$

or${|-ln⁡(Q_0-Q)|}_0^Q=\frac{1}{CR} |t|_0^t$$\\$

Or $-ln⁡(Q_0-Q)+ln⁡Q_0=\frac{t}{RC} \\$

Or $ln⁡(Q_0-Q)-ln⁡Q_0=-\frac{t}{RC} \\$

or $ln⁡(Q_0-Q)/{Q_0} =-\frac{t}{RC} \\$

or$\frac{Q_0-Q}{Q_0} =e^{(-t/R)} \\$

or $Q_0-Q=Q_0 e^{(-t/RC)} \\$

or $Q=Q_0 (1-e^{(-t/RC)} \\$

or $Q=Q_0 (1-e^{-t/τ} )$….........(4)$\\$

Where τ=RC$\\$Eqn (4) gives us the value of charge on the capacitor at any time during charging.$\\$

Time Constant:$\\$$\\$

The dimensions of CR are those of time. Further, if CR<<1,Q will attain its final value rapidly and if CR>>1, it will do so slowly. Thus, CR determines the rate at which the capacitor charges (or discharges) itself through a resistance. It is for this reason that the quantity CR is called the time constant or more appropriately, the capacitive time constant of the circuit.$\\$If t=τ, then from eqn. (4),$\\$

$Q=Q_0 (1-e^{-1} )=Q_0 (1-\frac{1}{e}) \\$

or $Q=Q_0 (1-\frac{1}{2.718})=Q_0$ (1-0.368)=0.632Q_0$\\$

=63.2% of Q_0$$\$

Time constant of a CR circuit is thus the time during which the charge on the capacitor becomes 0.632 (approx., 2/3 ) of its maximum value.$\\$

For the charge on the capacitor to attain its maximum value($Q_0$ ), i.e., for $Q=Q_0$,$\\$

$e^{-t/RC}$=0 or t=∞$\\$

Thus, theoretically, the charge on the capacitor will attain its maximum value only after infinite time.$\\$The current in the circuit during charging is given by $I=I_0 e^{-t/RC}$where$\\$$I_0=\frac{V_0}{R}$$\\$

Charging of a Capacitor:$\\$

When a charged capacitor (at t=0 ) is allowed to discharge through a resistance R, then the voltage across the capacitor and the current in the circuit falls exponentially as$\\$

$q=q_0 e^{(-t/RC)} \\$

$I=I_0 e^{(-t/RC)} \\$

$V=V_0 e^{(-t/RC)} \\$

(All the expressions for discharging can be derived by following similar process as we did while deriving charging of a capacitor.)$\\$

The rate of discharging depends on the time constant given by τ=RC.

Good to Know

POINTS:
$\\$

A capacitor works in AC circuits. It acts as a perfect insulator in DC circuit.

In a charged capacitor the energy resides on the electric field between the plates.

Net charge on the capacitor is zero.

Practically mica is used as dielectric in parallel plate capacitor since it has high dielectric strength and high dielectric constant.

Electric field intensity between the plates of capacitor is independent of distance between plates and thickness of dielectrics$\\$E=$\frac{σ}{ε_0 K}$$\\$

Energy stored in a capacitor is =$\frac{Q^2}{2C}$ or $\frac{1}{2}CV^2$ but energy supplied by the battery is $\frac{Q^2}{C}$ or $CV^2 \\$When air in a capacitor is replaced by dielectric of dielectric constant ‘K', then capacitance increases K times.$\\$$C_0=\frac{ε_0 A}{d}$And C'=$\frac{ε_0 KA}{d} \\$∴C'=K$C_o \\$

K or $ε_r=\frac{ε}{ε_0} =\frac{F_0}{F_mediun} = \frac{C_medium}{C_o} i.e. ε = ε_0 \space or \space ε_r \\$

To increase potential difference, capacitors are connected in series. To increase capacitance, capacitors are connected in parallel.$\\$

Introduction of dielectric between the plates of a parallel plate air capacitor doesn’t change the force between the plates (as E is constant.)$\\$[Note: Force between two charges F=$\frac{q_1.q_2}{4πε_0ε_r r}$ Force between two charges becomes F/K].$\\$

If force (F) acts on a charged particle kept between the plates of a charged capacitor, then on removing one of the plates, the force acting on the same particle is F/2

On increasing the surface area of a spherical conductor by K times, its capacitance becomes √K times.$\\$

The capacitance of earth is 712μF.

If one of the plates of a parallel plate capacitor is located horizontally submerged in a liquid while the other outside, the level of liquid rises on charging the plates.

To keep the capacity constant after inserting the slabs, the distance between the plates must be increased by a distance ’ x ’ given by$\\$$x=t(1-\frac{1}{K}) \\$where ‘t' is the thickness of slab.$\\$

The maximum electric field strength applied in a region of dielectric medium up to which it behaves as dielectric is called dielectric strength of that medium. The dielectric strength of of air at STP is 3×$10^6$ V/m.

If ‘n' drops of radius (R) and charge (q) are joined to form bigger drop, then$\\$(i) Radius of big drop, R=$n^{1/3}$ r$\\$(ii) Total charge on big drop Q=nq$\\$(iii) Total capacitance =$n^{1/3}$ C$\\$(iv) Total surface charge density =$n^{1/3}$ ρ$\\$(v) Total Potential V'=$n^{2/3}$ V$\\$(vi) Total Potential Energy U'=$n^{5/3}$U$\\$(vii) Total electric field =$n^{1/3}$ E When two capacitors are joined together, then charge flows from higher to lower potential until both acquire equal potential.$\\$V=$\frac{C_1V_1+C_2V_2}{C_1+C_2} \\$(viii) Ratio of final energies $\frac{U_1}{U_2} =\frac{\frac{1}{2}C_1V^2}{\frac{1}{2} C_2 V^2 } \\$(ix) If two identical capacitors are joined ($C_1=C_2$=C) then final voltage is half of net voltage i.e. V=$\frac{V_1+V_2}{2}$ and final charge will be half of net charge.$\\$

∴$\frac{U_1}{U_2}$ =$\frac{C_1}{C_2}$

If two spheres of capacitances $‘C_1' and ‘C_2’ and radii 'R_1' and ' R_2'$ are joined$\\$(i) Common Potential (V)=$\frac{C_1V_1+C_2V_2}{C_1+C_2} \\$∴V=$\frac{R_1V_1+R_2V_2}{R_1+R_2}$ ∵C=$4πε_0 R \\$(ii) Loss in Energy =$\frac{C_1C_2 (V_1-V_2 )^2}{2(C_1+C_2 )} =\frac{4πε_0 R_1 R_2 (V_1-V_2 )^2}{2(R_1+R_2}$$\\$

If a thin metal foil is introduced between the plates of the capacitor. Then the capacitance remains constant.$\\$C=$\frac{ε_0 A}{d-t+\frac{t}{R}}$ = $\frac{ε_0 A}{d-t+\frac{t}{∞}}$ = $\frac{ε_0 A}{d-t} \\$For thin metal foil t≈0$\\$∴ C remains constant.$\\$

If thin metal foil is inserted in the middle, then capacitance remains constant.$\\$C=$\frac{ε_0 A}{d} \\$After insertion,$\\$$C_1=\frac{ε_0 A}{d/2}=\frac{2ε_0 A}{d}=2C \\$$C_2=\frac{ε_0 A}{d/2}=\frac{2ε_0 A}{d}=2C \\$Since, the two capacitors will be in series. Thus equivalent capacitance $C_s$ is given by$\\$$C_s=\frac{C_1C_2}{C_1+C_2 }=C \\$

If a thin metal foil is inserted as in figure, the capacitance will become infinite.

When plates each of equal area ‘A’ with equal separation ‘d’ are connected alternatively, it forms n-1 capacitors in parallel and effective$\\$capacitance$C_s=(n-1)C, where C=\frac{ε_o A}{d}$ due to each capacitor.$\\$The four plates are alternatively connected and form three capacitors in parallel$\\$

∴ Net capacitance of combination$\\$$C_{PQ}=\frac{3ε_0 A}{d} \\$If six plates are connected alternatively$\\$$C_{eq}=(n-1)C=\frac{5ε_0 A}{d} \\$

The two capacitors in the middle are connected together and hence act as a single conductor. Thus effectively there are three plates which form two capacitors in parallel.$\\$Capacitance of each capacitor,$\\$C=$\frac{ε_0 A}{d} \\$∴ Net capacitance of combination,$\\$$C_{PQ}=2C=\frac{2ε_0 A}{d}$