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=$\frac{σ}{ε_o}$ =$\frac{Q}{Aε_o}$