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=VQ
- Its S.I unit is Farad and its cgs unit is Statfarad. Stat-farad =91×10−11 farad
- The dimensional formula of electrical capacitance is [M−1L−2T−4A−2]
- The capacitance of a spherical conductor is C=4πε0 R=9×109R
where R is the radius of the conductor.
- The capacitance of the earth is (6.37×106)/(9×109)=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 isC0=dε0.A Where A is the area of each plate and 'd’ is the distance between them.
With dielectric of dielectric constant εr or K. C0=dK.ε0.A
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 byC=d−t(1−K1)ε0.A
Capacitance of a parallel plate capacitor with a conducting slab of thickness (t<d) is given by C=d−t(1−∞1)ε0.A [∵K" for " Conductor = ∞] Hence C=d−tε0.A
If dielectric slab fills the entire space between the plates (instead of filling it partially) i.e. t=d, then C=Kdε0.A=Kd−tε0.A= KC0 If the conducting slab entirely fills the space between plate i.e. t=d then C=d−tε0.A=C=d−dε0.A = ∞
The capacitance of a parallel plate capacitor having a number of slabs of thickness t1,t2,t3,………. and dielectric constants K1,K2,K3,……… respectively in
d=t_{1}+t_{2}+t_{3}+\cdots \ldots \ldotsd=t1+t2+t3+⋯…… 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}Ceq=K1d/2+K2d/2ε0A=dεoA(K1+K22K1K2) Similarly, Keq =K1t1+K2t2+⋯⋅Knt0d For two dielectrics of equal thickness (t1=t2=d/2) arranged in series
K_{e q}=\frac{2 K_{1} K_{2}}{K_{I}+K_{2}}Keq=KI+K22K1K2 Dielectrics in parallelWhen a number of dielectric slabs of same thickness (d) and different areas of cross-section A1,A2,A3…… having dielectric constants K1,K2,K3…… respectively are placed between the plates of a parallel plate capacitor. Its capacitance is given by C=dε0( K1 A1+K2 A2+K3 A3+⋯……)
(1) When two dielectrics of equal areas are arranged in parallel i.e. A1=A2
\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}Ceq Similarly, Keq=dε0A(2K1+K2)=AK1A1+K2A2+⋯... (2) For two dielectrics of equal areas (A1=A2=2A) arranged in parallel
\mathbf{K}_{\mathrm{eq}}=\frac{K_{1}+\mathrm{K}_{2}}{2}Keq=2K1+K2
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=c1.v1=c2.v2=c3.v3The total potential difference across combination is:
V=V1+V2+V3
\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}V=C1Q+C2Q+C3QQV=C11+C21+C31 The ratio QN is called as the equivalent capacitance C between point a and b. The equivalent capacitance C is given by: c1=c11+c21+c31 The potential difference across C1 and C2 is V1 and V2 respectively, given as follows:
V_{1}=\frac{C_{2}}{C_{1}+C_{2}} ; V_{2}=\frac{C_{1}}{C_{1}+C_{2}} VV1=C1+C2C2;V2=C1+C2C1V In case of more than two capacitors, the relation is: C1=C11+C21+C31+C41+⋯.…
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.Ceq=Vq=C1+C2+C3Ceq=∑j=1nCj(n capacitors in parallel )
The total charge is Q given as: Q=Q1+Q2+Q3,Q=C1V+C2V+C3V,VQ=C1+C2+C3Equivalent capacitance between a and b is:C=C1+C2+C3 The charges on capacitors is given as:Q1=C1+C2C1QQ2=C1+C2C2QIn case of more than two capacitors, C=C1+C2+C3+C4+C5+...……
Spherical CapacitorsTwo concentric spherical capacitors having radii ra and rb of inner sphere and outer sphere respectively(i) If outer sphere is earthed, C=(rb−ra)4πε0Krarb(ii) If inner sphere is earthed, C=(rb−ra)4πε0Kr52
Electric Potential Energy Stored in a Charged CapacitorThe capacitor stores charge as well as electric potential energy, which is given byU=2CQ2=21QV=21CV2The 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. (Ud). It is given byUd=21ε0E2=2ε0ρ2=2ε0A2Q2Where U is surface charge density on plates of capacitor. Force between the plates of capacitor given byF=dr−dU (i.e. Energy gradient)F=2Cdq2=2dε0A⋅dq2∴F=2ε0Aq2Again, F=drdU=2dCV2∴F=2dCV2Also, AF=2ε0ρ2=21ε0E2∴F=2ε0ρ2A=21ε0E2Thus,F=2ε0 Aq2=2 dCV2=2ε0ρ2 A=21ε0E2 A
Regrouping of Capacitors:When two capacitors of capacities C1 and C2 are charged with different potentials V1 and V2 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}C1 V1+C2 V2=(C1+C2)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}=C1+C2C1V1+C2V2 i.e V= total capacity total charge (iii) Initial Energy Stored,
U_{i}=\frac{1}{2} C_{1} V_{1}^{2}+\frac{1}{2} C_{2} V_{2}^{2}Ui=21C1V12+21C2V22 (iv) Final Energy stored,
U_{f}=\frac{1}{2}\left(C_{1}+C_{2}\right) V^{2}Uf=21(C1+C2)V2 (v) Loss in Energy (ΔU)=Uf−Ui
=\frac{1}{2} \frac{C_{1} C_{2}}{C_{1}+C_{2}}\left(V_{1}-V_{2}\right)^{2}=21C1+C2C1C2(V1−V2)2 (vi) Final charges in two capacitors, q1′=C1 V and q2′=C2 VNote: Two capacitors C1 and C2 charged with V1 and V2 are connected with unlike plates joined together, then(i) Common potential V=C1−C2c1V1−C2V2(ii) Loss in Energy =21C1+C2C1C2(V1+V2)2
Inserting Dielectric
Effect of Inserting Dielectric:If Battery is disconnected charge remains constant and if the battery is connected potential difference remains constant
When battery is connected:(i) V= Constantii) E=dxdv= constant [∵V is constant ]
(iii) C= dKε0 A=KC0⇒C increases by K times.
(iv) U=21CV2. As ′V′ is constant and ' C′' increases K′ times. So U′ also increases K times.
(v) Charge (q) =CV. As ′V’ is constant and C increases, so q also increases.
(vi)F=2 dCV2. As 'C' increases, ' 'F' also increases.
(Vii) ρ=A−Q. As ' Q ' increases, ' ρ ' 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)
When battery is disconnected :(i) Q= constant(ii) C=dKeeA⇒C′ increases ′K′ times i.e. ' C′=C0 K(iii) U=21CQ2. As Q is constant, C increases, U decreases.(iv)V=cq⇒V∝c1, As ' C ' increases, V decreases(v)E=dxdV. As ' V ' decreases, 'E' also decreases.(vi) F=2AeOKQ2⇒F′ decreases.(vii) ρ=AQ= constant [∵Q is constant ] The above relations can be concluded as.
- Charge remains constant.
- Surface charge density [\rho) remains constant. Capacitance [C) Increases K times. Potential, Potential energy, Electric field intensity, force decreases 1/K times.
When the plates of a parallel plate capacitor are moved apart(i,e, 'd' increases):
When battery is connected :(i) V= constant
(iii) Q=C.V=Q∝C. As C decreases, Q also decreases.
(iv) E=dxdV. As ′V′ is constant and 'd' is increased, 'E decreases.
(v) U=21CV2⇒U∝ C. As 'C decreases, 'U' also decreases.
(vi) F=2dcV2⇒F∝dc. As C decreases and d increases, F decreases.
(vil) ρ=A4⋅ As ′Q′ decreases, ρ 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.
When battery is disconnected :(i) Q= Constant
(ii) C=dε0.A ⇒ C∝ dA. As ‘d’ increases, ‘C ’ decreases
(iii) V=CQ⇒V∝C1⋅ As ' C' decreases, ' V' increases.
(iv) E=dV. As ‘V' and 'd' both are increased, ‘E’ remains constant. Also, E=εoσ =Aεo