Capacitors And Capacitance

Capacitance

(i) When a conductor is given a charge , it's potential rises in proportional to the charge given i. e.

Q $\propto$ V or Q = CV or C = Q/V

This C is a constant and is called the Capacitance of conductor

(ii) Electrical capacitance of a conductor is a measure of it's ability to hold electric charge.

(iii) A given conductor can be charged to a limit. Charging after the limit results into ionization of medium , and charge gets leaked into medium

(iv) C depends on

(1) Size and shape of conductor

(2) Surrounding medium

(3) Presence of other conductor near by

(v) Dimension of [C] = [M^–1L^–2T⁴A²]

(vi) CGS unit : 9 × 1011 statfarad = 1 Farad

Capacitance of an isolated spherical conductor

Q : Charge on the sphere

V : Potential at the surface of sphere

R : Radius of sphere

(i) V = $\frac{Q}{4\pi {\epsilon }_{0}R}$ also V = $\frac{Q}{C}$

hence C = 4$\pi {\in }_0$R

(ii) $\frac{Q}{V}$= C = 4$\pi {\in }_0$R (in MKS) = R (in CGS)

(iii) If permittivity of medium is K, then

Cm = 4$\pi {\in }_0$KR

(iv) C $\propto$ R , C $\propto$ K

(v) C does not depend upon the charge given to the conductor

(vi) Here , we see that C = 4$\pi {\in }_0$R

$\Rightarrow$${\in }_0$= $\frac{C}{4\pi R}$

$\Rightarrow$ units of ${\in }_0$= Farad/meter

Note: C2/N -m2 is another unit of e₀ .Thus F/m and C2/N -m2 both represent the same physical quantity.

Distribution of charges

(i) Two insulated conductors A and B of capacitances C₁ and C₂ are given charges q1 and q2 and raised to potential V1 and V2 respectively. Then

q1 = C₁V₁ and q₂ = C₂V₂

(ii) When , these conductors are joined by a thin wire, then positive charge will flow from the conductor at higher potential to conductor at lower potential till their potentials become equal.

(iii) Charge remains conserved in this process i.e . If q'₁ and q'₂ are charges after distribution and V the potential on each conductor then

q'₁ = C₁V

and q'₂ = C₂V

and C₁V₁ + C₂V₂ = C₁V + C₂V

(iv) V = $\frac{Totalch\arg e}{Totalcapaci\tan ce}$ = $\frac{C_1{V}_1+C_2{V}_2}{C_1+C_2}$

(v) On connecting two charged conductors their distributed charges on them are in the ratio of their capacitances

i.e. $\frac{q_1}{q_2}=\frac{C_1}{C_2}$

(vi) Loss of energy :

$\Delta$U = Uinitial – Ufinal

$\Delta$U = $\frac{1}{2}\frac{C_1{C}_2}{C_1+C_2}$ (V₁ – V₂)²

Capacitor/Condenser

As we know that capacitance of a conductor depends on the presence of other conductors nearby. This fact is used to make a capacitor.

(i) Capacitance $\begin{array}{c}C=\frac{q}{\Delta V}\end{array}$

Combination of capacitors

There are two possible combinations-

(1) Series (2) Parallel

Series

(1) Charge on each condenser is same i.e.

Q = C₁V₁ = C₂V₂ = C₃V₃ = ...........

(2) Potential difference across each condenser is inversely proportional to it's capacity

i.e. V $\propto$ 1/C

So, V₁ = $\frac{Q}{C_1}$, V₂ = $\frac{Q}{C_2}$...............

(3) Total potential difference (V) in the circuit is sum of the potential differences across each capacitor i.e.

V = V₁ + V₂ + V₂ + V₃ ...........

or V = $\frac{Q}{C_1}+\frac{Q}{C_2}+\frac{Q}{C_3}$+...............

$\Rightarrow \frac{V}{Q}=\frac{1}{C}=\frac{Q}{C_1}+\frac{Q}{C_2}+\frac{Q}{C_3}$+...............

Series combination of two capacitors

(1) C = $\frac{C_1{C}_2}{C_1+C_2}$

(2) V₁= $\frac{C_2}{C_1+C_2}V$& V₂ =$\frac{C_1}{C_1+C_2}V$

(3) Q = $\left(\frac{C_1{C}_2}{C_1+C_2}\right)V$= C₁V₁ = C₂V₂

Parallel Combination

(i) Potential difference across each capacitor is same and is equal to the potential difference applied across the circuit.

(ii) Charge on each capacitor is proportional to it's capacitance i.e.

Q₁ = C₁V, Q₂ = C₂V, Q₃ = C₃V,.........

$\Rightarrow$ Q $\propto$ C

(iii) Total charge in circuit

Q = Q₁ + Q₂ + Q₃ +......

(iv) If C is the total capacitance of the circuit then

C = C₁ + C₂ + C₃ +.............

For parallel combination of two capacitance-

(1) C = C₁ + C₂

(2) Q₁ = $\frac{C_1}{C_1+C_2}$ Q = C₁V

Q₂ = $\frac{C_2}{C_1+C_2}$Q = C₂V

3) The total energy stored in parallel combination of two capacitors is

U = U₁ + U₂ = $\frac{1}{2}$C₁V² + $\frac{1}{2}$C₂V²

= $\frac{1}{2}$ V²(C₁ + C₂)

Illustration1: In the above circuit, find the potential difference across AB.

Solution: Let us mark the capacitors as 1, 2, 3 and 4 for identifications. As is clear, 3 and 4 are in series, and they are in parallel with 2. Then 2, 3, 4 combine is in series with 1.

\begin{align}  {{C}_{34}}=\dfrac{{{C}_{3}}.{{C}_{4}}}{{{C}_{3}}+{{C}_{4}}}=4\mu f,{{C}_{2,34}}=8+4=12\mu f \\  {{C}_{eq.}}=\dfrac{8x12}{8+12}=4.8\mu f,q={{C}_{eq.}}V=4.8x10=48\mu C \\ \end{align}

The `q' on 1 is 48 mC, thus V₁=q/c=6v [v₁ = $\frac{48\mu c}{8\mu F}=6v$ ]

$\Rightarrow$ V_PQ = 10 - 6 = 4V

By symmetry of 3 and 4, we say, V_AB = 2V.