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`color{blue} ✍️` A capacitor is a system of two conductors separated by an insulator (Fig. 2.24).

`color{blue} ✍️` The conductors have charges, say `Q_1` and `Q_2,` and potentials `V_1` and `V_2`. Usually, in practice, the two conductors have charges `Q` and `– Q,` with potential difference `V = V_1 – V_2` between them.

`color{blue} ✍️` We shall consider only this kind of charge configuration of the capacitor. (Even a single conductor can be used as a capacitor by assuming the other at infinity.)

`color{blue} ✍️` The conductors may be so charged by connecting them to the two terminals of a battery. `Q` is called the charge of the capacitor, though this, in fact, is the charge on one of the conductors – the total charge of the capacitor is zero.

`color{blue} ✍️` The electric field in the region between the conductors is proportional to the charge `Q.` That is, if the charge on the capacitor is, say doubled, the electric field will also be doubled at every point. (This follows from the direct proportionality between field and charge implied by Coulomb’s law and the superposition principle.)

`color{blue} ✍️` Now, potential difference `V` is the work done per unit positive charge in taking a small test charge from the conductor 2 to 1 against the field. Consequently, `V` is also proportional to `Q,` and the ratio `Q/V` is a constant:

`bb (C = Q/V)`


`color{blue} ✍️` The constant `C` is called the capacitance of the capacitor. `C` is independent of `Q` or `V,` as stated above.

`color{blue} ✍️` The capacitance `C` depends only on the geometrical configuration (shape, size, separation) of the system of two conductors. [As we shall see later, it also depends on the nature of the insulator (dielectric) separating the two conductors.]

`color{blue} ✍️` The SI unit of capacitance is 1 farad `(=1 "coulomb volt"^(-1)) or 1 F = 1 C V^(–1)`.

`color{blue} ✍️` A capacitor with fixed capacitance is symbolically shown as

`color{blue} ✍️` Equation (2.38) shows that for large `C, V` is small for a given `Q`. This means a capacitor with large capacitance can hold large amount of charge `Q` at a relatively small `V`.

`color{blue} ✍️` This is of practical importance. High potential difference implies strong electric field around the conductors.

`color{blue} ✍️` A strong electric field can ionise the surrounding air and accelerate the charges so produced to the oppositely charged plates, thereby neutralising the charge on the capacitor plates, at least partly.

`color{blue} ✍️` The maximum electric field that a dielectric medium can withstand without break-down (of its insulating property) is called its dielectric strength; for air it is about `color{purple}(3 × 10^6 Vm^(–1.))`

`color{blue} ✍️` For a separation between conductors of the order of `1 cm` or so, this field corresponds to a potential difference of `color{purple}3 × 10^4 V` between the conductors.

`color{blue} ✍️` Thus, for a capacitor to store a large amount of charge without leaking, its capacitance should be high enough so that the potential difference and hence the electric field do not exceed the break-down limits.

`color{blue} ✍️` Put differently, there is a limit to the amount of charge that can be stored on a given capacitor without significant leaking. In practice, a farad is a very big unit; the most common units are its sub-multiples `color{purple}(μF = 10^(–6) F, 1 nF = 10^(–9) F, 1 pF = 10^(–12) F,)` etc.

`color{blue} ✍️`Besides its use in storing charge, a capacitor is a key element of most ac circuits with important functions, as described in Chapter 7.


`color{blue} ✍️` Parallel plate capacitor consists of two large plane parallel conducting plates separated by a small distance (Fig. 2.25).

`color{blue} ✍️` We first take the intervening medium between the plates to be vacuum.

`color{blue} ✍️` The effect of a dielectric medium between the plates is discussed in the next section.

`color{blue} ✍️` Let A be the area of each plate and d the separation between them. The two plates have charges `Q` and `–Q.`

`color{blue} ✍️` Since d is much smaller than the linear dimension of the plates `(d^2 << A),` we can use the result on electric field by an infinite plane sheet of uniform surface charge density (Section 1.15). Plate 1 has surface charge density `σ = Q//A` and plate 2 has a surface charge density `–σ`. Using

`\color{green} "✍️ Outer region I (region above the plate 1)"`,

`color{purple} (E = (sigma)/(2epsilon_0) - (sigma)/(2epsilon_0) = 0)` ........................2.39

`\color{green} "✍️ Outer region II (region below the plate 2)"`,

`color{purple}(E = (sigma)/(2epsilon_0) - (sigma)/(2epsilon_0) = 0)` ..............2.40

`color{blue} ✍️` In the inner region between the plates 1 and 2, the electric fields due to the two charged plates add up, giving

`color{purple}(E = (sigma)/(2epsilon_0) + (sigma)/(2epsilon_0) = Q/((epsilon_0A))` ............................. 2.41

`color{blue} ✍️` The direction of electric field is from the positive to the negative plate.

`color{blue} ✍️` Thus, the electric field is localised between the two plates and is uniform throughout. For plates with finite area, this will not be true near the outer boundaries of the plates.

`color{blue} ✍️` The field lines bend outward at the edges – an effect called ‘fringing of the field’. By the same token, `σ` will not be strictly uniform on the entire plate. [E and σ are related by Eq. (2.35).] However, for `d^2 < < A`, these effects can be ignored in the regions sufficiently far from the edges, and the field there is given by Eq. (2.41). Now for uniform electric field, potential difference is simply the electric field times the distance between the plates, that is,

`color{purple}(V = Ed = 1/(epsilon_0) (Qd)/A)` ......................2.42

`color{blue} ✍️` The capacitance `C` of the parallel plate capacitor is then

`color{purple}{C= Q/V = (epsi_0A)/d}` ....................................2.43

`color{blue} ✍️` which, as expected, depends only on the geometry of the system. For typical values like `color{purple}(A = 1 m^2, d = 1 mm,)` we get

`color{purple}(C = (8.85 xx 10^(-12) C^2 N^(-1) m^(-2)xx1m^2)/(10^(-3)m) = 8.85 xx 10^(-9)F)` ..........................2.44

(You can check that if `color{purple}(1F= 1C V^(–1) = 1C (NC^(–1)m)^(–1) = 1 C^2 N^(–1)m^(–1).)`

`color{blue} ✍️` This shows that `1F` is too big a unit in practice, as remarked earlier.

`color{blue} ✍️` Another way of seeing the ‘bigness’ of `1F` is to calculate the area of the plates needed to have `C = 1F` for a separation of, say 1 cm:

`color{purple}(A = (Cd)/(epsilon_0) = (1F xx10^(-2)M)/(8.85 xx 10^(-12) C^2 N^(-1) N^(-2)) = 10^9 m^2)` ....................2.45

which is a plate about `30` km in length and breadth.


`color{blue} ✍️` With the understanding of the behaviour of dielectrics in an external field developed in Section 2.10.

`color{blue} ✍️` let us see how the capacitance of a parallel plate capacitor is modified when a dielectric is present.

`color{blue} ✍️` As before, we have two large plates, each of area A, separated by a distance `d`. The charge on the plates is `±Q,` corresponding to the charge density `±σ` (with `σ = Q/A`). When there is vacuum between the plates,

`color{purple}(E_0 = sigma/(epsilon_0))`

and the potential difference `color{purple}(V_0` is `V_0 = E_0d)`

`color{blue} ✍️` The capacitance `C_0` in this case is

`color{purple}(C_0 = Q/(V_0) = epsilon_0 A/d)` ................2.46

`color{blue} ✍️` Consider next a dielectric inserted between the plates fully occupying the intervening region.

`color{blue} ✍️`The dielectric is polarised by the field and, as explained in Section 2.10, the effect is equivalent to two charged sheets (at the surfaces of the dielectric normal to the field) with surface charge densities `σp` and `–σp`.

`color{blue} ✍️` The electric field in the dielectric then corresponds to the case when the net surface charge density on the plates is `color{purple}(±(σ – σp).)` That is,

`color{purple}(E = (sigma-sigma_P)/(epsilon_0))` ..............2.47

so that the potential difference across the plates is

`color{purple}(V = Ed = (sigma-sigma_P)/(epsilon_0)d)` ................2.48

`color{blue} ✍️` For linear dielectrics, we expect σp to be proportional to `E_0,` i.e., to `σ`. Thus, `color{purple}((σ – σ_p))` is proportional to `σ` and we can write

`color{purple}((sigma-sigma_P)= (sigma)/K)` ...................... 2.49

where `K` is a constant characteristic of the dielectric. Clearly, `K > 1.` We then have

`color{purple}(V = (sigmad)/(epsilon_0K) = (Qd)/(Aepsilon_0K))` ..................2.50

`color{blue} ✍️` The capacitance `C`, with dielectric between the plates, is then

`color{purple}(C = Q/V = (epsilon_0KA)/d)` ........................2.51

`color{blue} ✍️` The product `ε_0K` is called the permittivity of the medium and is denoted by `ε`

`color{purple}(ε = ε_0 K)` .............2.52

For vacuum `K = 1` and `color{purple}(ε = ε_0; ε_0)` is called the permittivity of the vacuum. The dimensionless ratio

`color{purple}(K = ε/(ε_0))` ............... 2.53

is called the dielectric constant of the substance. As remarked before, from Eq. (2.49), it is clear that K is greater than 1. From Eqs. (2.46) and (2. 51)

`color{purple}(K = C/(C_0))` ..................... 2.54

`color{blue} ✍️` Thus, the dielectric constant of a substance is the factor `( > 1)` by which the capacitance increases from its vacuum value, when the dielectric is inserted fully between the plates of a capacitor.

`color{blue} ✍️` Though we arrived at Eq. (2.54) for the case of a parallel plate capacitor, it holds good for any type of capacitor and can, in fact, be viewed in general as a definition of the dielectric constant of a substance.
Q 3144880753

A slab of material of dielectric constant K has the same area as the plates of a parallel-plate capacitor but has a thickness (3/4)d, where d is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates?
Class 12 Chapter example 8

Let `E_0 = V_0/d` be the electric field between the plates when there is no dielectric and the potential difference is `V_0`. If the dielectric is now inserted, the electric field in the dielectric will be `E = E_0/K.` The potential difference will then be

`color{blue}{V = E_0 (1/4d) + E_0/K (3/4d)}`

`= E_0 d(1/4+3/4k) = V_0 (K+3)(4K)`

The potential difference decreases by the factor `(K + 3)/4K` while the
free charge `Q_0` on the plates remains unchanged. The capacitance
thus increases

`C = (Q_0)/(V) = (4K)/(K+3) (Q_0)/(V_0) = (4K)/(K+3) C_0`