Capacitors and Capacitance
Definition
A capacitor is an electronic component that stores electrical energy in an electric field. It consists of two conductive plates (two conductors) separated by a dielectric material. These conductors carry charges `Q_1` and `Q_2`, creating potentials `V_1` and `V_2`.
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| capacitor |
In general, two conductors have charges Q and – Q, with potential difference `V\ =\ V_1\ -\ V_2` between them. Though a capacitor has charges, its total charge is zero because the charges on the conductors are equal and opposite.
When charge Q is given to an isolated conductor of any shape or size, It develops a potential V such that `V\ \propto\ Q `. The ratio Q/V is a constant which is defined as the capacitance of the conductor.
`Q\propto\ V`
`Q\ =C\ V`
`C\ =\ \frac{Q}{V}`
Where Q is the charge stored and V is the applied voltage.
Capacitance C depends on the shape and size of the conductor, the presence of other conductors, and the surrounding medium. SI unit of capacitance is coulomb/volt (farad). Farad (F) is a very large unit.
The Parallel Plate Capacitor
Definition of Capacitance
Capacitance is a measure of a capacitor’s ability to store charge per unit voltage.
The formula of Capacitance
`C\ =\ \frac{Q}{V}`
Where Q is the charge stored on the capacitor, and V is the voltage (potential difference) across the capacitor.
The Parallel Plate Capacitor
A parallel plate capacitor consists of two conductive plates of area A and the separation between the plates is d with a dielectric material (often air or vacuum) between the plates.
Fringing Effect
If the plates are large compared to the separation between them, then the electric field between them is uniform and perpendicular to the plates except for a region near the edge. It is called the fringing effect.
Electric Field
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| The parallel plate capacitor |
The electric field at the outer region I (above the plate 1)
`E\ =\ \frac{\sigma}{2\ \varepsilon_0}\ -\ \frac{\sigma}{2\ \varepsilon_0}\ `
`E\ =\ 0\ `
The electric field at the outer region II (below the plate 2)
`E\ =\ \frac{\sigma}{2\ \varepsilon_0}\ -\ \frac{\sigma}{2\ \varepsilon_0}\ `
`E = 0`
The electric field between plates
In the inner region between the plates 1 and 2, the electric field due to the two charged plates add up, giving
`E\ =\ E_+\ +\ E –`
`E\ =\ \frac{\sigma}{2\ \varepsilon_0}\ \ +\ \frac{\sigma}{2\ \varepsilon_0}\ `
`E\ =\ \frac{2\ \sigma}{2\ \varepsilon_0}\ \ `
`E\ =\ \frac{\ \sigma}{\ \varepsilon_0}\ \ `
`E\ =\ \frac{\ Q}{\ \varepsilon_0\ A}\ \ `
The direction of the electric field is from the positive to the negative plate and it is uniform. The field lines bend outward at the edges – an effect called ‘fringing of the field’.
`\frac{V}{d}= \frac{ Q}{ \varepsilon_0\ A}` `{\because E = \frac{V}{d}}`
`\frac{V}{Q}=\ \frac{\ d}{\ \varepsilon_0\ A}\ \ `
`\frac{Q}{V}= \frac{ \varepsilon_0 A} { d}`
`C\ =\frac{\ \ \varepsilon_0\ A}{d}`
The formula for the Capacitance of a Parallel Plate Capacitor
Thus, the formula for the capacitance of a parallel plate capacitor is
`C\ =\frac{\ \ \varepsilon_0\ A}{d}`
Where
C = Capacitance
`\varepsilon_0\ `= permittivity of free space (approximately `8.85 \times\ {10}^{- 12}` F/m)
A = Area of each plate,
D = distance between the plates.
Effect of Dielectric on Capacitance
When there is a vacuum between the plates,
Electric Field `E_0 =\frac{ \sigma}{ \varepsilon_0}`
Potential difference `V_0\ =\ E_0\ d`
Capacitance `C_0\ =\ \frac{Q}{V_0}\ =\ \frac{\ \ \varepsilon_0\ A}{d}`
A dielectric is inserted between the plates fully occupying the intervening region. The dielectric is polarised by the field and its effect is equivalent to two charged sheets (at the surface of the dielectric normal to the field) with surface charge densities `\sigma_p` and `-\ \sigma_p`. The electric field in the dielectric is
`E\ =\ \frac{\sigma_p\ -\ \sigma_p}{\varepsilon_0}`
Potential differences across the plates are
`V\ =\ E\ d\ =\ \frac{\sigma_p\ -\ \sigma_p}{\varepsilon_0}\ \ d\ `
For linear dielectric, we expect `\sigma_p` to be proportional to `E_0`, i.e., to `\sigma`.
Thus, `\sigma_p\ -\ \sigma_p` is proportional to \sigma and we can write
`\sigma_p\ -\ \sigma_p\ =\ \frac{\sigma}{K}`
Where K is a constant characteristic of the dielectric. Clearly, K>1. We then have
`V\ =\ E\ d\ =\ \frac{\sigma_p\ -\ \sigma_p}{\varepsilon_0}\ \ d`
`V\ =\ E\ d\ =\ \frac{\sigma}{\varepsilon_0\ K}\ d`
`V=\ \frac{\sigma}{\varepsilon_0\ K}\ d`
`V=\ \frac{Q}{A\ \varepsilon_0\ K}\ d` `{\because\ \sigma\ =\ \frac{Q}{A}\ }`
The product of `\ \varepsilon_0\ K` is called the permittivity of the medium and is denoted by `\varepsilon`
`\varepsilon\ =\ \varepsilon_0\ K`
For vacuum K = 1 and `\varepsilon\ =\ \varepsilon_0 and \ \varepsilon_0` is called the permittivity of the vacuum.
The dimensionless ratio
`K\ =\ \frac{\varepsilon}{\varepsilon_0\ }`
Is called the dielectric constant of the substance. It is clear that K>1
`K\ =\ \frac{C}{\ C_0\ }`
`C_{medium}\ =K\ \ C_0`
Due to factor K>1, the Capacitance in the medium is greater than the capacitance in the vacuum.
`C_{medium}\ >{\ C}_0`
(1) When a dielectric slab of thickness t is inserted between the plates, then capacitance,
`C\ =\ \frac{\ \ \varepsilon_0\ A}{d\ -\ t\ +\ \frac{t}{K}}`
(2) If a metallic slab (K\ =\ \infty) of thickness t is placed between the plates of capacitor, then capacitance,
`C\ =\ \frac{\ \ \varepsilon_0\ A}{d\ -\ t\ +\ \frac{t}{K}}`
(3) If several slabs of dielectric constants `K_1,K_2,K_3,\ .......,K_n ` and respective thickness `t_1,t_2,t_3,\ .......,t_n` are placed in between the plates of a capacitor, the capacitance
`C\ =\ \frac{\ \ \varepsilon_0\ A}{d\ -\ (t_1+t_2+t_3+\ .......+t_n)\ +\ \frac{t_1}{K_1}\ +\ \frac{t_2}{K_2}\ +\ \frac{t_3}{K_3}\ +\ .......+\frac{t_n}{K_n}}`
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