What is a capacitor ? Define its capacitance. Explain the units of capacitance.

Explanation

Suppose a sinusoidally alternating emf is applied across the plates of capacitor of capacitance C. The instantaneous value of the emf is ltbr. `E=E_(0) sin omega t " " ` …(1)
where `E_(0)` is the peack emf and frequency, `f=omega//2pi.`
As the p.d. between the plates alternates, charge continually flows onto and off the plates. This means an alternating current exists in the connecting wires although there is no conducting path by which a direct current could pass.

The instantaneous charge q on the capacitor plates is
`q= CV=CE_(0) sin omega t " " `...(2)
Then, the current,
`I= (dq)/(dt) =omega CE_(0) cos omega t`
`=omega CE_(0) sin (omega t +(pi)/(2)) " " ` ...(3)
The experession shows that the current leads `90^(@)` on the emf (the emf lags behind teh current by `90^(@)`). This is shown in the phasor diagram, Fig. (b). Equation (3) also shows that the current is proportional to both the frequency and the capacitance.
The peak current is
`I_(0) =omega CE_(0) " " ` ...(4)
`therefore (E_(0))/(I_(0))=(E_("rms"))/(I_("rms"))=(1)/(omega C) " " ` ....(5)
since `E_("rms") =E_(0)//sqrt(2) and I_("rms")=I_(0)//sqrt(2). `
The quantity `(1)/(omega C)` is known as the capacitive reactance of the capacitor as it represents the resistance offered by the capacitor to the alternating current. It has the same dimensions as resistance. It is denoted by `X_(C)`.
`X_(C)=(1)/(omega C)=(1)/(2pi f C) " " ` ...(6)
The capacitive reactance is defined as the ratio of the rms value of the p.d. across the capacitor to the rms current through it.
SI unit : The ohm `(Omega)`.