Question

A parallel plate capacitor made of two circular plates each of radius 10 cm and separated by 4.0mm. The capacitor is being charged by an external source. The charging current is constant and equal to 0.10A. Calculate (i) the capacitance (ii) the rate of change of potential difference between the plates (iii) the displacement current.

Answer 1

Here, `R=10cm=0.10m,`
`d-4.0mm=4.0xx10^-3m`
Charging current `I=0.10A`
Plate area of capacitor, `A=piR^2=pixx(0.1)^2m^2`
(i) Capacitance,
`C=(in_0A)/d=((8.85xx10^-12)xxpi(0.1)^2)/(4.0xx10^-3)`
`=69.5xx10^-12F=69.5pF`
(ii) Charge on capacitor at an instant is
`q=CV`
Current, `I=(dq)/(dt)=d/(dt) (CV)=C (dV)/(dt)`
Rate of change of potential difference between
the plate is
`(dV)/(dt)=I/C=0.10/(69.5xx10^-12)=1.44xx10^9Vs^-1`
(iii) Displacement current=conduction current
`:. I_D=I=0.10A`