A vacuum parallel-plate capacitor has identical disk-shaped plates that are sepe

Question

A vacuum parallel-plate capacitor has identical disk-shaped plates that are seperated by 2.0mm. The area of each plate is 5.0cm^2. The capacitor is being charged by a constant chargig current of Ic = 1.8mA. At t=0, the charge on the plates is zero, Q0 = Q(0) = 0C.

A.) At time t=0.5 micro seconds, calculate: the charge Q(t) on the plates, the electric field E(t) between the plates, and the potential difference V(t) between the plates.

B.) Calculate dE/dt, the time rate of change of the electric field between the plates.

C.) Calculate the displacement current density Jd between the plates and the displacement current Id.

Explanation / Answer

a)

Q(t) =It =1.8*10^-3*0.5*10^-6

Q(t)=9*10^-10 C or 9uC

E(t) =q/eoA =9*10^-10/(8.85*10^-12)*(5*10^-4)

E(t) =203.4*10^3 V/m or 203.4 kV/m

V(t) =Ed =203.4*10^3*2*10^-3

V(t) =406.8 V

b)

dE/dt =I/eoA =1.8*10^-3/(8.85*10^-12)*(5*10^-4)

dE/dt =4.07*10^11 V/ms

c)

Jd =eo*(dE/dt)=8.85*10^-12*4.07*10^11

Jd=3.6 A/m^2

Jd =Id/A

Id=AJd =(5*10^-4)*3.6

Id =1.8*10^-3 A or 1.8 mA

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