Laws of Electromagnetism HW question?

Question

A 0.109-A current is charging a capacitor that has square plates 4.50 cm on each side. The plate separation is 4.00 mm.

(a) Find the time rate of change of electric flux between the plates.

___ V·m/s

(b) Find the displacement current between the plates.

___ A

Answer 1

♠ thus capacitance is C=ε0*A/d, where ε0=8.854e-12, area A=4.5cm*4.5cm =(4.5e-2)^2 m^2, d=4mm=4e-3m; so C=4.5e-12F;

♣ a constant current means accumulation of charge on capacitor with rate dq/dt=I=0.109A;

sorry, I continue to-morrow!