Question

show that during charging of parallel plate capacitor, the rate of change of charge on each plate is equal to eo times the rate change of electric flux.NAME THE CURRENT.

Answer 1

This relates to Gauss law of electrostatics.


We have the Gauss Law as : (we  take it without proof)  Proof of this requires knowledge of Direc Delta function and Divergence theorem.

$Electric\ Flux =\Phi = \int\limits^{}_{} {\vec{E}.} \, \vec{ds} =\frac{q}{\epsilon_0}$
           where q = charge enclosed in a Closed Gaussian surface

   E is perpendicular to the plates and is uniform inside the gap between the plates.  This is because the gap between the plates d is << A, area of the two plates.  We take a Gaussian surface with area of cross section same as A, and enclosing one of the plates of the capacitor.    E and dS are both parallel and hence,

                   Ф = E . A = Q/ε₀

     Hence,  I = current = dQ/dt = d(ε₀ Φ)/ dt = ε₀ dΦ/dt

===  Additional information related to Capacitor.
Capacitance is defined as capacity to hold electric charge:
      C  = Q / V 
   V = potential difference across plates,  Q = Charge on each plate
   Since  the distance between plates d << A, area of plates, the electric field E between the plates is uniform and E = V / d

       Hence Q = ε₀ E  A
       C = Q / V =  ε ₀ E A / (E d) = ε₀ A / d