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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



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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

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what is the name of this type of current..???
charging current or discharging current. It is perhaps also called as Transient response / transient current... After the capacitor is fully charged, the capacitor is open circuit.
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