# 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.

1
by jiya007 17.09.2015

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flux.NAME THE CURRENT.

1
by jiya007 17.09.2015

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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.

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

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

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