Question
Prove that $I _{ D }=\epsilon_0 \frac{d}{d t}\left(\phi_{ E }\right)$, when symbols have their usual meanings.
✓
Answer
We know that :
$ I_{D} =I$
Or $ I =\frac{d q}{d t} $
By Gauss's law,
$ E=\frac{q}{\epsilon_0 A}$
Or $q=E \in_0 A $
On putting values in eqn. $(2)$
$ I=\frac{d}{d t}\left(E \epsilon_0 A\right) $
From equation $(1),$
$\therefore I_{D}=\epsilon_0 A \frac{d}{d t}(E)$
$=\epsilon_0 A \frac{d}{d t}\left(\frac{\phi_{E}}{A}\right) \because \phi_{E}=EA$
$=\frac{\epsilon_0 A}{A} \frac{d \phi_{E}}{d t}$
or,
$I_{D}=\epsilon_0 \frac{d}{d t}\left(\phi_{E}\right) $
$ I_{D} =I$
Or $ I =\frac{d q}{d t} $
By Gauss's law,
$ E=\frac{q}{\epsilon_0 A}$
Or $q=E \in_0 A $
On putting values in eqn. $(2)$
$ I=\frac{d}{d t}\left(E \epsilon_0 A\right) $
From equation $(1),$
$\therefore I_{D}=\epsilon_0 A \frac{d}{d t}(E)$
$=\epsilon_0 A \frac{d}{d t}\left(\frac{\phi_{E}}{A}\right) \because \phi_{E}=EA$
$=\frac{\epsilon_0 A}{A} \frac{d \phi_{E}}{d t}$
or,
$I_{D}=\epsilon_0 \frac{d}{d t}\left(\phi_{E}\right) $
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