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