Question: Q. 11. A capacitor made of two parallel plates each of plate area $A$ and separation $d$, is being charged by an external $a c$ source. Show that the displacement current inside the capacitor is the same as the current charging the capacitor.
U] [O.D. I, II, III 2013]
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Solution:
Ans.
$$ \begin{aligned} i_{d} & =\frac{d \phi_{E}}{d t} \ & =\varepsilon_{0} \frac{d(E A)}{d t} \end{aligned} $$
Electric field between the plates of a capacitor
$$ \begin{aligned} E & =\frac{\sigma}{\varepsilon_{0}}=\frac{q}{\varepsilon_{0} A} \ \therefore \quad i_{d} & =\frac{\varepsilon_{0} A}{\varepsilon_{0} A} \frac{d q}{d t} \ i_{d} & =\frac{d q}{d t}=i \end{aligned} $$
Where $i$ is the charging current. This proves that the displacement current inside the capacitor is the same as the current charging the capacitor.
