Question: Q. 14. Considering the case of a parallel plate capacitor being charged, show how one is required to generalize Ampere’s circuital law to include the term due to displacement current.
U [O.D. I, II, III 2014]
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Solution:
Ans. Ampere circuital law,
$$ \oint \vec{B} \cdot \overrightarrow{d l}=\mu_{0} I $$
between the plates of capacitor,
$1 / 2$
Which is impossible, as there is magnetic field.
Modified ampere circuital law,
$$ \oint \vec{B} \cdot \overrightarrow{d l}=\mu_{0}\left(I_{C}+I_{D}\right) $$
where, $I_{D}=$ Displacement current $=\varepsilon_{0} \frac{d \phi_{E}}{d t}$.
