NEET-XII-Physics
40: Electromagnetic Waves
- #3A parallel-plate capacitor of plate-area A and plate separation d is joined to a battery of emf ε and internal resistance R at t = 0. Consider a plane surface of area A/2, parallel to the plates and situated symmetrically between them. Find the displacement current through this surface as a function of time.Ans : Electric field strength for a parallel plate capacitor,
`` E=\frac{Q}{{\in }_{0}A}``
Electric flux linked with the area,
`` {\varphi }_{E}\mathit{=}EA\mathit{=}\frac{Q}{{\mathit{\in }}_{\mathit{0}}A}\mathit{\times }\frac{A}{\mathit{2}}=\frac{Q}{2{\in }_{0}}``
Displacement current,
`` {I}_{\,\mathrm{\,d\,}}={\in }_{0}\frac{d{\varphi }_{\,\mathrm{\,E\,}}}{dt}={\in }_{0}\frac{d}{dt}\left(\frac{Q}{{\mathit{\in }}_{\mathit{0}}\mathit{2}}\right)``
`` {I}_{\,\mathrm{\,d\,}}=\frac{1}{2}\left(\frac{d\,\mathrm{\,Q\,}}{dt}\right)...\left(\,\mathrm{\,i\,}\right)``
Charge on the capacitor as a function of time during charging,
`` Q=\epsilon C\left[1-{e}^{\mathit{-}t\mathit{/}RC}\right]``
Putting this in equation (i), we get:
`` {I}_{\,\mathrm{\,d\,}}=\frac{1}{2}\epsilon C\frac{d}{dt}\left(1\mathit{-}{e}^{\mathit{-}t\mathit{/}RC}\right)``
`` {I}_{\,\mathrm{\,d\,}}=\frac{1}{2}\epsilon C\left(\mathit{-}{e}^{\mathit{-}t\mathit{/}RC}\right)\times \left(\mathit{-}\frac{1}{RC}\right)``
`` \,\mathrm{\,C\,}=\frac{A{\mathit{\in }}_{\mathit{0}}}{d}``
`` \Rightarrow {I}_{\,\mathrm{\,d\,}}=\frac{\epsilon }{2R}\times {e}^{\mathit{-}\frac{td}{{\epsilon }_{\mathit{0}}AR}}``
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