NEET-XII-Physics
40: Electromagnetic Waves
- #2A point charge is moving along a straight line with a constant velocity v. Consider a small area A perpendicular to the motion of the charge. Calculate the displacement current through the area when its distance from the charge is x. The value of x is not large, so that the electric field at any instant is essentially given by Coulomb’s law.
FigureAns : From Coulomb's law:
Electric field strength,
`` E\mathit{=}\frac{kq}{{x}^{\mathit{2}}}``
Electric flux,
`` {\varphi }_{\,\mathrm{\,E\,}}=EA``
`` {\varphi }_{\,\mathrm{\,E\,}}=\frac{kqA}{{x}^{2}}``
Displacement current = Id
`` {I}_{\,\mathrm{\,d\,}}=\left|{\in }_{0}\frac{d{\varphi }_{\,\mathrm{\,E\,}}}{dt}\right|``
`` {I}_{\,\mathrm{\,d\,}}=\left|{\in }_{0}\frac{d}{dt}\left(\frac{kqA}{{x}^{2}}\right)\right|``
`` {I}_{\,\mathrm{\,d\,}}=\left|{\in }_{0}kqA\frac{d}{dt}{x}^{-2}\right|``
`` {I}_{\,\mathrm{\,d\,}}=\left|{\mathit{\in }}_{\mathit{0}}\frac{\mathit{1}}{\mathit{4}\mathit{\pi }{\mathit{\in }}_{\mathit{0}}}\mathit{\times }\mathit{q}\mathit{\times }\mathit{A}\mathit{\times }\mathit{(}\mathit{-}\mathit{2}\mathit{)}{\mathit{x}}^{\mathit{-}\mathit{3}}\mathit{\times }\frac{\mathit{d}\mathit{x}}{\mathit{d}\mathit{t}}\right|``
`` {I}_{\,\mathrm{\,d\,}}=\left|\frac{qAv}{2\,\mathrm{\,\pi \,}{x}^{3}}\right|``
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