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Show that displacement current is equal to conduction current during chargeing of a capacitor. |
Answer» Solution :LET at an instant magnitude of charge on the plates of capacitor be Q. Area of each plate is A. Electric field between the plates of capacitor E = `(Q)/(A epsilon_(0))`........ (1) Flux of this passing through the suface between the plaates is ` phi_(E ) = E xx A = (Q)/(A epsilon_(0)) xx A ` `phi_(E ) = (Q)/(epsilon_(0))` ......... (ii) Displacement current `i_(d)` is `i_(d) = epsilon_(0) xx (d phi_(epsilon))/(dt ) = epsilon_(0) (d)/(dt) ((Q)/(epsilon_(0))) = epsilon_(0) xx (1dQ)/(epsilon_(0)dt)` `I_(d) = (DQ)/(dt)` ....... (iii) `(dQ)/(dt)` is the rate at which charge is reaching to POSITIVE plate of capacitor through conducting wire therefore `I_( c) = (dQ)/(dt)` ... (iv) From equations (iii) and (iv) `I_(d) = I_(c )` |
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