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A parallel plate capacitor with plate area A and plate separation d, is charged by a constant current I. Consider a plane surface of area `A//2` parallel to the plates and situated symmetrically between the plates. Determine the displacement current through this area.A. IB. `I/2`C. `2I`D. `I/3` |
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Answer» Correct Answer - B Let q be the charge on the capacitor plates at any instant and `sigma` be the surface charge density. Then the electric field between the plates, `E = (sigma)/(epsilon_0) = (q)/(A epsilon_(0)) "where" sigma = q/A` `:.` Flux through the area `A/2 = E xx Area` `phi_(E) = q/(A epsilon_(0)) xx A/2 = q/(2 epsilon_0)` `:.` The displacement current `=I_(D) = epsilon_(0) (d phi_(E))/(dt) = epsilon_(0)1/(2 epsilon_(0)) xx (dq)/(dt)` `:. I_(D) = I/2` where `I = (dq)/(dt)` = the rate at which the charge flows through the conducting wire. |
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