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A parallel plate capacitor of area of 50 cm^2 and plate separation 3.0mm is charged initially to 80muc. Due to a radioactive source nearby, the medium between the plates gets lightly conducting and the plate loses the charge initially at the rate of 1.5xx10^-8Cs^-1 (i) What is the megnitude and direction of displacement current ? (ii) What is the magnetic field between the plates? |
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Answer» SOLUTION :Conduction current within the plates is from the positive plate to negative plate of parallel plate capacitor. (i) Displacement current, `I_D=in_0 (dphi_E)/(dt)=in_0d/(dt) (EA)=in_0A(dE)/(dt)` `=in_0A d/(dt)(q/(in_0A))=in_0A1/(in_0A) (dq)/(dt)` or `I_D=(dq)/(dt)=1.5xx10^-18A` Since charge is DECREASING with time, so `(dq)/(dt)` and hence `(dE)/(dt)lt0`. It shows that the direction of `I_D` is opposite to that of electric field and hence opposite to the conduction current. But the magnitude of displacement current is same as that of conduction current. The net current between the plates is zero. So total current, `I'=I+I_D=0` (ii) USING Ampere's LAW, `ointvecB.vec(dl)=mu_0I'=0` Hence magnetic field within the plates is zero. |
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