A parallel plate capacitor of area `50cm^(2)` and plate separation `3.0 mm` is charged initially to `80 mu C`. Due to radio-active source nearby , the medium between the plates gets slightly conducting and the plate loses charge the rate of `1.5 xx 10^(-8) "Cs"^(-1)`. what is the magnitude and direction of displacement current ? What is the magnetic field between the plates

Explanation

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.