## Displacement Current and Ampere's Law

Q) A circular parallel plate capacitor is charging. What is the magnetic both everywhere between the plates, either within or outside the radius of the plates?The radius of the plates is R and the rate of change of the field is a constant given as (DE/Dt). Show that:

- i) when r

_{0}(DE/Dt)r

ii) when r>R B(r) = (μ_{0}/2)ε_{0}(DE/Dt)R^{2}/r

- I

_{D}=ε

_{0}(dΦ

_{E}/dt)]

I_{D}=ε_{0}πr^{2}(DE/Dt)

- B(r)*2πr = μ

_{0}I

_{D}

B(r) = (μ_{0}/2πr)ε_{0}(DE/Dt)$#960;r^{2}

B(r) = (μ_{0}/2)ε_{0}(DE/Dt)r

ii) Outside the capacitor we proceed identically except, the displacement current through the circle is constant, because the electric field is (to a good approximation) only changing in the space between the plates:

- I

_{D}=ε

_{0}πR

^{2}(DE/Dt)

B(r) = (μ_{0}/2πr)ε_{0}(DE/Dt)$#960;R^{2}

B(r) = (μ_{0}/2)ε_{0}(DE/Dt)R^{2}/r

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