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A dielectric of permittivity 3.5×1011F/m3.5 \times 10^{-11} \mathrm{F} / \mathrm{m} completely fills the volume between two capacitor plates. For t > 0 the electric flux through the dielectric is (8.0×103m/s3)t3\left(8.0 \times 10^{3} \vee \cdot \mathrm{m} / \mathrm{s}^{3}\right) t^{3}. The dielectric is ideal and nonmagnetic; the conduction current in the dielectric is zero. At what time does the displacement current in the dielectric equal 21μA21 \mu \mathrm{A}?

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We have a parallel plate capacitor, which is filled with a dielectric material which has a permittivity of ε=3.5×1011\varepsilon=3.5 \times 10^{-11} F/m. The electric flux through the dielectric material is given by,

ΦE=(8.0×103 Vm/s3)t3\begin{align}\Phi_{E}=\left(8.0 \times 10^{3} \mathrm{~V \cdot m /s ^{3}}\right) t^{3} \end{align}

the conduction current in the dielectric is zero. We need to find the time tt when the displacement current in the dielectric equal iD=21×106i_{\mathrm{D}}=21 \times 10^{-6} A. The displacement current is given by,

iD=εdΦEdti_{\mathrm{D}}=\varepsilon \frac{d \Phi_{E}}{d t}

substitute from (1) we get,

iD=(3.5×1011 F/m)(8.0×103 Vm/s3)ddt(t3)=3(3.5×1011 F/m)(8.0×103 Vm/s3)t2=(8.4×107 A/s2)t2\begin{align*} i_{\mathrm{D}}&=\left( 3.5 \times 10^{-11} \mathrm{~F/m}\right) \left(8.0 \times 10^{3} \mathrm{~V \cdot m /s ^{3}}\right) \frac{d }{d t}\left(t^{3} \right)\\ &=3\left( 3.5 \times 10^{-11} \mathrm{~F/m}\right) \left(8.0 \times 10^{3} \mathrm{~V \cdot m /s ^{3}}\right) t^{2}\\ &=\left( 8.4 \times 10^{-7} \mathrm{~A/s^{2}}\right) t^{2} \end{align*}

solve for tt we get,

t=iD8.4×107 A/s2t=\sqrt{\frac{ i_{\mathrm{D}}}{8.4 \times 10^{-7} \mathrm{~A/s^{2}}}}

substitute with iD=21×106i_{\mathrm{D}}=21 \times 10^{-6} A, we get,

t=21×106 A8.4×107 A/s2=5.0 st=\sqrt{\frac{21 \times 10^{-6} \mathrm{~A}}{8.4 \times 10^{-7} \mathrm{~A/s^{2}}}}=5.0 \mathrm{~s}

t=5.0 s\boxed{t=5.0 \mathrm{~s}}

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