## Related questions with answers

A dielectric of permittivity $3.5 \times 10^{-11} \mathrm{F} / \mathrm{m}$ completely fills the volume between two capacitor plates. For t > 0 the electric flux through the dielectric is $\left(8.0 \times 10^{3} \vee \cdot \mathrm{m} / \mathrm{s}^{3}\right) t^{3}$. 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 $21 \mu \mathrm{A}$?

Solution

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

$\begin{align}\Phi_{E}=\left(8.0 \times 10^{3} \mathrm{~V \cdot m /s ^{3}}\right) t^{3} \end{align}$

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

$i_{\mathrm{D}}=\varepsilon \frac{d \Phi_{E}}{d t}$

substitute from (1) we get,

$\begin{align*} i_{\mathrm{D}}&=\left( 3.5 \times 10^{-11} \mathrm{~F/m}\right) \left(8.0 \times 10^{3} \mathrm{~V \cdot m /s ^{3}}\right) \frac{d }{d t}\left(t^{3} \right)\\ &=3\left( 3.5 \times 10^{-11} \mathrm{~F/m}\right) \left(8.0 \times 10^{3} \mathrm{~V \cdot m /s ^{3}}\right) t^{2}\\ &=\left( 8.4 \times 10^{-7} \mathrm{~A/s^{2}}\right) t^{2} \end{align*}$

solve for $t$ we get,

$t=\sqrt{\frac{ i_{\mathrm{D}}}{8.4 \times 10^{-7} \mathrm{~A/s^{2}}}}$

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

$t=\sqrt{\frac{21 \times 10^{-6} \mathrm{~A}}{8.4 \times 10^{-7} \mathrm{~A/s^{2}}}}=5.0 \mathrm{~s}$

$\boxed{t=5.0 \mathrm{~s}}$

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