Question 1: EM waves

Question

A parallel plate capacitor with plate area $A$ and separation between plates $d$ is charged with constant current $I$. Consider a plane surface of area $\frac{5 A}{6}$ parallel to the plates and drawn between the plates. The displacement current through area is

Accepted Answer

The charge on the capacitor plates at time $t$ is given by $q=I t$
Electric field $=\varepsilon=\frac{\sigma}{\varepsilon_0}=\frac{q}{A \varepsilon_0}=\frac{I t}{A \varepsilon_0}$
Flux through area $\frac{5 \mathrm{~A}}{6}$ is $\phi_{\varepsilon}=\left(\frac{5 \mathrm{~A}}{6}ight) \varepsilon$
$$ \Rightarrow \quad \phi_{\varepsilon}=\frac{5 \mathrm{~A}}{6} 	imes \frac{I t}{A \varepsilon_0} \Rightarrow \phi_{\varepsilon}=\frac{5 I t}{6 \varepsilon_0} $$
Now, displacement current $=I_d=\varepsilon_0 \frac{d \phi_e}{d t}$
ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ ÃÂ $$ \begin{aligned} & I_d=\varepsilon_0 \frac{d}{d t}\left[\frac{5 I t}{6 \varepsilon_0}ight] \ & I_d=\frac{5}{6} I \end{aligned} $$

Answer

$\frac{1}{6} I$

Answer

$\frac{5}{6} I$

Answer

$6 I$

Answer

$\frac{6}{5} I$

EM waves

A parallel plate capacitor with plate area A and separation between plates d is charged with constant current I. Consider a plane surface of area \frac{5 A}{6} parallel to the plates and drawn between the plates. The displacement current through area is

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