The electric field in the region between two parallel plates at time t2 is shown below- At t1-0- E-0- The displacement current through the shown surfaces is id for the short interval dt-t2-t1- Then -

We know that, displacement current, i_d=ϵ_0dϕ_Edt For a given short interval, nbsp; i_d ∝ dϕ_E From figure, for the short interval (dt=t_2 t_1), dϕ_E is ...

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Byju's Answer

Standard XII

Physics

Wave Equation of EM Waves

The electric ...

Question

The electric field in the region between two parallel plates at time $t_{2}$ is shown below. At $t_{1} = 0$ , $\to E = 0$ . The displacement current through the shown surfaces is $i_{d}$ for the short interval $(d t = t_{2} - t_{1})$ . Then -

i_{d_{1}} = i_{d_{2}} = i_{d_{3}} = 0

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i_{d_{1}} < i_{d_{2}} < i_{d_{3}}

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i_{d_{1}} > i_{d_{2}} > i_{d_{3}}

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i_{d_{1}} = i_{d_{2}} = i_{d_{3}} \neq 0

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Solution

The correct option is C $i_{d_{1}} > i_{d_{2}} > i_{d_{3}}$

We know that, displacement current,

$i_{d} = ϵ_{0} \frac{d ϕ_{E}}{d t}$

For a given short interval,

$i_{d} \propto d ϕ_{E}$

From figure, for the short interval $(d t = t_{2} - t_{1})$ ,

$d ϕ_{E}$ is maximum through surface $1$ because maximum number of electric field lines passes through it and minimum through surface $3$ because minimum number of electric field lines passes through it.

Therefore, $i_{d_{1}} > i_{d_{2}} > i_{d_{3}}$

Hence, option $(C)$ is the correct answer.

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The electric field in the region between two parallel plates at time t2 is shown below. At t1=0, →E=0. The displacement current through the shown surfaces is id for the short interval (dt=t2−t1). Then -

The electric field in the region between two parallel plates at time $t_{2}$ is shown below. At $t_{1} = 0$ , $\to E = 0$ . The displacement current through the shown surfaces is $i_{d}$ for the short interval $(d t = t_{2} - t_{1})$ . Then -