Is displacement current real? Do charges cross a capacitor gap? Showing I_d = I for a charging capacitor

7 answers & replies

• [+][–]Arjun Nair12thSample· 29d ago

  The calculation is short. For a parallel plate capacitor of plate area $A$ carrying charge $Q$:

  $E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A},\qquad \Phi_E = EA = \frac{Q}{\varepsilon_0}$

  So

  $I_d = \varepsilon_0\frac{d\Phi_E}{dt} = \frac{dQ}{dt} = I$

  The displacement current in the gap exactly equals the conduction current in the wire. So with the Maxwell term, the right side of the law is $\mu_0(I_c + I_d)$, and both surfaces now give the same answer, because on the bulged surface $I_c = 0$ but $I_d = I$.

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• [+][–]Harsh Gupta12thSample· 29d ago

  I'd say the displacement current is the electrons jumping across the gap, like the current "hops" the vacuum. That's why it's called current. 😅

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• [+][–]Siddharth RaoDropperSample· 29d ago

  Harsh, no. Nothing crosses the gap. Between the plates there is vacuum (or a dielectric) and no charge flows through it. Displacement current is not a flow of charge; it's the term produced by a changing electric field, $\varepsilon_0\, d\Phi_E/dt$. It's called a current only because it has the same units and appears in the equation of the magnetic field in the same position as a real current.

  And it's not a mathematical trick either. A changing E-field really does produce a magnetic field. That symmetry with Faraday's law (changing B gives E) is what lets an EM wave sustain itself in empty space. Maxwell predicted the waves, Hertz found them.

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  • [+][–]Zoya Khan12thSample· 28d agoAuthor

    Thank you, that makes it feel like physics and not magic. "Changing E gives B, changing B gives E" is something I'll write in my diary, honestly. And thank you for correcting Harsh gently 🙂

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  • [+][–]Harsh Gupta12thSample· 28d ago

    fine fine, will keep to jokes and leave Maxwell alone 😭 thanks Siddharth

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• [+][–]Tushar PawarDropperSample· 28d ago

  Fully charged capacitor: $Q$ is constant, so $E$ is constant, $d\Phi_E/dt = 0$ and so $I_d = 0$. That's consistent with $I = 0$ in the wire. So at every instant the current in the wire and the displacement current in the gap match, which makes the "total current" continuous around the circuit. Good way to remember: total current $I = I_c + I_d$ has no break anywhere.

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• [+][–]Aman YadavDropperSample· 28d ago

  NEET point. In a good conductor, $I_c$ is huge and $I_d$ is negligible. In the gap of a capacitor (vacuum), $I_c = 0$ and only $I_d$ exists. In the same circuit, both together make it continuous.

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