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This question already has an answer here:

Kindly help me understand the difference between conduction current and displacement current. Quick searches explain that displacement current is associated with changing electric field, frankly speaking, it is too technical for me to understand.

Also, please relate it to why Capacitor allows AC and blocks DC. They say time varying voltage(AC) produces displacement current which can flow through the insulator while fixed voltage(DC) produces conduction current which cannot flow through the insulator. This is what made me put this question to you.

Thanks in advance.

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marked as duplicate by RedGrittyBrick, PeterJ, Matt Young, Anindo Ghosh, Dave Tweed Jan 20 '14 at 5:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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From a physics perspective, for a capacitor we have

$$Q = Cv$$

Where \$Q\$ is the amount of charge separated (\$Q\$ charge on one plate, \$-Q\$ charge on the other), \$C\$ is the capacitance and \$v\$ is the voltage across the capacitor.

Due to conservation of electric charge, if \$Q\$ is changing, there must be a current \$i\$ into one plate and out of the other thus

$$i = \frac{dQ}{dt} = C\frac{dv}{dt}$$

Note that when the voltage across a capacitor is constant, i.e., \$\frac{dv}{dt} = 0\$, the capacitor current is zero.

Also note that when the capacitor current is constant, then the rate of change of capacitor voltage is constant.

Now, with that review in mind, consider that the equations above do not imply that a capacitor "blocks DC".

Rather, they imply that, for a DC (constant) voltage, the capacitor current is zero.

And, for a DC (constant) current, the capacitor voltage steadily changes.

But, if the voltage is changing, there is a changing electric field and thus, a changing electric flux in the dielectric of the capacitor.

And, according to Maxwell's equations, a changing electric flux is a type of electric current and produces a magnetic field just as a conduction current does:

$$\nabla \times \vec H = \vec J + \frac{\partial \vec D}{\partial t} $$

where the first term on the right hand side is the conduction current density and the second term is the electric flux current density (displacement current density).

So, it is true that a time varying capacitor voltage is associated with a displacement current in the capacitor dielectric and, in fact, this must equal the conduction current into and out of the capacitor.

But, as I've pointed out, for a fixed capacitor voltage, there is no conduction or displacement current.

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