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Electrostatics: One of the job interview question:

Plate capacitor, with area of electrodes S, and distance between them d,has air as dielectric. Capacitor is connected to constant voltage source. Then, it is disconnected and dielectric is introduced between plates with relative permitivity Ɛr>1,

What happens to current between plates? does current drops (asuming to zero if there is no voltage), increases, or stays the same?

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    \$\begingroup\$ If the capacitor is disconnected from the rest of the circuit,why do you think there will be a current at all, regardless of what dielectric is used in the capacitor? \$\endgroup\$ – The Photon Dec 12 '16 at 15:51
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    \$\begingroup\$ Q=CV and Q remains fixed and ditto what the photon said. I guess you didn't get the job. \$\endgroup\$ – Andy aka Dec 12 '16 at 15:52
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    \$\begingroup\$ for 0==0 so "stays the same" is technically correct. \$\endgroup\$ – user3528438 Dec 12 '16 at 15:56
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    \$\begingroup\$ It's entirely possible that the original question was actually about the voltage between the plates, which is a perfectly legitimate question. @Andyaka's equation is the key here. What does inserting a dielectric change about it? \$\endgroup\$ – Dave Tweed Dec 12 '16 at 16:12
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    \$\begingroup\$ @DaveTweed, ok, ok, the microphone example might be wrong. However, the described effect of voltage change with variable capacitance can be seen in everyday life. When anyone pulls a plastic wrap from a roll, initially small voltages (due to initial charges from friction) gets vastly amplified when distance between the sheet and roll increases, and can get into kV range. This effect of charge conservation is responsible for most troubles with static electricity in plastic film production lines. \$\endgroup\$ – Ale..chenski Dec 12 '16 at 20:58
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It's not completely clear if you are talking about a floating capacitor with no external connections and a fixed charge Q, or if this cap is somehow being used in a circuit. Since you say "current" this becomes ambiguous because this implies the flow of charges.

From a electrostatics point of view, assume that the capacitor is not connected to anything but has a fixed initial charge Qo.

Q = CV so V = Q/C; and a voltage will be across the capacitor plates.

When you replace the dielectric, you will increase the capacitance. So the voltage across the plates will decrease (for fixed initial charge Qo).

But if you are asking how the charging current on such a capacitor would change for the same input voltage, then it would increase. This is because it now takes more charge to create a change in voltage on the plates (with the higher dielectric).

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