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We have a capacitor let's say of capacitance C and is charged by Voltage say V. Then the voltage is disconnected and a dielectric of dielectric constant say k is inserted fully between the plates of parallel plate capacitor. We are asked to find the change in charge stored by the capacitor and change in voltage. Now what I am not getting is why does charge stored in capacitor remain constant. The surface charge density decreases due to polarisation of dielectric and so the net charge on the plates should decrease yet we are considering charge to be constant. Please correct me😶😶.enter image description hereenter image description here

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why does charge stored in capacitor remain constant.

Because you disconnected the voltage source. It's meant to be implied that the capacitor is disconnected from all external circuits. Therefore there's nowhere for the charge to go. And since charge is a conserved quantity, that means the charge on the capacitor plate must remain constant.

The surface charge density decreases due to polarisation of dielectric and so the net charge on the plates should decrease yet we are considering charge to be constant.

The charge associated with the polarization only compensates for some of the charge on the plate, it doesn't remove it. The charge associated with polarization is in the dielectric, and the charge on the plate is on the plate.

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  • $\begingroup$ Not related to your answer, but do you know why the yellow highlight feature has been replaced by a vertical bar? $\endgroup$
    – Bob D
    Mar 9, 2020 at 18:08
  • $\begingroup$ But then why does the surface charge density decreases if the charge on plate and on surface of dielectric are seperate? $\endgroup$ Mar 9, 2020 at 19:01
  • $\begingroup$ @Legend can you share the exact wording of where you read that the surface charge is decreased? Edit your question to include this information. $\endgroup$
    – The Photon
    Mar 9, 2020 at 19:10
  • $\begingroup$ @The Photon ...... Doesn't the paragraph implies that the surface charge density decreases from its initial value and dosen't that implies that charge decreases.... $\endgroup$ Mar 10, 2020 at 5:30
  • $\begingroup$ The net charge on the surface decreases. But actually what's going on is there is (for example) a positive charge on the metal surface and a partially compensating negative charge on the dielectric surface. They're near enough together that we can treat them as one surface charge, but really the charge on the plate hasn't changed. $\endgroup$
    – The Photon
    Mar 10, 2020 at 15:32

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