Question about capacitors with dielectrics

AI Thread Summary
When a dielectric is introduced to one plate of a charged capacitor, the net charge density on both plates remains the same, as charge is conserved. The dielectric polarizes, effectively increasing the capacitor's capacitance, akin to reducing the distance between the plates. Consequently, the voltage across the capacitor decreases, leading to a drop in stored energy. The work done in inserting the dielectric corresponds to the energy change, similar to the work done if the plates were allowed to move closer together. Overall, the introduction of the dielectric alters the electric field without changing the total charge on the plates.
anon6912
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Say i have a capacitor and i supply it with a voltage V and after that it has charge density sigmainitial on each plate.

Now i remove the voltage source and then introduce a dilectric with area A that it equal to the area of one of the plates in the capacitor to the capacitor.

But this dielectric is only touching one plate.

My question is: is the net charge density on each plate(sigmainitial - sigmainduced) the same in both plates or different?

please explain...
 
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I think the answer is that the same charge is there but the molecules in the dielectric have all polarised a bit. This has increased the capacity ov the capacitor. Equivalent to moving the plates closer together. Now
Q=CV
so the voltage will have dropped. As for the energy difference, some work has been done in moving the dielectric into place- nothing violated either way, I think.
 
Hi Anon-
Sophie is correct. Charge is conserved, and the charge distribution is unchanged. The stored energy is

E = Q2/2C

So when the dielectric is pulled (not pushed) in, the stored energy drops, and work is done.

Bob S
 
Yes- the same work that the plates would do if they were allowed to come together and have the equivalent capacitance.
 
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