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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 dielectric 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. Can someone 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. 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. |
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One could say that the effective charge density on the plate touching the dielectric has decreased. Following the answer above, one imagines the dielectric as a collection of polarized molecules. While the charge density in the bulk of the dielectric is zero everywhere, there is non-zero charge on its surfaces. As a result, the surface touching the capacitor can be thought of as decreasing the charge of the capacitor. To be more specific, the effective charge on the plate touching the dielectric would go from $\sigma$ to $ \frac{\sigma}{\epsilon}$, where $\epsilon$ is the (unit-less) dielectric constant of the dielectric. |
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