Hello! Please help in answering the following question: In a charged plate capacitor is connected t

Hello! Please help in answering the following question: In a charged plate capacitor is connected to a DC power source, place the material with a dielectric constant "epsilon". How does the electric field in the capacitor? I ask because apparently the correct answer, that does not change, in my opinion, is reduced as many times as is the dielectric hydrochloride constant. Thank you for odgoror and best regards! hydrochloride soncek_3 Posts: 23 Joined: 20.10.2006 20:57
In fact, it does not change because the capacitor is connected to a voltage hydrochloride source. If it was, then it is true, the field strength in the condenser decreased. I assume that the dielectric completely fills the space between the plates. Zdravapamet Posts: 2840 Joined: 08/16/2004 18:41
"Electric field" is precisely in this case, poor definition issues. Electric field strength (E) does not change (E = U / d). The density of the electric field (D) is to be amended hydrochloride (D = epsilon * E). Aniviller Posts: 7263 Joined: 15.11.2004 18:16
Zdravapamet itself you imagine matter so that the capacitor nabiješ and then disengage from the voltage source, but in the style of treatment: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/dielec.html Aniviller, "the electric field "is hydrochloride probably meant to be a strength el. Field E. shrink Posts: 9731 Joined: 9/4/2004 17:45
Hold shrink. However, if the capacitor is constantly connected to the voltage hydrochloride V, then the sigma-modified, in the SCU wherein the surface of the panel, and C = \ varepsilon C_0 new capacity after insertion of the dielectric (the old C_0). In the end, you get E = E_0. Zdravapamet Posts: 2840 Joined: 08/16/2004 18:41
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