- #36
jtbell
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Perhaps the attached diagram will help. It shows the electric field before and after inserting a dielectric into a capacitor, and how the potential difference across the capacitor is calculated in both cases.
The arrows are electric field lines, not vectors. The strength of the electric field is proportional to the number of field lines, not to their length. The original electric field is [itex]E_0[/itex]. When the dielectric is inserted, the polarization induces an effective electric charge on the surfaces of the dielectric, which produces an additional induced electric field [itex]E_i[/itex] inside the dielectric. For this example I assume that [itex]E_i = E_0 / 2[/itex]. The net electric field inside the dielectric is [itex]E_0 - E_i[/itex].
The + and - represent the charge on the capacitor and the induced charge on the surfaces of the dielectric.
The final result is that the potential difference [itex]\Delta V[/itex] across the capacitor is reduced (in this example) from [itex]E_0 d[/itex] to [itex]E_0 d - 0.6 E_i d[/itex]. The 0.6 is a purely geometrical factor. It appears because I chose the dielectric to fill 0.6 of the space between the capacitor plates. If you expand the dielectric to fill more and more of the space, this factor increases accordingly, until it reaches 1.0 when the dielectric fills the space completely.
The arrows are electric field lines, not vectors. The strength of the electric field is proportional to the number of field lines, not to their length. The original electric field is [itex]E_0[/itex]. When the dielectric is inserted, the polarization induces an effective electric charge on the surfaces of the dielectric, which produces an additional induced electric field [itex]E_i[/itex] inside the dielectric. For this example I assume that [itex]E_i = E_0 / 2[/itex]. The net electric field inside the dielectric is [itex]E_0 - E_i[/itex].
The + and - represent the charge on the capacitor and the induced charge on the surfaces of the dielectric.
The final result is that the potential difference [itex]\Delta V[/itex] across the capacitor is reduced (in this example) from [itex]E_0 d[/itex] to [itex]E_0 d - 0.6 E_i d[/itex]. The 0.6 is a purely geometrical factor. It appears because I chose the dielectric to fill 0.6 of the space between the capacitor plates. If you expand the dielectric to fill more and more of the space, this factor increases accordingly, until it reaches 1.0 when the dielectric fills the space completely.