Figure 1

The dark regions are the plates of the capacitor and the lightly shaded region is the dielectric inserted between it. The capacitor is charged fully by an external battery and then battery is removed.

Why the electric field in the region between the dielectric and the plate is $E_0$? $E_0$ is the field without the dielectric, $E$ is the field in the dielectric. Shouldn't the field outside the dielectric change due to the induced charges due to polarization?

Figure 2

In this figure, net force on $q_1$ and $q_2$ increases due to induced charges in the dielectric. So, surely the magnitude of the field between $q_1$ and $\mathrm{A}$ must have increased. So, why does not electric field in a the first figure increase in a similar fashion?

  • $\begingroup$ Please define the problem fully. Is the capacitor held at constant voltage while the dielectric is inserted, or is it open circuit? $\endgroup$ – Rob Jeffries Jan 15 '17 at 18:15

Just look at the upper Gaussian surface. If the dielectric is inserted with the capacitor disconnected from any voltage source, then the charge on the plates remains unchanged.

If the charge inside the upper Gaussian surface is unchanged, then (using Gauss's law and the fact that the E-field is still perpendicular to the plates) the electric field $E_0$ also remains unchanged from its value before the dielectric was inserted.

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  • $\begingroup$ Yes, I get it. The capacitor plates and the dielectric layer, both are infinitely big. So, the field due to +ve polarization charges and -ve charges will be same and opposite. So, if the E-fields were not perpendicular, the field should change, shouldn't it? Like in the second figure. $\endgroup$ – Apoorv Potnis Jan 16 '17 at 2:15
  • $\begingroup$ Yes, the assumption is of an ideal capacitor. In practice, so long as the plate size is much greater than the plate separation, this will be ok, apart from the fields right at the edge of the plates. @ApoorvPotnis $\endgroup$ – Rob Jeffries Jan 16 '17 at 6:52

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