In Lab 3, we measured the dielectric constant of glass using a parallel plate capacitor. Suppose thata piece of flat glass panel with a linear dielectric constant ɛ, = 2.5 is inserted between the platesof such a capacitor. The thickness of the glass is d = 5mm. The glass is in contact with both plates.One of the plates of the capacitor is kept at +3V, and the other plate is grounded. Treat the platesas infinitely large and thus the electric field is uniform between the plates.a) Calculate the electric field between the plates.b) Calculate the (uniform) polarization created in the glass.c) Calculate the surface bound charge density on the side of the glass that is in contact withthe positive plate of the capacitor.d) Calculate the electric displacement between the plates.e) Calculate the free charge per unit area on the positive plate of the capacitor.

Given that, εr=2.5d=5 mmV=3 V The formula for the capacitance of a capacitor is given as C=εrεoAd…

Answered: In Lab 3, we measured the dielectric…

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