To measure the dielectric constant of a dielectric material a slab of material i

Question

To measure the dielectric constant of a dielectric material a slab of material is slowly inserted between a pair d parallel plate capacities, each have an area of 0.40 m^2 and are separated by a distance of 0.02 m. Before insertion, potential difference across the capacitors is 3.0 x 10^5 V.
A) during the insertion, the charge on the plates remains constant. After, the Liberian difference is 1.8 x 10^5 V. What is the value of the dielectric constant?
B) now assume the potberiak differenc of 1.8 x 10^5 V is constant but the charge can vary. The surface charge is defined as the difference in charge on the plates with and without the dielectric. Determine the surface charge using the constant from part a in this case.
C) what is the difference in energy density between the capacitor with the dielectric and without the dielectric? To measure the dielectric constant of a dielectric material a slab of material is slowly inserted between a pair d parallel plate capacities, each have an area of 0.40 m^2 and are separated by a distance of 0.02 m. Before insertion, potential difference across the capacitors is 3.0 x 10^5 V.
A) during the insertion, the charge on the plates remains constant. After, the Liberian difference is 1.8 x 10^5 V. What is the value of the dielectric constant?
B) now assume the potberiak differenc of 1.8 x 10^5 V is constant but the charge can vary. The surface charge is defined as the difference in charge on the plates with and without the dielectric. Determine the surface charge using the constant from part a in this case.
C) what is the difference in energy density between the capacitor with the dielectric and without the dielectric?
A) during the insertion, the charge on the plates remains constant. After, the Liberian difference is 1.8 x 10^5 V. What is the value of the dielectric constant?
B) now assume the potberiak differenc of 1.8 x 10^5 V is constant but the charge can vary. The surface charge is defined as the difference in charge on the plates with and without the dielectric. Determine the surface charge using the constant from part a in this case.
C) what is the difference in energy density between the capacitor with the dielectric and without the dielectric?

Explanation / Answer

initial Volatge Vo = 1.5 V

new Volatge V = Vnew

these two are relaetd by V = V0/k

so

K = 3 e 5/1.8 e 5

k = 1.67

-----------------------

part B :

Surface charge density sigma = E eo

E = V/d

so

sigma = 1.8 e 5 * 8.85 e-12/(0.02

sigma = 79.65 uC/m^2

----------------------

Eo = V/d = 3 e 5 /0.02 = 1.5 e 7 V/m

Enew = 1.8 e 5/0.02 = 0.9 e 7 V/m

so

eenrgy density with out dieelctric = Uo = 0.5 eo E^2

Uo = 0.5 * 8.85 e-12* 1.5 e 7 * 1.5 e 7

Uo = 995.625 /m^3

Unew = 0.5 * 8.85 e -12* 0.9 e 7 * 0.9 e7

Unew = 358.425 J/m^3

differecne of energy density = 995.625 - 358.425 = 637.2 J /m^3

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