A slab of dielectric of thickness t is inserted into a parallel plate capacitor

Question

A slab of dielectric of thickness t is inserted into a parallel plate capacitor of separation d and plate area A as shown in the figure below. The surfaces of the slab are parallel to the plate surfaces. Find D, E, and P as functions of x, and plot your results. (express them in terms of Q) Find the capacitance of this system. Verify that your result for C reduces to the proper value when t = 0 and t = d.

(please make the answer clear to see and show the work as to how you got the solution, it helps me learn the material. The answer should be

Explanation / Answer

The electric field intensity E at any point in the dielectric field is numerically equall to the force experienced by a unit positive charge placed at that point.The direction of E being the same as that of the field.

Dielectric polarization P is the electric dipole moment per unit volume is called as dielectric polarization P =Q/A

The electric displacement D , when a dielectric slab is placed between the plates of a parallel plate condesner, the medium is polarized .Now the induced surface charge appears . The charge is negative on the surface nearer to the positive palte of the condesner and positive charge nearer the negative plate.Let Q' be the induced surface charge .now the chargeq on the plates of the condesner and induced surface chargeQ' are related as

Q/keoA =Q/eoA- Q'/eoA

Q/eoA =Q/keoA +Q'/eoA

Q/A =eo(Q/keoA)+Q'/A

Q'/A =p and Q/keoA =E

and Q/A =D

We get

D =eoE+P

When p=0

Then Q/keoA =E

We can write D and P interms of E

E =Q/eokA and Q/A =eokE ==> D =keoE

Therefore

D =eoE+P

keoE =eoE+P

P =eo(k-1)(Q/keoA )

=[(k-1)Q]/kA

The capacitance of the capacitor C =Q/V =Q/(Q/eoA)[d-t(1-1/k)] =eoA/[d-t(1-1/k)]

C =eoAk/[kt-t(k-1)]

q

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