Electric Displacement D Assignment Help | Electric Displacement D Homework Help

Electric Displacement D

When a dielectric slab is placed in between the plates of a charged parallel plate capacitor, due to polarization of dielectric, charges are induced at the surface of dielectric. The charge is negative near the positive plate of the capacitor and positive near the negative plate. If q is the charge on the plates and q’ is the induced charge on the boundary of the dielectric, then we can write Eq. (3) as

        q/A = ε0E + q’/A                                … (1)

Here, E = electric field strength inside the dielectric,

A = cross-sectional area of the dielectric slab,
Now, q’/A = induced surface charge per unit area
        q/A = ε0E + P                                … (2)

The quantity q/A is called electric displacement D. Thus,
        D = ε0E    + P                                … (3)
Since E and P are vectors, D must also be a vector.
        D = ε0E + P                                … (4)

Clearly, D has the same direction as that of E or P.

Electric susceptibility

When a dielectric material is placed in an electric field, it becomes electrically polarized. For isotropic dielectric, the polarization vector P is proportional to the electric field E. Thus,
            P = ε0 χE                        … (5)

Where χ, the electric susceptibility, is a dimensionless quantity and is a characteristics of the medium

This means P is in the same direction as E and the magnitudes of these two vectors have the ratio χε0

For more help in Electric Displacement D please click the button below to submit your homework assignment.