#94 Suppose the field inside a large piece of dielectric

Suppose the field inside a large piece of dielectric - physics

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Suppose the field inside a large piece of the dielectric is E0, so that the electric displacement is 
(a) Now a small spherical cavity (Fig. 4.19a) is hollowed out of the material. Find the field at the center of the cavity in terms of E₀ and P. Also find the displacement at the center of the cavity in terms of D₀ and P. Assume the polarization is “frozen in,” so it doesn’t change when the cavity is excavated.
(b) Do the same for a long needle-shaped cavity running parallel to P (Fig. 4.19b).
(c) Do the same for a thin wafer-shaped cavity perpendicular to P (Fig. 4.19c). Assume the cavities are small enough that P, E₀, and D₀ are essentially uniform. [Hint: Carving out a cavity is the same as superimposing an object of the same shape but opposite polarization.]
Figure 4.19

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Free Chegg Answer

Electric displacement is a topic discussed in electromagnetism that measures the response of dielectric insulating materials that applies to the electric field. A conducting material remains neutral before applying an electric field on it. It means there is no movement in charge of particles inside the material.
When electric field is applied to the material, the free charges start responding to the force of field. The whole material becomes polarized as because the positive charges shift towards the direction of the applied field (electric) and the negative charges move in the opposite direction of the electric field. Thus, polarization P takes place within the material.

Here,  is the dielectric field.
So, Displacement is defined as follows:

(a)
From Gauss’s law, it is found that surface integral of displacement current is

Here,  is the charge is enclosed on the surface and is the small area.
We know, field due to electricity inside a uniformly polarised sphere is,
 â€¦â€¦ (1)
Here,  is permittivity in free space.
If  is the electric field inside the dielectric material, then total field at the center inside the cavity is  â€¦â€¦ (2)
From equation (1) and (2) we get,
 â€¦.. (3)
From Maxwell’s differential equation,
Electric displacement, 
Substituting the value of E from equation (3), in (a)

Therefore,
 â€¦â€¦ (4)
We know, electric field displacement

Substituting  in equation (4),

(b)
The electric field inside the cavity of the needle

In the above equation, the field due to charges is negligible due to a very small magnitude.
So, we can write.
Thus, it can be said that the electric field inside the cavity (needle-shaped) is equal to the electric field due to dielectric.
But we know that,

Therefore,
 â€¦â€¦ (5)
From the electric field displacement equation, .
Now substituting the above value in equation (5)

Thus, the electric displacement that parallels to the needle with cavity shape is 

(c)
The electric field due to the cavity is

Therefore,
 â€¦â€¦ (6)
From Maxwell’s equation, .
From equation (6), substituting the value of E in D,

Thus,
 â€¦â€¦ (7)
But 
Substituting the  in equation (7),

Hence, the electric displacement which is parallel to the thin cavity (with wafer shape).