Question 4.16: Suppose the field inside a large piece of dielectric is E0, ...

Suppose the field inside a large piece of dielectric is E _{0}, so that the electric displacement is D _{0}=\epsilon_{0} E _{0}+ P

(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 E _{0} and P. Also find the displacement at the center of the cavity in terms of D _{0} . Assume the polarizationis “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, E _{0}, and D _{0} are essentially uniform. [Hint: Carving out a cavity is the same as superimposing an object of the same shape but opposite polarization.]

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(a) Same as E _{0} minus the field at the center of a sphere with uniform polarization P. The latter (Eq. 4.14)

E =-\nabla V=-\frac{P}{3 \epsilon_{0}} \hat{ z }=-\frac{1}{3 \epsilon_{0}} P , \quad \text { for } \quad r<R                                (4.14)

is – P / 3 \epsilon_{0} . \text { So } \mathbf { E } = \mathbf { E } _ { 0 } + \frac { 1 } { 3 \epsilon _ { 0 } } \mathbf { P } . D =\epsilon_{0} E =\epsilon_{0} E _{0}+\frac{1}{3} P = D _{0}- P +\frac{1}{3} P , \text { so } D = D _{0}-\frac{2}{3} P .

(b) Same as E _{0} minus the field of ± charges at the two ends of the “needle” but these are small, and far away, so E = E _{0} . \quad D =\epsilon_{0} E =\epsilon_{0} E _{0}= D _{0}- P , \text { so } D = D _{0}- P .

(c) Same as E0 minus the field of a parallel-plate capacitor with upper plate at σ = P. The latter is -\left(1 / \epsilon_{0}\right) P, \text { so } E = E _{0}+\frac{1}{\epsilon_{0}} P . \quad D =\epsilon_{0} E =\epsilon_{0} E _{0}+ P , \text { so } D = D _{0} .

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