# Introduction to solid state physics: instructor's manual by Charles Kittel

By Charles Kittel

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**Example text**

E2 x 3 ⋅ = eE; ex = r 3 E = p; α = p E = r 3 = a H . r r 4π 4π 3 P = 0 inside a conducting sphere. Thus p = a P = a 3E 0 , and 3 3 α = p E0 = a 3 . 2. E i = E 0 − 3. Because the normal component of D is continuous across a boundary, Eair = εEdiel, where Eair = 4πQ/A, with Q the charge on the boundary. The potential drop between the 1⎞ ⎛ two plates is E air qd + E diel d = E air d ⎜ q + ⎟ . For a plate of area A, the capacitance is ε⎠ ⎝ C= A . 1⎞ ⎛ 4πd ⎜ q + ⎟ ε⎠ ⎝ It is useful to define an effective dielectric constant by 1 1 = +q .

E i = E 0 − 3. Because the normal component of D is continuous across a boundary, Eair = εEdiel, where Eair = 4πQ/A, with Q the charge on the boundary. The potential drop between the 1⎞ ⎛ two plates is E air qd + E diel d = E air d ⎜ q + ⎟ . For a plate of area A, the capacitance is ε⎠ ⎝ C= A . 1⎞ ⎛ 4πd ⎜ q + ⎟ ε⎠ ⎝ It is useful to define an effective dielectric constant by 1 1 = +q . εeff ε If ε = ∞, then εeff = 1/q. We cannot have a higher effective dielectric constant than 1/q. For q = 10–3, εeff = 103.

2 Solve this set of equations for exx: C − α 2 ( C11 + 2C12 ) e xx = B1 12 . ( C11 − C12 )( C11 + 2C12 ) 2 Similarly for eyy, ezz, and by identical method for exy, etc. 12-2 5a. U ( θ ) = K sin 2 θ − Ba M s cos θ Kϕ2 − Ba M s 1 2 ϕ , for θ = π + ϕ 2 and expanding about small ϕ . 1 Ba M s . ). For minimum near ϕ = 0 we need K > b. If we neglect the magnetic energy of the bidomain particle, the energies of the single and bidomain particles will be roughly equal when M s d 3 ≈ σ w d 2 ; or d c ≈ σ w M s .