Solutions. V in = ρ 0. r 2 + a r 2 + b, where a and b are constants. The potential at the center of the atom has to be finite, so a = 0. r 2 + b.

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1 Solutions. Plum Pudding Model (a) Find the coesponding electostatic potential inside and outside the atom. Fo R The solution can be found by integating twice, 2 V in = ρ 0 ε 0. V in = ρ 0 6ε a 2 + b, whee a and b ae constants. The potential at the cente of the atom has to be finite, so a = 0. Finally, Fo > R V in = ρ 0 6ε b. 2 V out = 0. The solution can be found by integating twice, V out = c + d. But fo lage, the potential has to go to zeo, theefoe d = 0. Using the bounday condition that the potential has to be continuous at = R: ρ 0 6ε 0 R 2 + b = c R.

2 (b) Find the electostatic vecto-field inside and outside the atom. Fo R: Fo R: Using the continuity of E at = R: In summay: E out = V out = c 2 ˆ. E in = V in = ρ 0 3ε 0. c = ρ 0R R 2 3ε 0 c = ρ 0R 3 3ε 0 E in = ρ 0 3ε 0, E out = ρ 0R 3 ˆ 3ε 0 = Q total 2 4πε 0 ˆ, b = ρ 0R 2 2 6ε 0, and V in = ρ ( 0 6ε R 2) (c) Electons inside the Plum Pudding Model atom will oscillate. Explain why! The potential enegy of an electon inside the atom is and the foce on the electon is: (d) Find the fequency of this oscillation. Use Using we can find U = ev ()= eρ 0 6ε 0 ( 2 + R 2) = A 2 + const F = q E in = eρ 0 R 3ε 0 m 2 t = eρ 0 2ε 0 2 t = 2ε 0 eρ 0 = 0 e iωt ω = ρ0 e 3mε 0 2

3 2. Dielectic Sphee (a) Detemine the bounday conditions fo the given setup. The bounday conditions can be summaized as follows. lim V ( )= E 0 cosθ 2. lim 0 V ( ) emains finite. 3. V (R) in = V (R) out 4. ε,in V in V out = 0 (b) Find the potential inside and outside the dielectic sphee. Explain you appoach! Use the method of sepaation of vaiables in spheical coodinates to detemine the potential inside and outside the sphee. In combination with the fist two bounday conditions, you will find that V in = A l l D P l (cosθ) and V out = E 0 cosθ l P l l l+ l (cosθ) Using the emaining bounday conditions, Fouie s Tick and the othogonality of the Legende Polynomials, we can find: V in = 3E 0 ε +2 z and V out = E 0 cosθ + ε ε +2 R3 E 0 cosθ 2 (c) Find the electic field E and polaization P inside the dielectic sphee. and fo a linea dielectic: E in = V in = z V in ẑ = 3E 0 ε +2ẑ P = ε 0 χ e E = ε ε +2 ε 0χ e E 0 ẑ 3

4 (d) Sketch the electic field lines fo all egions of this setup. (e) Find the bound volume chage density ρ b, and all the bound suface chage densities σ b. Since thee ae no fee chage inside the sphee ρ b = 0. The bound suface chage is σ b = P ˆn = E 0(ε ) ε +2 ε 0 χ e cosθ. 4

5 3. Inductance (a) In the quasistatic appoximation, find the induced electic field as a function of distance s fom the wie. In the quasistatic appoximation, the magnetic field of a wie is B = µ 0I s. Using Faaday s law, we can find fo the electic field: Ed l = E(s 0 )l E(s)l = dt d B d a = µ 0I di dt s ds [ s 0 E(s)= µ0 I 0 ω ]ẑ lnssin(ωt)+k = [ sin(ωt)+k ] ẑ. s (b) Is you answe valid fo the limit s? Explain you answe. No, since lns diveges fo lage s The quasistatic appoximation only holds fo s ct. (c) Find the self-inductance L of the ectangula coil. The magnetic field inside a tooid is given by So, the flux though a single tun is B = µ 0NI s. Φ = B d a = µ 0NI h b a s ds = µ 0NIh ln b a. Which means that the total flux is N times this, and the self-inductance is Φ = LI L = µ 0N 2 h ln b a. (d) In the quasi-static appoximation, what emf is induced in the coil? In the quasistatic appoximation: So, B = µ 0 s ˆφ. φ = µ 0I b a s hds = µ 0Ih ln b a. 5

6 This is the flux though only one tun, so the total flux is N times Φ : So, Φ = µ 0Nh ln b a I 0 cos(ωt). E = dφ dt = µ 0Nh ln b a I 0ω sin(ωt)= sin(ωt) V, using ω = 377 s. (e) Find the cuent I(t) in the esisto R. I = E R = sinωt A. (f) Calculate the back emf in the coil, due to the cuent I(t). The back emf E b is given by: E b = L di dt ; Now use the self-inductance of the squae coils calculated in pat (a): L = µ 0N 2 h ln b a = H. Theefoe, E b = cos(ωt) V. 6

7 4. Momentum of Electomagnetic Fields (a) What is the Poynting vecto S? The Poynting vecto is given by S = µ 0 E 0 B 0 The electic field of a paallel plate capacito is given by E = σ ε 0 ẑ So S = σb 0 c 2 µ 0 ε 0 ˆx (b) What is the momentum density of the electomagnetic field? The momentum density is given by p = c 2 E 0 H 0 p = σb 0 ˆx (c) What is the total momentum of the electomagnetic field? The total momentum is given by P total = pdv = QB 0 h ˆx (d) What is the impulse p in time t expeienced by each plate, as deived fom the induced electic field? How does this compae to the field momentum deived in pat c)? Now, we conside a closed loop in the x z-plane of width a and height h. The magnetic flux though this loop is: B 0 d a = B b ahŷ and fo the induced electic field: E ind d l = t B 0 d a = B 0ah t. Taking the closed loop integal: 7

8 The foces on the plates ae now: E ind d l = E ind 2a = B 0ah t E ind = B 0h 2 t F = q E with E ind = E ind ˆx at the top plate and E ind = E ind ˆx at the bottom plate. So, the foce on both plates with Q on the top plate and Q on the bottom plate, is given by: To find the momentum p: F = Q B 0h 2 t ˆx p = F dt = QB 0h 2 ˆx So, each plate eceives half of the momentum stoed in the fields. 8

9 5. Dipole Radiation (a) Find the electic field E(,t) and magnetic field B(,t) to leading ode in powes of ( ) in tems of p 0 (t ) [i.e. the second time deivative of p 0 (t) evaluated at the etaded time, t ]. To find the electic field, use E = V t A Since we only conside tems of powes ( ) in tems of p0 (t ), we only deal with the last tem of V (,t). Now, V = 4πε 0 ˆ p c ˆ = = 4πε 0 4πε 0 c [ (ˆ p(t ) )] ˆ ˆ c t (ˆ p(t ) ) t = 4πε 0 ˆ p c 2 ˆ ( ) t A = µ 0 4π p(t ) t = µ 0 p(t ) 4π Finally, E = (ˆ p)ˆ p(t ) 4πε 0 c 2 The magnetic field B can be found, as use the vecto identity Now, B = A = µ 0 4π p(t ) ( ) ( f v = f v = v ) f ( ) p = p p ( ) and ( ) = 2 ˆ,which can be neglected. p = ˆpṗ = ˆp ṗ with ṗ = p (t )= p ( c ˆ ) 9

10 So, finally: B = µ 0 4π c ( ˆp pˆ)= µ ) 0 4πc (ˆ p (b) Assume that p(t )=p 0 (t)ẑ. Show that E(,t)=E(,t) ˆθ and B(,t)=B(,t) ˆφ. Find expessions fo E(,t) and B(,t). using E = p((ˆ ẑ)ˆ ẑ) 4πε 0 c 2 = p cosθ ˆ ẑ 4πε 0 c 2 ẑ = cosθ ˆ sinθ ˆθ, we finally find fo the electic field: Similaly fo the magnetic field: E = p sinθ ˆθ 4πε 0 c 2 B = µ 0 4πc p(ˆ ẑ)= µ 0 4πc psinθ ˆφ. (c) Find the powe adiated to infinity by this time dependent chage distibution. The total powe is given by: P = Sd a (d) Calculate the Poynting vecto S. So, we need to find the Poynting vecto: So, the total powe is then: ( ) S = µ 0 E B = µ 0 p 2 6π 2 0 sin 2 θ ˆ c 2 P = Sd a = µ 0 p 2 0 sin 2 θ 6π 2 c 2 = µ 0 p 2 0 8πc = µ 0 p 2 0 6πc sin 3 θ dθ 2 sinθ dθdφ 0