Prof. Anchordoqui Problems set # 12 Physics 169 May 12, 2015 1. Two concentric conducting spheres of inner and outer radii a and b, respectively, carry charges Q. The empty space between the spheres is half-filled by a hemispherical shell of dielectric (of ± dielectric constant /0), as shown in Fig. 1. (i) Find the electric field everywhere between the spheres. (ii) Calculate the surface-charge distribution on the inner sphere. (iii) Calculate the polarization-charge density induced on the surface of the dielectric at r = a. Solution This is a somewhat curious problem. It should be obvious that without any dielectric ~ Q r^ the electric field between the spheres would be radial E = 2 . We cannot expect this to be 4π0 r unmodified by the dielectric. However, we note that the radial electric field is tangential to the interface between the dielectric and empty region. Thus the tangential matching condition E1k = E2k is automatically satisfied. At the same time there is no perpendicular component to the interface, so there is nothing to worry about for the D1? = D2? matching condition. This suggests that we guess a solution of the radial form E = Ar=r^ 2, where A is a constant to be determined. This guess is perhaps not completely obvious because one may have expected the field lines to bend into or out of the dielectric region. However, we could also recall that parallel fields do not get bent across the dielectric interface. We may use the integral form of Gauss law in a medium to determine the above ~ 0A 2 A 2 Q ~ Q r^ constant A, i.e., D ndA^ = Q 2 2πr + 2 2πr = Q, or A = . Hence, E = 2 . v · ) r r 2π(+0) 2π(+0) r Note that ( + 0)=2 may be viewed as the average permittivity in the volume between the spheres. (ii) The surface-charge density is given by σ = D , where either D = E or D = E ?jr=a ? 0 ? ? ? depending on region. This gives Q 2 ; dielectric side +0 2πa σ = Q : (1) 2 ; empty side ( 0+0 2πa Note that the total charge obtained by integrating σ over the surface of the inner sphere gives Q as expected. (iii) The polarization charge density is given by ρ = ~ P~ , where P~ = pol −r · χ E~ = ( )E~ . Since the surface of the dielectric at r = a is against the inner sphere, we 0 e − 0 can take the polarization to be zero inside the metal (\outside" the dielectric). Gauss' law in this 0 Q case gives σpol = P ? r=a = ( 0)E? r=a = − 2 . Note that when this is combined − j − − j − +0 2πa 0 Q with (1), the total (free and polarization) charge density is σ = σ + σ = 2 on either tot pol +0 2πa half of the sphere. Since this is uniform, this is why the resulting electric field is radially symmetric. 2. A coaxial capacitor of length l = 6 cm uses an insulating dielectric material with /0 = 9, see Fig. 2. The radii of the cylindrical conductors are 0.5 cm and 1 cm. If the voltage applied across the capacitor is V (t) = 50 sin(120πt) what is the displacement current? Solution To find the displacement current, we need to know E~ in the dielectric space between the cylindrical conductors. Using Gauss law we obtain E~ = Q r^ and so V = Q ln(a=b). − 2πrl 2πl Therefore, E~ = V r^ = 50 sin(120πt) r^ = 72:1 sin(120πt)^r V=m. This implies that D~ = E = − r ln(a=b) r ln 2 − r 12 72:1 5:75 10−9 2 κ E = 9 8:85 10 sin(120πt)^r = × sin(120πt)^r C=m . The displacement 0 − · × − · r − r current flows between the conductors through an imaginary cylindrical surface of length l and radius r. The current flowing from the outer conductor to the inner conductor along r^ crosses @D~ @ 5:75 10−9 − surface S~ where S~ = 2πrlr^. Therefore, I = S~ = × sin(120πt) r^ ( r^2πrl) = − d @t · − @t r · − 5:75 10 9 120π 2πl cos(120πt) = 0:82 cos(120πt) µA: Alternatively,h since the coaxiali capacitor is × − · · lossless, its displacement current has to be equal to the conduction current flowing through the wires 2πl connected to the voltage sources. The capacitance of a coaxial capacitor is given by C = ln(a=b) . The current is I = C dV = 2πl [120π 50 cos(120πt)] = 0:82 cos(120πt) µA; which is the same dt ln(b=a) × answer we obtained before. 3. The parallel-plate capacitor shown in Fig. 3 is filled with a lossy dielectric material of relative permittivity κ and conductivity σ. The separation between the plates is d and each plate is of area A. The capacitor is connected to a time-varying voltage source V (t). (i) Obtain an expression for Ic, the conduction current flowing between the plates inside the capacitor, in terms of the given quantities. (ii) Obtain an expression for Id, the displacement current flowing inside the capacitor. (iii) Based on your expressions for parts (i) and (ii), give an equivalent-circuit representation for the capacitor. (iv) Evaluate the values of the circuit elements for A = 4 cm2, d = 0:5 cm, κ = 4, 3 1 σ = 2:5 (S=m), and V (t) = 10 cos(3π10 t) V. [Hint: 1 S = 1Ω− ,(S stands for siemens)] d V V σA Solution The resistance is R = σA , and so Ic = R = d . (ii) The electric field is E = V=d and @D @E A @V so Id = @t A = A @t = d @t . (iii) The conduction current is directly proportional to V , as char- acteristic of a resistor, whereas the displacement current varies as @V=@t, which is characteristic of d A 0:5 10−2 a capacitor. Hence, R = σA and C = d . The circuit is shown in Fig. 3. (iv) R = 2:5 4× 10−4 = 5 Ω 4 8:85 10−12 4 10−4 12 · × and C = · ×0:5 10−· 2× = 2:84 10− F. × × 4. Figure 4 shows a plane electromagnetic sinusoidal wave propagating in the x-direction. Sup- pose that the wavelength is 50 m, and the electric field vibrates in the xy plane with an amplitude of 22 V/m. Calculate (i) the frequency of the wave and (ii) the magnitude and direction of the magnetic field when the electric field has its maximum value in the negative y-direction. (iii) Write an expression for the magnetic field with the correct unit vector, with numerical values for Bmax, k, and !, and its magnitude in the form B = B cos(kx !t). max − Solution (i) During one oscillation (one period) the wave moves by a distance equal to the wavelength of the wave. Therefore T = λ/c. If one oscillation take time T , in one second the 1 3 108m=s number of oscillations will be f = T − = c/λ = ×50 m = 6 MHz. (ii) The magnetic flux through the vertical, differential surface indicated on Fig. 4 is proportional to the magnitude of the mag- netic field. Therefore the rate at which this magnetic flux changes is related only to the rate at dΦB d @B which the magnetic field varies. The flux is dt = dt (B dx dy) = @t dx dy. The linear integral of the electric field vector can be related directly to the electric field strength at the location of the differential surface E~ d~s = 0+E(x+dx)dy +0 E(x)dy = E(x) + @E dx E(x) dy = @E dx dy. · − @x − @x Therefore, from Faraday'sH law (using the general form forh a sinusoidal wave), thei above equa- tion requires that E k sin(kx !t) = @E = @B = B ! sin(kx !t), from which − max − @x − @t − max − k 2π Emax 22 V=m the magnitude of the magnetic field is Bmax = ! Emax = 2πfλ Emax = c = 3 108 m=s = 73:3 nT. Using the right-hand rule, when the y-component of the electric field is× negative, the z-component of the magnetic field is negative too; hence B~ = 73:3 nTk^. We can arrive − at the same solution from Ampere-Maxwell's law. From the definition, the displacement cur- rent through the horizontal, differential surface is related to the rate of change in the electric dΦE d @E flux Id = 0 dt = 0 dt (Edxdz) = 0 @t dxdy. The linear integral of the magnetic field vector can be related directly to the magnetic field strength at the location of the differential surface B~ d~s = 0 B(x + dx)dz + 0 + B(x)dz = B(x) @B dx + B(x) dz = @B dxdz. Ampere- · − − − @x − @x h@B @E i HMaxwell's law yields Bmaxk sin(kx !t) = @x = µ00 @t = µ00Emax! sin(kx !t), from ! − Emax − − − − which Bmax = µ00 k Emax = c that leads to the same answer. (iii) From the frequency of the wave, its angular frequency is ! = 2πf = 2π 6 MHz = 37:7 MHz. From the given wavelength, · 2π 2π 1 the value of the x-component of the propagation vector is k = λ = 50 m = 0:126 m− . (The peak value of the magnetic field was found in part (ii)). Hence B(x; t) = B cos(kx !t)k^ = max − 73:3 nT cos(0:126 m 1 x 37:7 MHz t)k^. − · − · 5. Some science fiction writers have described solar sails that could propel interstellar spaceships. Imagine a giant sail on a spacecraft subjected to radiation pressure from our Sun. (i) Explain why this arrangement works better if the sail is highly reflective rather than highly absorptive.