Department of Physics IIT Kanpur, Semester II, 2017-18 PHY103A: Physics II Solution # 3 Instructors: AKJ & SC Solution 3.1: Finding potential, given a charge distribution (Griffiths 3rd ed. Prob 2.26) z h b x h r’ r a y x FIG. 1: The potential V (a) at point a is given by (see Fig. 1) p 1 Z 2h σ2πx V (a) = dr 4π0 0 r p we have x = r= 2. Therefore we get, p 2πσ Z 2h 1 2πσ 1 p σh V (a) = p dr = p ( 2h) = 4π0 0 2 4π0 2 20 The potential V (b) at point b is given by (see Fig. 1) p 1 Z 2h σ2πx V (b) = 0 dr 4π0 0 r p p p we have x = r= 2 and r0 = h2 + r2 − 2hr. Therefore we get, p p ! 1 Z 2h σ2πx 2πσ 1 Z 2h r V (b) = dr = p dr 0 p p 4π0 0 r 4π0 2 0 h2 + r2 − 2hr p σ q p h q p p 2h 2 2 2 2 = p h + r − 2hr + p ln 2 h + r − 2hr + 2r − 2h 2 20 2 0 σ h p p h p = p h + p ln(2h + 2 2h − 2h) − h − p ln(2h − 2h) 2 20 2 2 p ! p ! σ h h p p i σh 2 + 2 σh 2 + 2 = p p ln(2h + 2h) − ln(2h − 2h) = ln p = ln p 2 20 2 40 2 − 2 40 2 − 2 p ! σh (2 + 2)2 σh p = ln = ln 1 + 2 40 2 20 Thus we get the required potential difference to be σh h p i V (a) − V (b) = 1 − ln 1 + 2 20 1 Solution 3.2: Finding field and charge density, given an electric potential, (Griffiths 3rd ed. Prob 2.46) e−λr (a) The electric potential is V (r) = A . Therefore, the electric field E(r) can be written as r @ e−λr r(−λ)e−λr − e−λr E = −rV = −A ^r = −A ^r @r r r2 ^r = Ae−λr(1 + λr) r2 (b) The corresponding charge density ρ(r) can be calculated by using the differential form of Gauss's Law ρ = 0r · E. Using the product rule for divergence, r · (fA) = f(r · A) + A · (rf), we obtain ^r ^r ρ = r · E = Ae−λr(1 + λr)r · + A · r e−λr(1 + λr) 0 0 r2 0 r2 Next we use the properties of the Dirac-delta function and the formula for gradient in spherical coordinates to get ^r Ae−λr(1 + λr)r · = Ae−λr(1 + λr)4πδ3(r) = A4πδ3(r) 0 r2 0 0 ^r ^r @ A · r e−λr(1 + λr) = A · e−λr(1 + λr)^r 0 r2 0 r2 @r ^r = A · −λe−λr(1 + λr) + e−λrλ^r 0 r2 ^r = A · −λ2re−λr^r 0 r2 λ2 = − A e−λr 0 r Therefore, we get the charge density ρ(r) as λ2 ρ = A 4πδ3(r) − e−λr 0 r (c) The total charge Q can now be calculated to be Z Q = ρdτ Z Z 1 −λr 3 2 e 2 = 0A4π δ (r)dτ − 0Aλ 4πr dr 0 r Z 1 2 −λr = 0A4π − 0Aλ 4π re dr 0 1 = A4π − Aλ24π 0 0 λ2 = 0 Therefore the total charge is zero. 2 Solution 3.3: Verifying Poisson's Equation (Griffiths 3rd ed. Prob 2.29) z ½(r’) r = r-r’ r’ r y x FIG. 2: The potential V (r) at r due to the localized charge distribution is 1 Z ρ(r0) V (r) = dτ 0; 4π0 r We note that the charge distribution has been represented in the (r0; θ0; φ0) coordinates. We take the Laplacian of the potential in (r; θ; φ) coordinates. Therefore, we get 1 Z ρ(r0) r2V (r) = r2 dτ 0 4π0 r 1 Z ρ(r0) = r2 dτ 0 4π0 r 1 Z 1 = ρ(r0) r2 dτ 0 4π0 r 1 Z 1 = ρ(r0) r · r dτ 0 4π0 r Z r 1 0 −^ 0 = ρ(r ) r · 2 dτ 4π0 r 1 Z = −ρ(r0)4πδ3(r)dτ 0 4π0 1 Z = −ρ(r0)4πδ3(r − r0)dτ 0 4π0 1 = − ρ(r) 0 Thus, we see that the given potential satisfies the Poisson's equation. 3 Solution 3.4: Electrostatic energy of two spherical shells (Griffiths 3rd ed. Prob 2.34) 1 q (a) The electric field due to the two shells is given by E(r) = 2 ^r, for (a < r < b), and E(r) = 0, otherwise. 4π0 r The energy of this configuration is: Z 2 Z b 2 2 0 2 0 q 1 2 q 1 1 W = E dτ = 2 4πr dr = − 2 2 4π0 a r 8π0 a b (b) Let's us first calculate the energy of the individual shells. The electric filed due to the shell of radius a is 1 q Ea(r) = 2 ^r, for (r > a), and Ea(r) = 0, otherwise. The electric filed due to the shell of radius b is 4π0 r 1 q Eb(r) = − 2 ^r, for (r > b), and Eb(r) = 0, otherwise. Therefore, the energy of the first spherical shell is: 4π0 r Z 2 Z 1 2 2 0 2 0 q 1 2 q Wa = Eadτ = 2 4πr dr = 2 2 4π0 a r 8π0a Similarly, the energy of the second spherical shell is Z 2 Z 1 2 2 0 2 0 q 1 2 q Wb = Eb dτ = 2 4πr dr = 2 2 4π0 b r 8π0b The interaction energy of this system is therefore: Z q2 q2 Wint = 0 Ea · Ebdτ = W − Wa − Wb = − − 8π0b 8π0b q2 = − 4π0b Solution 3.5: Electrostatic Force (a) z (b) z Q R da Q R dτ θ θ FIG. 3: 1 Q (a) The electric filed due to the metal sphere of radius R is given by E(r) = 2 ^r, for (r ≥ R), and E(r) = 0, 4π0 r otherwise. From the symmetry of the problem, it is clear that the total electrostatic force on northern hemisphere will be in the z direction. Now, the electrostatic force per unit area in the z-direction at the area 4 element da, as shown in Fig. 3(a), is: E(r) Q 1 1 Q Q2 fz = σEother · ^z = σ · ^z = 2 2 cos θ = 2 4 cos θ 2 4πR 2 4π0 r 32π 0R Therefore, the total repulsive force on the northern hemisphere is Z Z π=2 Z 2π 2 Q 2 Fz = fzda = 2 4 cos θ R sin θdθdφ θ=0 φ=0 32π 0R Q2 Z π=2 = 2 2 2π cos θ sin θdθ 32π 0R θ=0 Q2 1 = 2 2 2π 32π 0R 2 Q2 = 2 32π0R 1 Qr (b) The electric filed inside a uniformly charged sphere of radius R and charge Q is given by E(r) = 3 ^r. 4π0 R From the symmetry of the problem, it is clear that the total electrostatic force on the northern hemisphere will be in the z direction. Now, the electrostatic force per unit volume in the z-direction on the volume element dτ, as shown in Fig. 3(b), is: 3Q 1 Qr 3Q2 fz = ρE(r) · ^z = 3 3 cos θ = 2 6 r cos θ 4πR 4π0 R 16π 0R Therefore, the electrostatic force on the northern hemisphere is Z Z R Z π=2 Z 2π 2 3Q 2 Fz = fzdτ = 2 6 r cos θ r sin θdθdφ 0 θ=0 φ=0 16π 0R 2 Z R Z π=2 Z 2π 3Q 3 = 2 6 r cos θ sin θdθdφ 16π 0R 0 θ=0 φ=0 2 Z R Z π=2 Z 2π 3Q 3 = 2 6 r dr cos θ sin θdθ dφ 16π 0R 0 θ=0 φ=0 3Q2 R4 1 = 2 6 × × × 2π 16π 0R 4 2 3Q2 = 2 64π0R Solution 3.6: Capacitance of coaxial metal cylinders (Griffiths 3rd ed. Prob 2.39) Suppose that for a length L, the charge on the inner cylinder is Q and the charge on the outer cylinder is −Q. Using Q 1 the Gaussian surface as shown in Fig. 4, it can be shown that the field in between the cylinders is E(s) = ^s. 2π0L s The potential difference between the cylinders is therefore, Z b Q Z b 1 Q b V (b) − V (a) = − E · dl = − ds = − ln : a 2π0L a s 2π0L a Q b We see that a is at a higher potential. So, we take the potential difference as V = V (a) − V (b) = ln . The 2π0L a 5 capacitance C of this configuration is therefore given by Q 2π C = = 0 V b ln a b -Q s a Q L FIG. 4: 6.