Assume a thin semi-circular washer of dimensions is as shown in Fig. Q.3 and its edges...

  

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Assume a thin semi-circular washer of dimensions is as shown in Fig. Q.3 and its edges are biased as illustrated. Further, the semi-circular washer is made of two dielectric materials. Region I is a dielectric medium of permittivity, &1 = Eo x Erl; and, Region II is another dielectric medium of permittivity, 82= Eo&r2. With reference to the boundary conditions involved and their types in the 1-D electrostatic system shown, solve the electrostatic potentials (as functions of o- direction in cylindrical coordinates), namely vi() in Region I and vi() in Region II respectively and answer relevant questions indicated below. Note: Notations: (eo → Eo), (eox er →→ Eo × Er), (pi→ = 3.1416, phi questions Laplace equation in cylindrical coordinate system: V²V(r, o, z) = (1/r) × (1/²) × ²V/²) + a²V/az²) = 0. X = Gradient of V(r, o, z) in cylindrical coordinate system: VV(r, o, z) avlab + (az) avəz Fig. Q.3 +V volt OFF 11/ P = π 4=0 Assume the following data: = a b → 6) are used in Canvas quiz [rx@V/ar]/ or + (a)@V/or + (a)(1/r) 4= 3π/4 v(6) Determine the following: Entities V₂ Free-space permittivity Dielectric constant of Region I Dielectric constant of Region II (a, b) Choices Table Q.3 1 (i) vi() (at (ii) VII (b) (at (iii) E, (at any (iv) D (at any (v) NBC (at = 3/4; a<r<b) 2 3 4 5 (1) Entities Free-space permittivity Dielectric constant of Region I Dielectric constant of Region II Vo (a, b) Required Answers 5.94 5.86 = π/4 in Region I) = 57/8 in Region II) and a < (r= (a + b)/2) < b) and a < (r= (a + b)/2) < b) 5.88 5.93 5.96 volt (ii) volt (iii) N/C 5.75 5.55 5.85 5.33 5.45 –83.91 a(phi) –83.51 a(phi) –83.81 a(phi) –83.91 a(phi) –87.41 a(phi) Values Values &o = [1/(36× π)] × 10 ⁹F/m eo = [1/(36 x pi)] × 10-⁹ F/m 3 4 12volt (6, 12) cm (iv) C/m² -2.7 x 10-8 a(phi) -2.8 x 10-8 a(phi) -2.8 × 10-8 a(phi) (v) phi = (3pi/4); a <r<b Reg I Reg 1 NBC at Reg I 23.01 = Reg II 23.01 22.15 = Reg II 22.15 23.13 = Reg II 23.13 -2.9 x 10-8 a(phi) Reg 1 22.66 = Reg II 22.66 -2.9 × 10-8 a(phi) Reg | 23.14 = Reg II 23.14 Assume a thin semi-circular washer of dimensions is as shown in Fig. Q.3 and its edges are biased as illustrated. Further, the semi-circular washer is made of two dielectric materials. Region I is a dielectric medium of permittivity, &1 = Eo x Erl; and, Region II is another dielectric medium of permittivity, 82= Eo&r2. With reference to the boundary conditions involved and their types in the 1-D electrostatic system shown, solve the electrostatic potentials (as functions of o- direction in cylindrical coordinates), namely vi() in Region I and vi() in Region II respectively and answer relevant questions indicated below. Note: Notations: (eo → Eo), (eox er →→ Eo × Er), (pi→ = 3.1416, phi questions Laplace equation in cylindrical coordinate system: V²V(r, o, z) = (1/r) × (1/²) × ²V/²) + a²V/az²) = 0. X = Gradient of V(r, o, z) in cylindrical coordinate system: VV(r, o, z) avlab + (az) avəz Fig. Q.3 +V volt OFF 11/ P = π 4=0 Assume the following data: = a b → 6) are used in Canvas quiz [rx@V/ar]/ or + (a)@V/or + (a)(1/r) 4= 3π/4 v(6) Determine the following: Entities V₂ Free-space permittivity Dielectric constant of Region I Dielectric constant of Region II (a, b) Choices Table Q.3 1 (i) vi() (at (ii) VII (b) (at (iii) E, (at any (iv) D (at any (v) NBC (at = 3/4; a<r<b) 2 3 4 5 (1) Entities Free-space permittivity Dielectric constant of Region I Dielectric constant of Region II Vo (a, b) Required Answers 5.94 5.86 = π/4 in Region I) = 57/8 in Region II) and a < (r= (a + b)/2) < b) and a < (r= (a + b)/2) < b) 5.88 5.93 5.96 volt (ii) volt (iii) N/C 5.75 5.55 5.85 5.33 5.45 –83.91 a(phi) –83.51 a(phi) –83.81 a(phi) –83.91 a(phi) –87.41 a(phi) Values Values &o = [1/(36× π)] × 10 ⁹F/m eo = [1/(36 x pi)] × 10-⁹ F/m 3 4 12volt (6, 12) cm (iv) C/m² -2.7 x 10-8 a(phi) -2.8 x 10-8 a(phi) -2.8 × 10-8 a(phi) (v) phi = (3pi/4); a <r<b Reg I Reg 1 NBC at Reg I 23.01 = Reg II 23.01 22.15 = Reg II 22.15 23.13 = Reg II 23.13 -2.9 x 10-8 a(phi) Reg 1 22.66 = Reg II 22.66 -2.9 × 10-8 a(phi) Reg | 23.14 = Reg II 23.14