ELEC3115 Electromagnetic Engineering T1, 2019 Midsession Test (Part A: Low frequency EM) Total Marks: 60 Duration: 100 mins (1 hour and 40 mins) It will contribute 12.5% toward your final marks. You must return this question paper with your answer sheets. Question 1 [20 marks] (a)[4 marks] Discuss the two fundamental postulates of the electrostatic field. (b)[3 marks] Starting from Gauss’s law show that electric field at an arbitrary radial distance r from a long, uniform line charge density ρl surrounded by a dielectric of permittivity ε is V/m2 l rE ar . (c) A coaxial cable operates at 66 kV and the inner conductor diameter is 2 cm. The cable has two insulating layers: rubber with relative dielectric constant 4rr and polystyrene with relative dielectric constant 5.2rp . The dielectric breakdown strengths of these two materials are: Rubber 25 MV/m Polystyrene 20 MV/m (i) [2 marks] Which dielectric is to form the inner layer and why (ii) [8 marks] Calculate the thickness of the two insulation layers so that the maximum electric field intensities will not exceed 25% of the respective breakdown levels. Given 12 0 8.854 10 (F/m) . (iii) [3 marks] Find the ratio of the E fields across the rubber-polystyrene boundary. Taking help of the boundary conditions of two dielectric media explain why the E field jumps by a factor at the rubber-polystyrene boundary. Question 2 [20 marks] (a) [4 marks] Discuss polarization of a dielectric material. (b) [6 marks] The relation of the bound charge density of a dielectric material ρp and the polarization vector P is given as p P . Use this relation and the first postulate to Name: ____________________________ Student ID: ________________________ Page 2 of 2 show that 0P D E , where D and E are the density and intensity of an applied electric field in a dielectric material. (c) [10 marks] A parallel plate capacitor is made of two plates 100 mm by 100 mm in size, separated by distance of 1mm. The design calls for three layers of insulation between the two conductors. The first layer is rubber ( r = 4), next layer is plastic ( r = 9) and third layer is foam ( r = 1.5). The thickness of first layer is 0.2mm, second layer is 0.6mm and the third layer is 0.2mm. A potential of 450V is applied between the two plates of the capacitor. Calculate the magnitude values of (i) electric field E, (ii) flux density D, (iii) polarisation vector P, at the first layer rubber and (iv) also calculate the total capacitance of the capacitor. Question 3 [20 marks] (a) [10marks] Show that the electric field E at an arbitrary radial distance r along the line joining the centres of the two parallel overhead lines of bare, smooth-surface conductors of radius a, at a centre to centre separation distance d is given by V 1 1E( r ) d r d r2 ln a V/m where, l 0 dV ln a is the potential difference between the conductors. Sketch the E field along the shortest line joining the conductors indicating where will the electric field is maximum and minimum. (b) [4 marks] If the breakdown dielectric strength of air is 3 MV/m, calculate the maximum voltage that can be used, without breakdown of air, between the two parallel overhead lines forming a two-wire transmission line. The two lines are each of bare, smooth-surface conductors of diameter 10 mm. The separation between the conductors is 1.0 meter. Assume that the ground beneath has no effect. (c) [6 marks] The capacitance per length between the two wires of the transmission line is oC dln a and if stored electrostatic energy in this capacitance is We , find the magnitude and the direction of the electrostatic force between the two wires when they operated at the maximum voltage found in (b). Note that the force between the conductors at a fixed potential is given by eF W . You may use 2 1 1 ln( / ) ln / d dx x a x x a End of the paper
