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ENGINEERING PHYSICS II DIELECTRICS PROBLEMS 1. Calculate the dielectric constant of a material which when inserted in parallel condensor of area 10mm x 10mm and distance of separation of 2 mm, gives a capacitance of 10-9 F. Solution: βπ βπ π΄ πΆ= π πΆπ βπ = βπ π΄ = 10β9 × 2 × 10β3 8.854 × 10β12 × 10β6 βπ« = ππππ β΄ The relative permittivity of the material is βπ =2259 (no unit) 2. A crystal subjected to an electric field of 1000 Vm-1 and the resultant polarisation is 4.3 ×10-8 C/m2. Calculate the relative permittivity of the crystal. Solution π =βπ (βπ β 1) πΈ π βπ β 1 = βπ πΈ βπ = π βπ πΈ + 1 4.3 × 10β8 βπ = + 1 8.85 × 10β12 × 1000 βπ = 4.8588 + 1 βπ = 5.8588 β΄ The relative permittivity of the crystal is βπ =5.8588 (no unit) 75