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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:
βˆˆπ‘œ βˆˆπ‘Ÿ 𝐴
𝐢=
𝑑
𝐢𝑑
βˆˆπ‘Ÿ =
βˆˆπ‘œ 𝐴
=
10βˆ’9 × 2 × 10βˆ’3
8.854 × 10βˆ’12 × 10βˆ’6
∈𝐫 = πŸπŸπŸ“πŸ—
∴ The relative permittivity of the material is βˆˆπ’“ =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
𝑃
=βˆˆπ‘œ (βˆˆπ‘Ÿ βˆ’ 1)
𝐸
𝑃
βˆˆπ‘Ÿ βˆ’ 1 =
βˆˆπ‘œ 𝐸
βˆˆπ‘Ÿ =
𝑃
βˆˆπ‘œ 𝐸
+ 1
4.3 × 10βˆ’8
βˆˆπ‘Ÿ =
+ 1
8.85 × 10βˆ’12 × 1000
βˆˆπ‘Ÿ = 4.8588 + 1
βˆˆπ‘Ÿ = 5.8588
∴ The relative permittivity of the crystal is βˆˆπ’“ =5.8588 (no unit)
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