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If the charges are situated in a medium of permittivity ε, then the magnitude of the force between them will be, If the charges are situated in a medium of permittivity ε, then the magnitude of the force between them will be,
Fm = 1/4∏€ X q1q2/r2 -------------------------------------(2)
Dividing equations 1 and 2 we get F/Fm = € / €0 = € r. This ratio is called as relative permittivity or dielectric constant. This is equal to 1 in air. Hence the force between two charges depends on the medium in which they are situated.
Some wire of cross-sectional area 1mm 2 has a resistance of 20 ?. Determine (a) the resistance of a wire of the same length and material if the cross-sectional area is 4mm 2 ,
The helicopter view in fig. p3.29
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Michlson interferometer
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