InterviewSolution
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InterviewSolution
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Explain dielectric constant with expressions . |
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Answer» Solution :When two point charges `q_(1)andq_(2)` are at a distance r from each other in air . The COULOMB force between them `F_(1) = (1)/(4piepsilon_(0))(q_(1)q_(2))/(Kr^(2))` Where `epsilon_(0)` is the permittivity of THEFREE space . Ifa medium of DIELECTRIC constant K is FILLED in the space between them Coulomb force `F_(2)=(1)/(4piepsilon_(0))(q_(1)q_(2))/(r^(2))` `therefore(F_(1))/(F_(2))=K` Therefore the ratio of the forces between the charges when placed in air and when placed in the medium at the same distance is known as the dielectric constant of that medium. We know that `EPSILON=epsilon_(0)K` `K=(epsilon)/(epsilon_(0))` orDielectric constant `=("Permittivity of the medium")/("Permittivity of the free space")` |
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