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    Rotation of an electric dipole in a field and emission of photon

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    A point dipole with p0 = 10-12 [C m] is fixed at the origin of a Cartesian coordinate system and points up. Four, N = 4, little dipoles with p = 10-15 [C m] each and pointing radially away from p0 are uniformly distributed on a horizontal circle of radius R = 1 [cm] as shown in the Figure below. The little dipoles are allowed to rotate until they assume the equilibrium orientation. During this process each little dipole emits a photon. Your task is to determine:
    a)
    The initial and final dipole moments 푝퐴beforeand 푝퐴afterof the little dipole at location A.
    b)
    The frequency, f, of each of the emitted photons.
    c)
    The total emitted energy, 퓔 total.

    © BrainMass Inc. brainmass.com April 4, 2020, 1:29 am ad1c9bdddf
    https://brainmass.com/physics/gauss-law/rotation-electric-dipole-field-emission-photon-617536

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    Distance of the little dipole A from P0 is
    r=R/sin⁡〖〖45〗^0 〗 =1/(1/√2)= √2 cm=1.414*〖10〗^(-2) m
    The magnitude of the electric field at distance r from a point dipole P in the direction  with the dipole is given by
    E=P/(4πϵ_0 r^3 ) √(1+3 cos^2⁡θ )
    Thus magnitude of electric field at position of dipole A due to dipole P0 will be
    E=P_0/(4πϵ_0 r^3 ) √(1+3 cos^2⁡〖〖45〗^0 〗 )
    Or E=(〖10〗^(-12)*9*〖10〗^9)/(1.414*〖10〗^(-2) )^3 √(1+3 cos^2⁡〖〖45〗^0 〗 ) = 3.18*103*1.58 = 5.02*103 N/C
    This is the magnitude of the field at position of dipole A and its direction is making angle ...

    Solution Summary

    A point dipole is generating an electric field and a four little dipoles are placed radially. The little dipole rotates when released and emits a photon. The frequency of photon, energies, fields and their directions are calculated.

    $2.19

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