an electric dipole having a dipole moment of 4 × 10-9 c m is placed in a uniform electric field such that the dipole is in stable equilibrium. if the magnitude of the electric field is 3 × 103 n/c, what is the work done in rotating the dipole to a position of unstable equilibrium?

an electric dipole having a dipole moment of 4 × 10-9 c m is placed in a uniform electric field such that the dipole is in stable equilibrium. if the magnitude of the electric field is 3 × 103 n/c, what is the work done in rotating the dipole to a position of unstable equilibrium?

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Nature

An electric dipole with a dipole moment p=4\times 10^{-9},\text{C·m} is placed in a uniform electric field of magnitude E=3\times 10^{3},\text{N/C} such that the dipole is in stable equilibrium. Stable Equilibrium Condition:

  • The dipole is in stable equilibrium when its dipole moment \mathbf{p} is aligned parallel to the electric field \mathbf{E}, i.e., the angle \theta =0^\circ between \mathbf{p} and \mathbf{E}

Work Done in Rotating Dipole from Stable to Unstable Equilibrium:

  • The potential energy of a dipole in an electric field is given by:

U=-pE\cos \theta

  • At stable equilibrium (\theta =0^\circ ):

U_{\text{stable}}=-pE

  • At unstable equilibrium (\theta =180^\circ ):

U_{\text{unstable}}=-pE\cos 180^\circ =pE

  • The work done W to rotate the dipole from stable to unstable equilibrium is the change in potential energy:

W=U_{\text{unstable}}-U_{\text{stable}}=pE-(-pE)=2pE

  • Substituting values:

W=2\times (4\times 10^{-9})\times (3\times 10^{3})=2.4\times 10^{-5},\text{J}

Summary:

  • The dipole is stable when aligned with the field (\theta =0^\circ ).
  • The work done to rotate it to the unstable equilibrium (\theta =180^\circ ) is 2.4\times 10^{-5},\text{J}

Thus, the dipole in stable equilibrium aligns parallel to the electric field, and the work done to rotate it to the unstable position is 2.4\times 10^{-5},\text{J}.

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