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Electric PE as a dipole changes its orientation in an electric field

  1. Feb 23, 2014 #1
    1. The problem statement, all variables and given/known data
    The ammonia molecule (NH3) has a dipole moment of 5.0×10−30C⋅m. Ammonia molecules in the gas phase are placed in a uniform electric field E⃗ with magnitude 1.0×106N/C.
    What is the change in electric potential energy when the dipole moment of a molecule changes its orientation with respect to E⃗ from parallel to perpendicular?


    2. Relevant equations
    V = U/q
    ΔU = -integral([itex]qE dl[/itex])
    ΔV = -integral([itex]E dl[/itex])
    E(r) = -∇V(r)
    I have no idea what the upside down delta is...

    3. The attempt at a solution

    I don't even know where to start. :confused:
     
  2. jcsd
  3. Feb 23, 2014 #2
    Potential energy of a dipole is given by this formula: $$U = -\stackrel{\rightarrow}{p} \cdot \stackrel{\rightarrow}{E}$$Where U is defined to be zero when the dipole is perpendicular to the field. It's derived from the formula for the torque on the dipole: [itex]\tau = \stackrel{\rightarrow}{p} \times \stackrel{\rightarrow}{E}[/itex], using the relationship [itex]U = -W = -\int^{\theta_{f}}_{\theta_{i}}\tau d\theta[/itex]
     
    Last edited: Feb 23, 2014
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