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A neutral water molecule (H(2)O) in its ...

A neutral water molecule `(H_(2)O)` in its vapor state has an electric dipole moment of magnitude `6.2xx10^(-30)C*m`.
How much work must an external agent do to rotate this molecule by `180^(@)` in this field, starting from its fully aligned position , for which `theta=0`?

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AI Generated Solution

To find the work done by an external agent to rotate a neutral water molecule (H₂O) by 180 degrees in an electric field, we can use the formula for the work done on an electric dipole in an electric field. ### Step-by-Step Solution: 1. **Understanding the Dipole Moment**: The electric dipole moment \( P \) of the water molecule is given as \( P = 6.2 \times 10^{-30} \, \text{C m} \). 2. **Work Done Formula**: ...
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