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A water molecule has an electric dipole ...

A water molecule has an electric dipole moment of `6.3xx10^(-30)` cm . A sample contains `10^(22)` water molecules with all the dipole moments aligned parallel to the external electric field of magnitude `3xx10^(5)NC^(-1)` . How much work is required to rotate all the water molecules from `theta=0^(@) "to" 90^(@)` ?

Text Solution

Verified by Experts

When the water molecules are aligned in the direction of the electric field it has minimum potential energy. The work done to rotate the dipole from `theta = 0^(@) "to" 90^(@)` is equal to the potential energy difference between these two configurattions
`W=DeltaU=U(90^(@))-U(0^(@))`
As we know U = `-pE cos theta ` Next we calculate the work done to ratate one water molecule from `theta=0^(@)"to " 90^(@)`
For one water molecule `W=-pE cos 90^(@)+pE cos 0^(@)=pE`
`W=6.3xx10^(-30)xx3xx10^(5)=18.9xx10^(-25)` J
For `10^(22)` water molecules the total work done is `W_("tot")=18.9xx10^(-25)xx10^(22)=18.9xx10^(-3)J`
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