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A water molecule has an electric dipole moment of 6.3xx10^(-30) cm . A sample contains 10^(22) watermolecules 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^(@) ? |
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Answer» Solution :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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