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^(@)` ?

Explanation

When the water molecules are aligned in the direcction of the electric field.it has minimum potential energy.The work done to rotate dipole from `theta=0^(@)` to `90^(@)` is equal to the potential energy difference between these two configurations.
`W=triangleU=U(90^(@))-U(0^@)`
As we know `U-pE cos theta`Next we calculate the work done to rotate one water molecule From `theta=0^(@)` to `90^(@)`
For one waqter 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")=1811.9xx10^(-25)xx10^(22)=18.9xx10^(-3)J`