A water molecule has an electric dipole moment of 6.3 x 10 Cm. A sampl

1.	A water molecule has an electric dipole moment of 6.3 x 10 Cm. A sample contains 10 water molecules, with all the dipole moments aligned parallel to the external electric field of magnitude 3 x 105 NC-1 . How much work is required to rotate all the water molecules from θ = 0° to 90°?
Answer» 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 θ = 0° to 90° is equal to the potential energy difference between these two configurations. W = ΔU = U(90°) – U(0°) As we know, U = -pE cos θ, Next we calculate the work done to rotate one water molecule from θ = 0° to 90°, For one water molecule, W = – pE cos 90° + pE cos 0° = pE W = 6.3 x10^-30 x 3 x 10⁵ = 18.9 x 10^-25 For 10²² water molecules, the total work done is W_tot = 18.9x 10²⁵ x 10²² = 18.9 x 10³ J

A water molecule has an electric dipole moment of 6.3 x 10 Cm. A sample contains 10 water molecules, with all the dipole moments aligned parallel to the external electric field of magnitude 3 x 105 NC-1 . How much work is required to rotate all the water molecules from θ = 0° to 90°?

Answer»

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 θ = 0° to 90° is equal to the potential energy difference between these two configurations.

W = ΔU = U(90°) – U(0°)

As we know, U = -pE cos θ, Next we calculate the work done to rotate one water molecule from θ = 0° to 90°,

For one water molecule, W = – pE cos 90° + pE cos 0° = pE

W = 6.3 x10^-30 x 3 x 10⁵ = 18.9 x 10^-25

For 10²² water molecules, the total work done is

W_tot = 18.9x 10²⁵ x 10²² = 18.9 x 10³ J

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