A water molecule perpendicular to an electric field has 1.40×10−21 J more potential energy than a water molecule aligned with the field. The dipole moment of a water molecule is 6.2×10−30Cm.

What is the strength of the electric field?

Respuesta :

To solve this problem we will apply the concepts related to the potential energy in the molecules and obtain its electric field through the relationship given by the dipole moment. The change in potential energy from one state to another is given by,

[tex]U_2-U_1 = 1.4*10^{-21} J[/tex]

From this difference we can identify that [tex]U_1[/tex] is equivalent to the potential energy when it is perpendicular to the electric field. At the same time, the potential energy [tex]U_2[/tex] would be equivalent when it is aligned with the electric field.

From there the relationship between energy, the dipole moment and the electric field would be subject to

[tex]U_2-U_1 = pE[/tex]

Here,

[tex]p = \text{Dipole moment} = 6.2*10^{-30} C \cdot m[/tex]

Rearranging to find the electric field,

[tex]E = \frac{(U_2-U_1)}{p}[/tex]

[tex]E = \frac{(1.4*10^{-21})}{(6.2*10^{-30})}[/tex]

[tex]E =2.25*10^8 N/C[/tex]

Therefore the electric field is [tex]2.25*10^8N/C[/tex]