contestada

The permanent electric dipole moment of the water molecule (h2o) is 6.2×10−30c⋅m. part a what is the maximum possible torque on a water molecule in a 3.3×108 n/c electric field

Respuesta :

alimo3az

The torque produced on an electric dipole by an electric field is defined as

[tex]tourqr=p*E[/tex]

where:

  • p: the dipole moment in (c.m)
  • E: electric field in (N/c)

Thus

[tex]tourqr=6.2*10^{-30}*3.3*10^{8}=20.46*10^{-22}=2.046*10^{-21} N.m[/tex]

note that the unit of the torque is N.m

The maximum possible torque on a water molecule is [tex]20.46\times 10^{-22}[/tex] N.m.

Electric Dipole

An electric field that is subjected to external torque is known as an electric dipole.

It is given by the formula:

[tex]\rm{dipole\ Moment=\dfrac{Torque}{Electric\ field}[/tex]

Given to us

  • the permanent electric dipole moment of the water molecule (h2o) is 6.2×10−30c⋅m.
  • electric field = 3.3×108 n/c

Solution

We know the formula for electric dipole,

[tex]\rm{dipole\ Moment=\dfrac{Torque}{Electric\ field}[/tex]

Substituting the values we get

[tex]6.2\times10^{-30} =\dfrac{\tau}{3.3\times 10^8}\\\\ 6.2\times10^{-30} \times 3.3\times 10^8= \tau\\ \tau = 20.46 \times 10^{-22}[/tex]

Hence, the maximum possible torque on a water molecule is [tex]20.46\times 10^{-22}[/tex] N.m.

Learn more about Electric Dipole:

https://brainly.com/question/24209752

Otras preguntas