Question

A natural water molecule (H2O) in its vapor state has an electric dipole moment of magnitude, p = 6.2 x 10-30 C.m. (a) Find the distance of the positive and negative charge centers of the molecule. Note that there are 10 electrons and 10 protons in a natural water molecule. (b) If the molecule is placed in a uniform electric field, E = 2 x 10' N/C find the maximum torque acting on the molecule. (c) How much work is needed to rotate this molecule by 180° in this field starting from the initial position, for which 0 = 0? Hint: 0 is the angle between the electric dipole moment and the electric field​

Answer 1

Answer:

a    $D = 3.9 *10^{-12} \ m$

b    $\tau_{max} = 1.24 *10^{-25} \ N\cdot m$

c   $W = 2.48 *10^{-25} J$

 

Explanation:

From the question we are told that

   The magnitude of electric dipole moment is  $\sigma = 6.2 *10^{-30} \ C \cdot m$

     The electric field is $E = 2*10^{4} \ N/C$

   

The distance between the positive and negative charge center is mathematically evaluated as

     $D = \frac{\sigma }{10 e}$

Where  e is the charge on one electron which has a constant value of  $e = 1.60 *10^{-19} \ C$

  Substituting values

     $D = \frac{6.20 *10^{-30}}{10 * (1.60 *10^{-19})}$

      $D = 3.9 *10^{-12} \ m$

The maximum torque is mathematically represented as

       $\tau_{max} = \sigma * E * sin (\theta)$

Here  $\theta = 90^o$

This because at maximum the molecule is perpendicular to the field

    substituting values

       $\tau_{max} = 6.2 *10^{-30} * 2*10^{4} sin ( 90)$

       $\tau_{max} = 1.24 *10^{-25} \ N\cdot m$

The workdone is mathematically represented as

      $W = V_{(180)} - V_{0}$

where   $V_{(180)}$ is the potential energy at 180° which is mathematically evaluated as

     $V_{(180) } = - \sigma * E cos (180)$

Where the negative signifies that it is acting against the  field

   substituting values

     $V_{(180) } = - 6.20 *10^{-30} * 2.0 *10^{4} cos (180)$

      $V_{(180) } = 1.24*10^{-25} J$

and

     $V_{(0)}$ is the potential energy at 0° which is mathematically evaluated as

            $V_{(0) } = - \sigma * E cos (0)$

   substituting values

     $V_{(0) } = - 6.20 *10^{-30} * 2.0 *10^{4} cos (0)$

      $V_{(0) } =- 1.24*10^{-25} J$

So $W = 1.24 *10^{-25} - [-1.24 *10^{-25}]$

    $W = 2.48 *10^{-25} J$