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

a
`D = 3.9 *10^(-12) \ m`

b
`\tau_(max) = 1.24 *10^(-25) \ N\cdot m`

c
`W = 2.48 *10^(-25) J`

Step-by-step 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 = (\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 = (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`