Question

A proton is located at the origin and an electron is located at (1.0, 1.0) mm:

(a) Determine the electric dipole moment of these two particles in unit vector notation.

(b) If we put this electric dipole moment in an external electric field vector(E) = 300vector(i) (N/C), calculate the work done by the electric field to rotate the dipole so that it becomes in the same direction as the field.

Answer 1

(a).
`{\vec{p} =(1.6*10^(-22)\bold{i}+1.6*10^(-22)\bold{j})m \cdot C.}`

(b).
`U = 4.8*10^(-20)J.`

Step-by-step explanation:

(a).

The electric dipole moment of the charges is

`\vec{p} = q \vec{r}`

In our case

`\vec{r} = (1.0*10^(-3)\bold{i}+1.0*10^(-3)\bold{j})m`

and

`q =1.6*10^(-19)C`;

therefore, the dipole moment is

`\vec{p} =1.6*10^(-19)C *(1.0*10^(-3)\bold{i}+1.0*10^(-3)\bold{j})m`

`\boxed{\vec{p} =(1.6*10^(-22)\bold{i}+1.6*10^(-22)\bold{j})m \cdot C.}`

(b).

The work done
`U` by an external electric field
`\vec{E}` is

`U = -\vec{p}\cdot \vec{E}`

`U = [1.6*10^(-22)\bold{i}+1.6*10^(-22)\bold{j}] \cdot[300\bold{i}]`

`\boxed{U = 4.8*10^(-20)J.}`