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

Answer:

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

(b). $U = 4.8*10^{-20}J.$

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

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

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

$\boxed{U = 4.8*10^{-20}J.}$