Solved Problem on Electric Field
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Determine the torque that acts on an electric dipole in a uniform electric field and determine the potential
energy of this dipole.
Problem diagram:
Figure 1
Solution
Under the action of the electric force, this system rotates where torque is given by
The torque will be
we can write
Problem diagram:
The dipole consists of two charges of equal magnitude and opposite signs, +q and
−q, placed in an external electric field E. In the positive charge acts an
electric force FE in the same direction as the electric field, and in the negative
charge acts an electric force of the same magnitude and opposite direction to the electric field
(Figure 1).
We chose as reference the negative charge, the position vector r points towards the positive charge, where p is the dipole moment of the system, and θ is the angle between the segment that connects the dipole charges and the electric field.
We chose as reference the negative charge, the position vector r points towards the positive charge, where p is the dipole moment of the system, and θ is the angle between the segment that connects the dipole charges and the electric field.
Figure 1
Solution
Under the action of the electric force, this system rotates where torque is given by
\[ \bbox[#99CCFF,10px]
{\mathbf{N}=\mathbf{r}\times{\mathbf{F}}}
\]
the only force acting in the system is the electric force FE
\[
\begin{gather}
\mathbf{N}=\mathbf{r}\times{\mathbf{F}}_{E} \tag{I}
\end{gather}
\]
The electric force is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{{\mathbf{F}}_{E}=q\mathbf{E}} \tag{II}
\end{gather}
\]
substituting the expression (II) into expression (I)
\[
\begin{gather}
\mathbf{N}=\mathbf{r}\times q\mathbf{E}\\
\mathbf{N}=q\mathbf{r}\times{\mathbf{E}} \tag{III}
\end{gather}
\]
The dipole moment is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\mathbf{p}=q\mathbf{r}} \tag{IV}
\end{gather}
\]
substituting the expression (IV) into expression (III)
\[
\mathbf{N}=\mathbf{p}\times{\mathbf{E}}
\]
Using the definition of the Cross Product
\[
\mathbf{c}=\mathbf{a}\times{\mathbf{b}}
\]
\[
|\;c\;|=|\;a\;||\;b\;|\operatorname{sen}\theta
\]
The torque will be
\[ \bbox[#FFCCCC,10px]
{N=pE\operatorname{sen}\theta}
\]
The work of a force is given by
\[ \bbox[#99CCFF,10px]
{W=\int F\;dr}
\]
substituting the force with torque, F = N, and the displacement with angular displacement
dr = dθ
\[
\begin{gather}
W=\int N \;d\theta \\[5pt]
W=\int pE\operatorname{sen}\theta \;d\theta \\[5pt]
W=pE\int_{\theta _{0}}^{\theta}\operatorname{sen}\theta \;d\theta \\[5pt]
W=pE\left.\left(-\cos \theta\right)\right|_{\;\theta_{0}}^{\;\theta}\\[5pt]
W=-pE\left(\cos \theta-\cos \theta _{0}\right)
\end{gather}
\]
The work is stored in the form of the electric potential energy in the electric field
\[
\begin{gather}
W=\Delta U\\
\Delta U=-pE(\cos \theta -\cos \theta_{0})
\end{gather}
\]
choosing the initial situation
\( \theta_{0}=\frac{\pi}{2}\Rightarrow \cos \theta_{0}=0 \)
\[ \bbox[#FFCCCC,10px]
{U=-pE\cos \theta}
\]
Using the definition of Dot Product
\[
c=|\;a\;||\;b\;|\cos \theta
\]
\[
c=\mathbf{a}\cdot {\mathbf{b}}
\]
we can write
\[ \bbox[#FFCCCC,10px]
{U=-{\mathbf{p}}\cdot {\mathbf{E}}}
\]
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