2
$\begingroup$

In the following equation

$$V = -\mathbf{P}\cdot \mathbf{E}$$

why I have to take Potential Energy as negative. Is there any simple reason behind this ?

I am preparing for my high school examination.

$\endgroup$
  • $\begingroup$ I'm assuming you mean V=potential energy, P=electric dipole moment, E=electric field? I.e. $V = \vec{p}\cdot\vec{E}$? $\endgroup$ – Michael Brown Sep 12 '13 at 5:34
  • $\begingroup$ yes, sorry for confusion b/w P and V :D $\endgroup$ – Shushant Sep 12 '13 at 5:36
  • 1
    $\begingroup$ No worries. Check out the edit to see how formatting equations works here. Anyway. I assume you know that the electric field and the dipole moment are vectors, and that the dot product of two vectors is positive if they are pointing in the same direction and negative if they are pointing in opposite directions. Is that ok? Then think about the two cases where $P$ and $E$ are pointing in the same or opposite direction. In which case is $V$ greater? What does that mean in terms of dipoles wanting to align with/against an electric field? $\endgroup$ – Michael Brown Sep 12 '13 at 5:45
  • $\begingroup$ i think if they are moving in same direction the V will be greater and can please tell explain me about your last question is this relate to Earnshaw's_theorem ? $\endgroup$ – Shushant Sep 12 '13 at 5:53
  • $\begingroup$ If $P$ and $E$ are in the same direction then $P\cdot E > 0$, so $V < 0$, but if $P$ and $E$ are in opposite directions $P\cdot E <0$, so $V > 0$. So the configuration with $P$ in the same direction as $E$ has the lower energy, hence the dipole will tend to align with the field. This doesn't really have much to do with Earnshaw's theorem, which says that there are no stable equilibrium configurations for a set of static point charges and permanent magnets. $\endgroup$ – Michael Brown Sep 12 '13 at 12:55
3
$\begingroup$

Neoh's answer is very thorough and mathematical. Here is a less mathematical, but hopefully more intuitive way to look at it.

Take the two exterme cases of $\theta = 0, 180^\circ$. Then $V = \pm PE$, and you just need to fix the sign. To do that, recall the convention that systems move from higher states of potential energy to lower states of potential energy (eg, balls roll downhill). The state where the dipole is pointing in the same direction as the electric is lower than that where the two vectors are anti-parallel, so you want the $\theta = 0^\circ$ state to be lower than the $\theta = 180^\circ$ state. So you insert the negative sign to enforce that.

$\endgroup$
3
$\begingroup$

The only real reason is because of how they derive the expression from physical equations. Verbal arguments simply won't make the cut here.

Consider an electric dipole consists of two opposite charges of equal magnitude q, separated by a distance of d as figure below:

enter image description here

The vector d is directed from negative charge to positive charge. When placed in a static electric field at a distance r from the source of electric field, the electric potential energy of this dipole can be written as \begin{equation} V = \frac{kqQ}{|\boldsymbol{r}+\boldsymbol{d}/2|}-\frac{kqQ}{|\boldsymbol{r}-\boldsymbol{d}/2|} \tag{1} \end{equation}

where Q is the charge that gives rise to the electric field. This is just adding up the potential energy of energy experienced by each positive and negative charge q only.

Using the simple cosine rule, \begin{equation} |\boldsymbol{r} +\boldsymbol{d}/2| = \left[{r}^2 + {\left(\frac{d}{2}\right)^2-2r\left(\frac{d}{2}\right)cos\theta}\right]^{1/2} \end{equation} \begin{equation} |\boldsymbol{r} -\boldsymbol{d}/2| = \left[{r}^2 + {\left(\frac{d}{2}\right)^2+2r\left(\frac{d}{2}\right)cos\theta}\right]^{1/2} \end{equation}

Substituting them into (1):

\begin{align} V &= \frac{kqQ}{|\boldsymbol{r}+\boldsymbol{d}/2|}-\frac{kqQ} {|\boldsymbol{r}-\boldsymbol{d}/2|} \\ &= \frac{kqQ}{\left[{r}^2 + {\left(\frac{d}{2}\right)^2-2r\left(\frac{d}{2}\right)cos\theta}\right]^{1/2}} - \frac{kqQ}{\left[{r}^2 + {\left(\frac{d}{2}\right)^2+2r\left(\frac{d}{2}\right)cos\theta}\right]^{1/2}}\\ &= \frac{kqQ}{r\left[1 + {\left(\frac{d}{2r}\right)^2-2\left(\frac{d}{2r}\right)cos\theta}\right]^{1/2}} - \frac{kqQ}{r\left[1 + {\left(\frac{d}{2r}\right)^2+2\left(\frac{d}{2r}\right)cos\theta}\right]^{1/2}}\\ &={\frac{kqQ}{r}\left( \left[1 + {\left(\frac{d}{2r}\right)^2-2\left(\frac{d}{2r}\right)cos\theta}\right]^{-1/2} - \left[1 + {\left(\frac{d}{2r}\right)^2+2\left(\frac{d}{2r}\right)cos\theta}\right]^{-1/2}\right)} \tag{2}\\ \end{align}

The square brackets are simplified using Taylor expansion approximation to first order:

\begin{equation} (1+x)^{-1/2} \approx 1 - \frac{1}{2}x \end{equation}

Thus (2) becomes

\begin{align} V &\approx \frac{kqQ}{r} \left( 1- \frac{1}{2}\left(\frac{d}{2r}\right)^2 +\frac{d}{2r}cos\theta -1+ \frac{1}{2}\left(\frac{d}{2r}\right)^2 +\frac{d}{2r}cos\theta \right) \\ &= \frac{kqQd cos\theta}{r^2} \end{align}

Observe that $dcos\theta$ is the scalar product $-\boldsymbol{d. \hat r}$. Thus

\begin{align} V &\approx -\frac{kqQ\boldsymbol{d. \hat r}}{r^2} \\ &= -\frac{kQ}{r^2} \boldsymbol{\hat r .} q\boldsymbol{d} \\ &= - \boldsymbol{E.p} \end{align}

That's why you have the negative sign.

By now you should be aware that the equation is just an approximation. It is valid only if the source is far away, or to be exact $\textbf{d} \ll \textbf{r}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.