Why potential energy of a dipole in an electric field has a negative sign? 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.
 A: 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.
A: 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:

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 $\textbf{d} \ll \textbf{r}$.
