When we derive the formula for potential energy caused by the torque of a dipole in uniform electrical field we get $U = -pE \cos \theta$. And my textbook tells me that the when the dipole is kept parallel to the electric field, the angle made is zero $\cos\theta = 1$, thus the potential energy $U$ is $-pE$. The textbook also tells me that this is the minimum energy attained by a dipole in external electric field ($U = -pE$) and the configuration is stable. But since energy is a scalar quantity, it isn't supposed to have direction, So, what does the negative symbol signify?


1 Answer 1


Absolute values of the potential energies of systems do not have any physical meaning. It is the change in potential energy that has a physical meaning.

When the potential energy of a dipole system is derived, the following approach is used: $$U_f - U_i = \int dW = \int_{\theta_i}^{\theta_f} \tau\cdot d\theta$$ $$U_f - U_i = pE \int_{\theta_i}^{\theta_f}\sin\theta\cdot d\theta,$$ and consequently, $$\boxed{\Delta U= -pE(\cos\theta_f - \cos\theta_i).}$$

From here, a reference configuration is chosen for the dipole system such that $U_i = 0 \ \text{at} \ \theta_i = 90^{\circ}$, purely for convenience, which gives the formula $U = -pE\cdot\cos\theta$.

Naturally, any positive or negative signs arise only due to this choice of convention and do not possess any meaning as such.

Hope this helps.

  • $\begingroup$ So negative PE signifies the decrease in potential energy from the chosen reference configuration right? $\endgroup$ May 29, 2022 at 9:09
  • $\begingroup$ Yes. For example, $U = mgy$ is a formula obtained only when you define $U = 0$ at $y = 0$ (usually the ground). If you go below $y= 0$, $U$ is negative only relative to the chosen configuration of $y = 0$. $\endgroup$
    – Cross
    May 29, 2022 at 9:14
  • $\begingroup$ Wow, thank you so much :) $\endgroup$ May 29, 2022 at 9:33

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