How and why potential energy is stored in a dipole in uniform electric field?

Question

I learnt that potential energy is stored in a dipole in uniform electric field when it is rotated from theta = pi/2 to any other theta, and the magnitude of stored potential energy can be found out by P.E. = -PEcos(theta), but i am thinking what causes the potential energy to be stored in the dipole and will potential energy be stored if there is no electric field, and how and where it is stored, the last part of question might sound silly, but i am seriously stuck in this thought.

Answer 1

A compass is a needle (magnetic dipole) which is free to rotate in a magnetic field.
When you use a compass to wait for the needle to align itself with the Earth's magnetic field.
In this position the potential energy of the dipole is a minimum, that is why it aligned itself with the magnetic field.
If the compass needle is moved from that position and released, the needle will after undergoing some damped oscillations return to a position where it is aligned with the Earth's magnetic field.
Why is that?
It is because by moving the needle from the aligned position the needle has had an increase it the potential energy you having done work moving the needle.

The case with an electric dipole in an electric field is similar with the dipole having a minimum potential energy position when it is aligned with the external electric field.
Moving the dipole from that minimum potential energy position requires work to be done and thus the potential energy of the dipole is increased with the energy stored in the electric field. With no external electric field there is no potential energy stored and the dipole does not have a preferred direction (minimum potential energy) of alignment.

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Consider the following two state, a and b, of the dipole and external field, $E$, system.

The magnitudes of all the forces are the same.

In state a there is no net force and no torque on the dipole so it is in static equilibrium.

In state b there is no net force but there is a torque on the dipole so if it is released it will move towards state b and in doing so will do work, forces are being displaced and that work is done at the expense of the electric potential energy of the system.
Thus in moving from state b to state a the electric potential energy of the system decreases and is a minimum at state $a$.

Note that I have never defined the zero of electric potential energy as there are two camps for this.
One assumes the zero when the dipole is aligned with the electric field and the other assumes the zero when the dipole is broadside to the electric field.