**What is Potential Energy?**

As we know, an object can store energy due to its position. In the case of a bow and an arrow, when the bow is drawn, it stores some amount of energy, which is responsible for the kinetic energy it gains when released.

Similarly, in the case of a spring, when it is displaced from its equilibrium position, it gains some amount of energy which we observe in the form of stress we feel in our hands upon stretching it. We can define potential energy as a form of energy that results from the alteration of its position or state.

**Potential Energy Formula**

Where:

= | gravitational energy | |

= | mass | |

= | gravitational field | |

= | height |

**Potential Energy Unit**

Gravitational potential energy has the same units as kinetic energy: **kg m ^{2} / s^{2}**

**Types of Potential Energy**

- Gravitational Potential Energy
- Elastic Potential Energy

**Gravitational Potential Energy**

The gravitational potential energy of an object is defined as the energy possessed by an object that rose to a certain height against gravity.

W = m×g×h = mgh

**Elastic Potential Energy**

Elastic potential energy is stored in objects that can be compressed or stretched, such as rubber bands, trampolines and bungee cords.

- A twisted rubber band that powers a toy plane
- An archer’s stretched bow
- A bent diver’s board just before a diver dives in
- The coil spring of a wind-up clock

Where:

*U*is the elastic potential energy*k*is the spring force constant*x*is the string stretch length in m

**Potential Energy in an External Field**

**The potential energy of a single charge:**

The external electric field E and the corresponding external potential V may vary from point to point. By definition, V at a point P is the work done in bringing a unit positive charge from infinity to the point P.

Work done in bringing a charge q from infinity to the point P in the external field is qV. This work is stored in the form of potential energy of q. If the point P has position vector r relative to some origin,

**The potential energy of a system of two charges in an external field:**

Work done in bringing the charge q_{1} from infinity to r_{1} is q_{1} V(r_{1}). Consider the work done in bringing q_{2} to r_{2}. In this step, work is done not only against the external field E but also against the field due to q_{1}.

Work done on q_{2} against the external field

Work done on q_{2 }against the field due to q_{1}

By superposition principle for fields, add up the work done on q_{2} against the two fields. Work done in bringing q_{2} to r_{2}

Thus, Potential energy of the system = the total work done in assembling the configuration

**The potential energy of a dipole in an external field:**

Consider a dipole with charges q_{1} = +q and q_{2} = -q placed in a uniform electric field E.

In a uniform electric field, the dipole experiences no net force; but experiences a torque τ given by which will tend to rotate it (unless p is parallel or antiparallel to E).

Suppose an external torque τ_{ext} is applied in such a manner that it just neutralizes this torque and rotates it in the plane of paper from angle θ_{0} to angle θ_{1} at an infinitesimal angular speed and *without angular acceleration. *The amount of work done by the external torque will be given by

This work is stored as the potential energy of the system. We can then associate potential energy U(θ) with an inclination θ of the dipole. Similar to other potential energies, there is a freedom in choosing the angle where the potential energy *U *is taken to be zero. A natural choice is to take θ_{0} = π / 2.

Here, **r _{1} **and

**r**denote the position vectors of +

_{2}*q*and –

*q*. Now, the potential difference between positions

**r**and

_{1}**r**equals the work done in bringing a unit positive charge against field from

_{2}**r**to

_{2}**r**.

_{1}The displacement parallel to the force is 2*a*cosθ. Thus,

We note that *U*′(θ) differs from *U*(θ ) by a quantity which is just a constant for a given dipole. Since a constant is insignificant for potential energy.

We can now understand why we took θ_{0} = π/2. In this case, the work done against the *external *field **E **in bringing +*q *and *– q *are equal and opposite and cancel out, i.e., *q *[*V *(**r**_{1}) – *V *(**r**_{2})] = 0.