## 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.

## How useful was this post?

Click on a star to rate it!

Average rating / 5. Vote count:

## We are sorry that this post was not useful for you!

Let us improve this post!

Thanks for your feedback!