Potential energy of a system of two charges in an external field
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In this context, we have to consider two charges q1, q2 located at r1 and r2 respectively in an external field E. The work done in bringing charge q1 from infinity to r1 is given by q1 V(r1). Similarly, the work done in bringing q2 to r2, the work done is not only against the external field but also against the field due to q1.
Work done against the external field = q
2 V (r
2) and
Work done against the field due to charge q1
where 'r12' is the distance between charge 'q1' and 'q2'.
By the superposition principle for field, we add the work done on q
2 against the two fields.

As the path is independent of work, the potential energy of two charges q
1, q
2 located at r
1 and r
2 in an external field is given by

Potential energy of a dipole
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Potential energy of a dipole in a uniform external field
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In case of a dipole, the charge q1 = +q and q2 = -q. Let the external field 'E' be along the x-direction and origin be the centre of a dipole, then the potential energy of the dipole is

The potential difference between q1 and q2 equals to the work done in bringing a unit positive charge against the field from q2 to q1 and is given by
V(r
1) - V(r
2) = - E . 2a cos
q
Negative sign indicates the decrease in potential in the direction of the field. Thus, potential energy of a dipole in a uniform external field 'E' is given by


Then,
V = -q.2a E cosq
V = - p.E
Dipole has minimum potential energy when aligned with the field, and also remember that a dipole in a uniform field experiences a torque (
t=p.E).
Thus, if the dipole can fritter away its potential energy, the torque will align the dipole in the direction of the external field, and bringing its potential to a minimum.