Week 1: Discrete Charge and the Electrostatic Field 49
that you have seen across product, if you have started to forget it needless to say it is a very good
idea to backtrack to the math section of this textbook and review its pictorial representation, its
algebra and geometry, and of course the good old right hand rule!
Associated with this torque is the following potential energy which is clearly minimized when
the dipole moment aligns with the applied field. We look at the picture above, and consider the
amount of work done by only the component of the force perpendicular to the arc of motion as we
twist the dipole from a position at right angles to the field (where we define the potential energy
to be zero) to an arbitrary angle. A bit of consideration and a good picture (see homework) should
convince you that:
U = −∫
Ftds (or =−∫
τ dθ)= −
∫θπ/ 2(−qEsin(θ))ℓdθ= −pEcos(θ)(Note well! The force/torque has theoppositesign to the angleθ!) or
U=−~p·E~ (28)Note thatU(θ) is minimum (negative) when the dipole is aligned with the field, maximum (positive)
when antialigned.
This expression is only generally exact if~pis a “point dipole”, since it assumes thatE~is at least
approximately the same at the two ends of the dipole so the forces form a couple and the energy is
strictly due to the torque. More practically, however, it isusable(and quite accurate) whenever the
dipole is short relative to the scale over whichE~varies, so that the value ofE~“at the position of
the dipole” is a well-defined quantity. From this and our general knowledge of intro-level mechanics,
we can see that the force on the dipole in a more generalnon-uniformfieldshouldbe:
F~=−∇~U=∇~(~p·E~) (29)which can be difficult to compute but is easy to understand^31.
In our simple model for the dipole above, if the field isnotuniform then it will in generalnot
be equal at the locations of the two charges. In fact, if we letE~be the field at (say) the location of
the negative charge andE~
′
=E~+ ∆E~at the location of the positive charge, we have:F~ = −qE~+qE~′
= −qE~+qE~+q∆E~
= q∆E~
= ∇~(~p·E~) (30)where the last step, in very rough terms, results from letting~p=q∆~l (a very short point-like
dipole) then ∆E~≈∆~l·∇~E~is basically the first term of a Taylor series expansion ofE~, where the
gradient has to be applied to each component of the field separately. This will be explored further
in homework problems.
(^31) Students who have never seen the gradient operator∇~before and who are not potential physics or math majors
will not betestedon this, but are still advised to read andstudyit and to try to understand it, because it actually
explains a lot of things very compactly that otherwise (as wehave seen and will see further below) are actuallymore
difficult to derive and evaluate than the gradient.