# Potential energy of a dipole in an electric field

When the dipole makes an angle with an electric field, its potential energy is given as the work done in bringing the dipole from the position, when it is aligned with the electric field direction, to a desired position, i.e

W = pE [-cosθf -(-cosθi)] ....eq(1)

p = moment of the dipole
θi = initial position which is when the dipole is aligned with electric field i.e θi = 0
θf = final position of the dipole.

i.e eq(1) becomes,

W= pE(1-cosθf) ....eq(2)

But for convenience, θi is taken when the dipole is perpendicular to electric field, i.e θi= 90°
. This is where I didn't understand. In book, the explanation is just that , it is convenient to take θi=90°, as cos 90°=0, but why and how there is no explanation.

i.e eq(1) now becomes,

W = -pE cosθf ...eq(3)

eq(2) and eq(3) gives different values, so how could they both give the same potential energy of the dipole in an electric field.

Related Other Physics Topics News on Phys.org
Delta2
Homework Helper
Gold Member
Well equation (3) is simpler and more convenient to use than (2). You are right that the two equations dont give the same value however it is important that the difference between the two values is only a constant pE (as long as the electric field and the dipole moment remain constant). And because we usually are interested in differences of the potential energy and not in the absolute value, that constant is eliminated anyway when we take the difference of the potential energy between two positions.

1 person
Is this right?

eq(2) deduce the direct value of potential energy at θ i.e it is work done from θ=0 to θ. With Eq(3), you have to deduce the value of potential energy at θ=0 and at θ. And then there difference is the potential energy at θ.

Delta2
Homework Helper
Gold Member
It is correct, their difference is the potential energy at θ, considering as zero potential energy the potential energy at θ=0. You are free to choose where the potential energy is set to zero thats the whole point here.

their difference is the potential energy at θ, considering as zero potential energy the potential energy at θ=0.
What am I missing?

Why are we taking difference of the potential energy to get potential energy.

To get potential energy at θ, we find the potential energy at θ=0 and at θ. And their difference is the potential energy at θ.

And because we usually are interested in differences of the potential energy and not in the absolute value, that constant is eliminated anyway when we take the difference of the potential energy between two positions.
These two position has to be θ=0 and θ, to know potential energy at θ.

You are free to choose where the potential energy is set to zero thats the whole point here.
Can you explain please, how is it, that?

sophiecentaur
Gold Member
These two position has to be θ=0 and θ, to know potential energy at θ.

Can you explain please, how is it, that?
The orientation of the dipole is relevant as well as the 'position'. The reference (start) position and orientation (0° or whatever) are both relevant to the work needed to move to the test condition. Both of these are arbitrary.

Delta2
Homework Helper
Gold Member
The potential energy at a position F is the work done by the field when the dipole is moved from an initial position I to the position F. What makes you think that the choosing of initial position I is not arbitratry? You think that the position where the dipole moment is aligned with the electric field is the "only rightfull king" of initial positions?

The initial position can be any position. Ofcourse by changing the initial position we ll find a different expression for the potential energy. But any two expressions for the PE differ by a constant, like equation (2) and (3) differ by a constant.