Consider a dipole ($\vec{p}$) in an electric field ($\vec E$) making an angle $\theta$ with the field.

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We can see that $V_1-V_2=Ed\cos\theta$
In books, the derivation for the interaction energy of the dipole is given as

Interaction energy, $U=-qV_1+qV_2$
$\implies U=-q(V_1-V_2)$
$\implies U=-qEdcos\theta$
$\implies U=-pEcos\theta$
$Hence, U=-\vec{p}.\vec{E}$

We know that interaction energy of the dipole by definition means that the work done by the external agent in assembling the dipole in the electric field.
Suppose there is a uniform electric field $\vec{E}$ in the space.
The work done in placing the charge $-q_1$ at the position where potential due to electri field is $V_!$ is $-qV_1$
The work done in placing the charge $+q$ at a distance $d$ from $-q$ charge at angle $\theta$ with electric field is $q$(potential due to electric field + potential due to -q)
Work done in placing $+q$ charge is $q(V_2-\frac{kq_1}{d}$.
So, net interaction energy, $U=$ Work done in placing $-q$ and $+q$ charge in electric field $E$.
$\implies U=-qV_1+qV_2-\frac{kq_1}{d}$
$\implies U=-q(V_1-V_2)-\frac{kq_1}{d}$
$\implies U=-\vec{p}.\vec{E}-\frac{kq_1}{d}$

I am not able to understand why I get extra $-\frac{kq_1}{d}$ term in $U$. If we go by the basic definition of interaction energy, then I get different expression as that given in books.
Why this is so? What is the mistake in my derivation.
Please clarify the doubt.


1 Answer 1


The extra term is included in $V_1$, since $V_1$ stands for the total potential at that point. The way you have posted your question,

$U = -qV_1 + qV_2 - kq_/d = -q (V_1 + k/d) + q V_2 = -q V_1' + q V_2 = -q \,(V_1' - V_2) = - \mathbf{p \cdot E}.$

  • $\begingroup$ But in the derivation which is given in books $V_1$ is the potential only due to electric field ($E$). And they have not taken into account $-kq/d$ term. My question is why is it so? $\endgroup$
    – Iti
    May 9, 2021 at 9:41
  • $\begingroup$ I don't know what books you are talking about. All the books I have seen derive the formula differently -- by considering a dipole initially aligned with the field and then rotated infinitesimally until the desired orientation is reached. In the part you have cited in your post, $V_1$ precludes the potential due to one of the charges. $\endgroup$
    – Yejus
    May 9, 2021 at 14:57

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