# Doubt in the interaction energy of the dipole in an electric field

Consider a dipole ($$\vec{p}$$) in an electric field ($$\vec E$$) making an angle $$\theta$$ with the field. 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.
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,
$$U = -qV_1 + qV_2 - kq_/d = -q (V_1 + k/d) + q V_2 = -q V_1' + q V_2 = -q \,(V_1' - V_2) = - \mathbf{p \cdot E}.$$
• 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? – Iti May 9 at 9:41
• 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. – Yejus May 9 at 14:57