0
$\begingroup$

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

enter image description here
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.

$\endgroup$
0
$\begingroup$

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,

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

$\endgroup$
2
  • $\begingroup$ 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 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 at 14:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.